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CLASS XII (ASSIGNMENTS)
Matrices and Determinants
Properties of determinants :-
1. The value of a determinant remains unchanged if its rows and
columns are interchanged.
2. The sign of value of a determinant is changed if its any two rows
or columns are interchanged.
3. The value of a determinant is zero if its any two rows or columns
are identical.
4. If a determinant is multiplied by a scalar (number) , its only one
row or column gets multiplied by that constant.
5. If any two row or column of a determinant are proportional, its
value becomes zero.
6. If all elements of a row or column are expressed as sum of two
or more elements ,the whole of the determinant can be
expressed in sum of two or more determinants.
7. If some multiple of one row or column is added or subtracted to
another row or column (elementwise), its value remains
unchanged.
Tips to solve properties based problems:-
1. If a determinant is of order ,we can apply only n-1 propertis
at a time to it.
2. The format of applicationof properties is :-Row affected Row
affected n (Rowused)
Ex.
3. The format for interchanging Rows or columns :-
4. You can never multiply a number to Row affected, it is always
multiplied to Row used.
5. First always try to make elements of any one row or column
identical
so that you could take out common from that row or column. It
makes all the elements of that Row or column unity(1) and then
you
make at the most two elements of that row or columns zero
(0).Now
expand that the determinant by that row or column.
EXAMPLE:
We shall apply
and
Taking out common b from
Expand by =
Important Questions (ASSIGNMENT)
Q. 1. Construct a 3X4 matrix, whose elements are given by
aij=
Q. 2. Construct a 2x3 matrix A = [aij] whose elements are given by aij
= , i ≠ j = |i + 2j|, i = j
Q.3. Construct a 3 × 3 matrix whose elements aij are givenby
Q.4. Howmany number of matrices are possible of order 3 × 3 with
each entry 0 or 1.
Q. 5. Express the matrix
A= as the sum of symmetric and skew-symmetric matrix
Q. 6. Obtain the inverse of the matrix
A =
using elementary transformations.
Q. 7. If
f(x)= Prove that f(x). f(y) = f(x + y)
8. Find x,y,z if
[x 3 2] =[0 0 1]
Q. 9. If A and B are invertible matrices of the same order, then prove
that (AB)-1 = B-1A-1
10. Express as a sum of a symmetric and a skew – symmetric
matrix.
Q. 11. If
A =
Show that A2 -5A + 7I = 0, Use this to find A-1.
Q. 12. Express the matrix
A =
as the sum of a symmetric and a skew-symmetric matrix.
Q. 13. Find the values of x, y, z if the matrix
A =
satisfy the equation A'A = I3.
Q. 14. Show that :
=
Q. 15. Find the inverse of
, if a2 + b2 + c2 + d2 = 1.
Q. 16.(i) Find x such that:
(ii) find x, p, y &q
(iii) Find the value of x, if
Q. 17. For what value of k the matrix A = has no inverse.
Q. 18. Prove that the product of matrices
and
is the null matrix, when and differ by an odd multiple of .
Q. 19. A matrix X has a + b rows and a + 2 columns while the matrix Y
has b + 1 rows and a + 3 columns. Both matrices XY and YX exist. Find
a and b. Can you say XY and YX are of the same type? Are they equal.
Q. 20. Find the matrix A satisfying the matrix equation
A =
21. If A = , B = , c = Find a matrix D such that CD -
AB = 0.
22. Let A = , Verify that.
23. If A = ,find k so that.
24.Find B if .
25. Find A = , find a and b such that such that where I
is unit matrix of order 2.
Q. 26.Prove that every square matrix can be uniquely expressed as the
sum of a symmetric matrix and a skew-symmetric. Hence represent
as above.
27. If , prove that ,n N
28. If A = ,Using principle of mathematically induction prove
that
29. If A = , B = , c = Find a matrix D such that CD -
AB = 0.
30. Let A = , Verify that.
Q. 31.(i)Prove that every square matrix can be uniquely expressed as
the sum of a symmetric matrix and a skew-symmetric. Hence
represent
as above.
(ii) Express the matrix as the sum of symmetric
and skew symmetric matrix.
Q.32(i).Using elementary row transformation find the inverse of the
matrix
(ii) Using the method of reduction (i.e elementary row
transformations),find the inverse of
(iii)IfA =
; showthat A2 = A-1 . (Without using elementary
transformations)
(iv) Using elementary transformation , find the inverse of the
matrix.
Q. 33.(i) Showthat the matrix A = satisfies the equation x2 –
4x –5=0. Hence find A-1
(ii) Solve the equations by matrix method
(iii) Solve using matrices :
(iv) Solve by matrix method:
2x + y + z = 1
x - 2y – z = 3/2
3y - 5z = 9
34.(i) If A = and B = find the product AB and
use this result to solve the following system of equations:
(ii) If find A-1 and use it solve the system of
equations:
(iii) Determine the product
and use it solve the system ofequations :
x-y+z=4
x-2y-2z=9
2x+y+3z=1
(iv) If
A =
, find A-1, using A solve the following system of linear equations.
2x – y + z + 3 = 0
3x – z + 8 = 0
2x + 6y – 2 = 0
35.(i) For what value of a and b ,the following system of equations is
consistent?
(ii) Find whether the following system ofequations is consistent or
not, find the solutionofthe system also.3x-y+2z=3, x-2y-z = 1, 2x+y +
3z = 5.
(iii) Using matrix methed, solve the following system ofequations:
(iv) Solve the following system oflinear equation
(v)The sum of three numbers is 6. If we multiply third number by 3
and add secondnumber to it, we get 11. By adding first and third
numbers, we get double of the secondnumber. Represent it
algebraically and find the numbers using matrix method.
(vi) The management committee of a residential colony decided to award
some of its members for honesty, some for helping others and some others
for supervising the workerstokeepthe colony neat and clean. The sum of all
the awardees is 12. Three times the sum of awardees for cooperation and
supervisionaddedtotwo times the number of awardees for honesty is 33. If
the sum of the number awardees for honesty and supervision is twice the
number of awardees for helping others, using matrix method, find the
number of awardees of each category. Apart from above values, suggest one
more value which the management of the colony must include for awards.
(vii) Showthat the following system ofequations is consistent 2x – y
+ 3z = 5, 3x + 2y – z = 7, 4x + 5y – 5z = 9. Also,find the solution.
(viii)Using matrix method,solve the following system of equations for
x, y and z :
Determinants
Q. 1. (i)Find the equation of the line joining A(1,3) and B(0,0) using
determinants and find if D (K, 0) is a point such that area of a triangle
ABD is 3 square units.
(ii) if |
𝒙 + 𝟏 𝒙 − 𝟏
𝒙 − 𝟑 𝒙 + 𝟐
| = |
𝟒 −𝟏
𝟏 𝟑
|, then write the value of x.
Q. 2. Using determinants,find the area ofthe triangle whose vertices
are (1, 4), (2, 3), (-5, 3). Are the givenpoints collinear.
Q. 3. If the points (a1, b1), (a2, b2) and (a1 + a2,(b1 + b2) are collinear,
Showthat a1b2 = a2b1.
Q. 4. If a, b and c are real numbers,and
, show that either a + b + c = 0 or a = b = c.
Q.5. Using properties ofdeterminants, prove
that = 0.
Q. 6. Without expanding showthat
Q. 7. Prove that :
Q. 8. Solve :
Q. 9. If a, b, c are all positive and are pth , qth and rth terms ofG.P., then
showthat
Q. 10. If
= 0, then Prove that a, b, c are in G.P or x, y, z are in G.P
Q. 11. If x, y, z are different and
Q. 12. Showthat points A (a, b + c), B (b,c + a),C (c, a + b) are collinear.
Q.13. Using properties of determinants,prove that:
Q.14. Using properties ,prove that:
Q. 15. Using properties of determinants solve for x:
Q16. Using properties prove that :
Q.17. Prove using properties of determinants.
Q. 18. Showthat :
Q. 19. Prove that :
Q. 20.
ASSIGNMENT
ASSIGNMENT(matrices)
Qoestion.1 Using matrices, solvethe following systemof equations
(i) x+2y+z = 1 , 2x – y+z = 5 , 3x+y – z = 0.
[Hint use AX= B ⇨ X = A-1 B, |A|=15≠0 meansA is invertible.Adj(A)=
[
𝟎 𝟑 𝟑
𝟓 −𝟒 𝟏
𝟓 𝟓 −𝟓
] ,A-1 =
𝒂𝒅𝒋(𝑨)
|𝑨|
Ans. x =1, y=-1, z=2]
(ii) 2x+y – 3z = 13, x + y – z = 6 , 2x – y+4z = -12.
[ Ans. |A| = 9, adj(A) = [
𝟑 −𝟏 𝟐
−𝟔 𝟏𝟒 −𝟏
−𝟑 𝟒 𝟏
] , x=1, y=2, z=-3.]
(iii) 2x+y+z = 1 , x – 2y – z = 3/2 , 3y – 5z = 9.
[Hint |A| = 34, adj (A) = [
𝟏𝟑 𝟖 𝟏
𝟓 −𝟏𝟎 𝟑
𝟑 −𝟔 −𝟓
], x=1, y= 1/2., z=-3/2.]
Question.2 Use the product [
−𝟒 𝟒 𝟒
−𝟕 𝟏 𝟑
𝟓 −𝟑 −𝟏
] [
𝟏 −𝟏 𝟏
𝟏 −𝟐 −𝟐
𝟐 𝟏 𝟑
] to solve the
equations x – y+z = 4, x – 2y – 2z = 9, 2x+y+3z = 1.
[Hint take product of above two matrices,we get identity matrix,then
use AB=BA = I means B is the inverse of A
Or A is the inverse of B.
⇨ [
−𝟒 𝟒 𝟒
−𝟕 𝟏 𝟑
𝟓 −𝟑 −𝟏
] [
𝟏 −𝟏 𝟏
𝟏 −𝟐 −𝟐
𝟐 𝟏 𝟑
] = 8I3 ,
according to above equation let A
let B (1/8) B is the inverse of A. Ans. x=3, y=-2, z=-1.]
Question.3 Solve the following system ofhomogenous equations:
2x+3y – z = 0, x – y – 2z = 0, 3x+y+3z = 0.
Solution: system of homogenous equations can be writtenas AX = O ,
A = [
𝟐 𝟑 −𝟏
𝟏 −𝟏 −𝟐
𝟑 𝟏 𝟑
] ,|A| = -33
So, the system has only the trivial solutiongivenby x=y=z=0. If|A| = 0
then system has non-trivial solution.]
Question.4 Showthat system ofequations x+y – z = 0, x – 2y+z = 0,
3x+6y – 5z = 0 has non-trivial solution.Find sol.Answer: |A| = 0, it
has infinitely many solutions ∴ let z = k a arbitrary x+y = z = k , x – 2y
= -z = -k , 3y = 2k i.e, y = 2k/3 ⇨ x = k/3from first equation by putting
the values of x, y & z in thirdequation, we get 0 which is true. The
requiredsolutionis z = k, y = 2k/3, x = k/3 where k is arbitrary.
Question.5 Showthat system ofequations3x+2y +7 z = 0, 4x – 3y - 2z
= 0, 5x+9y +23z = 0 has non-trivial solution.Find the solution. [Hint x
= -k, y = -2k, z = k]
Question.6 The system ofequations 2x+3y = 7 , 14x+21y = 49 has (a)
only one solution (b) finitely many solution (c) no solution (d)
infinitely many solution . [give reason]
Question.6 Find the inverse (using elementary transformations) of
following matrices: (i) A = [
𝟎 𝟏 𝟐
𝟏 𝟐 𝟑
𝟑 𝟏 𝟏
]
(ii) A = [
𝟑 𝟒 𝟐
𝟎 𝟐 −𝟑
𝟏 −𝟐 𝟔
] [Hint: A-1 = [
𝟑 −𝟏𝟒 −𝟖
−𝟑/𝟐 𝟖 𝟗/𝟐
−𝟏 𝟓 𝟑
] ,R1↔R3,R3→R3
– 3R1,R1→R1+R2,R2→1/2R2,R3→R3 – 10R2, R3→-R3,R1→R1–
3R3,R2→R2+3/2R3]
(iii) A = [
𝟏 𝟑 −𝟐
−𝟑 𝟎 −𝟓
𝟐 𝟓 𝟎
] [Hint: A-1 = [
𝟏 −𝟐/𝟓 −𝟑/𝟓
−𝟐/𝟓 𝟒/𝟓 𝟏𝟏/𝟐𝟓
−𝟑/𝟓 𝟏/𝟐𝟓 𝟗/𝟐𝟓
],
Question. IfA is singular matrix then under what conditionset of
equations AX = B may be consistent. [answer if (adjA)B = O ,then
eqns. Will have infinitly many sols.Hence consistent.]
Question. If A is a square matrix of order 3 such that |adjA| = 289, find
|A|. [ |A| = ±17 ∵ |adjA| = |A|n-1.]
Q. The management committee of a residential colony decided to award
some of its members for honesty, some for helping others and some others
for supervising the workers to keep the colony neat and clean. The sum of
all the awardees is 12. Three times the sum of awardees for cooperation
and supervisionadded to two times the number of awardees for honesty is
33. If the sum of the number awardees for honesty and supervision is
twice the number of awardees for helping others, using matrix method,
find the number of awardees of each category. Apart from above values,
suggest one more value which the management of the colony must include
for awards.
[HINT: x+y+z = 12, 3(y+z) + 2x = 33, x – 2y +z = 0, |A|= 3, Adj A =
𝟗 −𝟑 𝟎
𝟏 𝟎 −𝟏
−𝟕 𝟑 𝟏
X=3, Y=4 &Z=5. Regularity, sincerity.]
ASSIGNMENT( WITH HINTS)(determinant)
Question: (i) Let ∆ =
𝑨𝒙 𝒙² 𝟏
𝑩𝒚 𝒚² 𝟏
𝑪𝒛 𝒛² 𝟏
and ∆₁ =
𝑨 𝑩 𝑪
𝒙 𝒚 𝒛
𝒛𝒚 𝒛𝒙 𝒙𝒚
, then ∆ - ∆′₁
= 0
[Hint ∆₁ =
𝒙𝒚𝒛
𝒙𝒚𝒛
𝑨𝒙 𝑩𝒚 𝑪𝒛
𝒙² 𝒚² 𝒛²
𝟏 𝟏 𝟏
]
(ii) If f(x) =
𝟎 𝒙 − 𝒂 𝒙 − 𝒃
𝒙 + 𝒂 𝟎 𝒙 − 𝒄
𝒙 + 𝒃 𝒙 + 𝒄 𝟎
, then which is correctf(a)=0,
f(b)=0, f(0)=0 and f(1)=0 [ Hint f(0)=0 det.(skewsymm.matrix)=0].
**(iii) Let f(t) =
𝒄𝒐𝒔𝒕 𝒕 𝟏
𝟐𝒔𝒊𝒏𝒕 𝒕 𝟐𝒕
𝒔𝒊𝒏𝒕 𝒕 𝒕
, then 𝐥𝐢𝐦
𝒕→𝟎
𝒇(𝒕)
𝒕²
is equal to 0,1,2,3.
[Hint 0,
𝒇(𝒕)
𝒕²
=
𝒄𝒐𝒔𝒕 𝒕 𝟏
𝟐𝒔𝒊𝒏𝒕
𝒕
𝟏 𝟐
𝒔𝒊𝒏𝒕
𝒕
𝟏 𝟏
→
𝟏 𝟎 𝟏
𝟐 𝟏 𝟐
𝟏 𝟏 𝟏
as t→∞].
(iv) There are two values of a which makes determinant ∆ =
𝟏 −𝟐 𝟓
𝟐 𝒂 −𝟏
𝟎 𝟒 𝟐𝒂
= 86, then sum of these numbers is 4,5,-4,9. [Hint a=-
4, operate R2 – 2R1]
Question.1 Prove that the points P (a, b+c), Q(b, c+a), R(c, a+b) are
collinear.
Answer : If P,Q and R are collinear then
𝒂 𝒃 + 𝒄 𝟏
𝒃 𝒄 + 𝒂 𝟏
𝑪 𝒂 + 𝒃 𝟏
= 0
By applying C2 → C2+C1
𝒂 𝒂 + 𝒃 + 𝒄 𝟏
𝒃 𝒂 + 𝒃 + 𝒄 𝟏
𝒄 𝒂 + 𝒃 + 𝒄 𝟏
= (a+b+c)
𝒂 𝟏 𝟏
𝒃 𝟏 𝟏
𝒄 𝟏 𝟏
=0 (
∵ C2, C3 are identical)
Question.2 Find the value of k if the areaof the triangle with vertices
(-2,0),(0,4) and (0,k) is 4 square units.
Answer: Area of ∆ = ½
−𝟐 𝟎 𝟏
𝟎 𝟒 𝟏
𝟎 𝒌 𝟏
= 4
⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨ the absolute value of (k
– 4) = 4 ⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8, 0.
Question.3 Without expanding, showthat
(i)
𝒃 − 𝒄 𝒄 − 𝒂 𝒂 − 𝒃
𝒄 − 𝒂 𝒂 − 𝒃 𝒃 − 𝒄
𝒂 − 𝒃 𝒃 − 𝒄 𝒄 − 𝒂
= 0
Operating C1 → C1+C2+C3, we get
𝟎 𝒄 − 𝒂 𝒂 − 𝒃
𝟎 𝒂 − 𝒃 𝒃 − 𝒄
𝟎 𝒃 − 𝒄 𝒄 − 𝒂
= 0.
(ii)
𝟎 𝒂 −𝒃
−𝒂 𝟎 −𝒄
𝒃 𝒄 𝒐
= 0 Taking out (-1) from C1,C2 and C3, we get, ∆ = (-
1)(-1)(-1)
𝟎 −𝒂 𝒃
𝒂 𝟎 𝒄
−𝒃 −𝒄 𝒐
= -1
𝟎 𝒂 −𝒃
−𝒂 𝟎 −𝒄
𝒃 𝒄 𝒐
=
- ∆ (by interchanging rows and columns) 2∆ = 0 ⇨ ∆ = 0
(iii)
𝒃²𝒄² 𝒃𝒄 𝒃 + 𝒄
𝒄²𝒂² 𝒄𝒂 𝒄 + 𝒂
𝒂²𝒃² 𝒂𝒃 𝒂 + 𝒃
= 0 ⇨
𝒂𝒃𝒄
𝒂𝒃𝒄
𝒃²𝒄² 𝒃𝒄 𝒃 + 𝒄
𝒄²𝒂² 𝒄𝒂 𝒄 + 𝒂
𝒂²𝒃² 𝒂𝒃 𝒂 + 𝒃
=
𝟏
𝒂𝒃𝒄
𝒂𝒃²𝒄² 𝒂𝒃𝒄 𝒂𝒃 + 𝒂𝒄
𝒃𝒄²𝒂² 𝒃𝒄𝒂 𝒃𝒄 + 𝒃𝒂
𝒄𝒂²𝒃² 𝒄𝒂𝒃 𝒄𝒂 + 𝒄𝒃
𝒂𝒃𝒄.𝒂𝒃𝒄
𝒂𝒃𝒄
𝒃𝒄 𝟏 𝒂𝒃 + 𝒂𝒄
𝒄𝒂 𝟏 𝒃𝒄 + 𝒃𝒂
𝒂𝒃 𝟏 𝒄𝒂 + 𝒄𝒃
= abc
𝒃𝒄 𝟏 𝒂𝒃 + 𝒃𝒄+ 𝒂𝒄
𝒄𝒂 𝟏 𝒂𝒄 + 𝒃𝒄 + 𝒃𝒂
𝒂𝒃 𝟏 𝒂𝒃 + 𝒄𝒂 + 𝒄𝒃
( Operating C3 → C3+C1)
abc(ab+bc+ac) x 0 = 0 ( two cols.Are identical)
(iv)
𝟏 𝒂 𝒂² − 𝒃𝒄
𝟏 𝒃 𝒃² − 𝒄𝒂
𝟏 𝒄 𝒄² − 𝒂𝒃
= 0 ⇨
𝟏 𝒂 𝒂²
𝟏 𝒃 𝒃²
𝟏 𝒄 𝒄²
-
𝟏 𝒂 𝒃𝒄
𝟏 𝒃 𝒄𝒂
𝟏 𝒄 𝒂𝒃
=
𝟏 𝒂 𝒂²
𝟏 𝒃 𝒃²
𝟏 𝒄 𝒄²
-
𝟏
𝒂𝒃𝒄
𝒂 𝒂² 𝒂𝒃𝒄
𝒃 𝒃² 𝒃𝒄𝒂
𝒄 𝒄𝟐 𝒄𝒂𝒃
𝟏 𝒂 𝒂²
𝟏 𝒃 𝒃²
𝟏 𝒄 𝒄²
-
𝒂 𝒂² 𝟏
𝒃 𝒃² 𝟏
𝒄 𝒄𝟐 𝟏
= 0 (PassC3overthefirsttwocolumns.)
(v)
𝟏 𝒂 𝒂𝟐
𝟏 𝒃 𝒃𝟐
𝟏 𝒄 𝒄𝟐
=
𝟏 𝒃𝒄 𝒃 + 𝒄
𝟏 𝒄𝒂 𝒄 + 𝒂
𝟏 𝒂𝒃 𝒂 + 𝒃
R.H.S.
𝟏
𝒂𝒃𝒄
𝒂 𝒂𝒃𝒄 𝒂𝒃 + 𝒂𝒄
𝒃 𝒂𝒃𝒄 𝒃𝒄+ 𝒃𝒂
𝒄 𝒂𝒃𝒄 𝒄𝒂 + 𝒄𝒃
=
𝟏 𝒂 𝒂𝒃 + 𝒂𝒄
𝟏 𝒃 𝒃𝒄 + 𝒃𝒂
𝟏 𝒄 𝒄𝒂 + 𝒄𝒃
( applying C1 ↔ C2)
= -
𝟏 𝒂 −𝒃𝒄
𝟏 𝒃 −𝒄𝒂
𝟏 𝒄 −𝒂𝒃
(apply C3 → C3 – (ab+bc+ca))C1)
=
𝟏
𝒂𝒃𝒄
𝒂 𝒂² 𝒂𝒃𝒄
𝒃 𝒃² 𝒂𝒃𝒄
𝒄 𝒄𝟐 𝒂𝒃𝒄
=
𝟏 𝒂 𝒂𝟐
𝟏 𝒃 𝒃𝟐
𝟏 𝒄 𝒄𝟐
(apply C2 ↔ C3 and C1 ↔ C2)
Q. If a,b,c are +ve and are the pth,qth and rth terms resp.Of a
G.P.,show without expanding that
** (vi)
𝐥𝐨𝐠 𝒂 𝒑 𝟏
𝐥𝐨𝐠 𝐛 𝒒 𝟏
𝐥𝐨𝐠 𝐜 𝒓 𝟏
= 0 (put a=xyp-1 ,b=xyq-1 ,c=xyr-1 , apply C1
→C1-logx.C3 ,C1→C1+C3)
(vii)
𝟏 𝒂 𝒃𝒄
𝟏 𝒃 𝒄𝒂
𝟏 𝒄 𝒂𝒃
=
𝟏 𝒂 𝒂²
𝟏 𝒃 𝒃²
𝟏 𝒄 𝒄²
(same methodas given below) (viii)
𝒂 𝒂² 𝒃𝒄
𝒃 𝒃² 𝒄𝒂
𝒄 𝒄𝟐 𝒂𝒃
=
𝟏 𝒂² 𝒂³
𝟏 𝒃² 𝒃³
𝟏 𝒄² 𝒄³
( Multiply by abc as R1with a,R2 by b and R3by c then divide with
abc )
Find the values of: (ix)
𝟏 𝒂𝒃
𝟏
𝒂
+
𝟏
𝒃
𝟏 𝒃𝒄
𝟏
𝒃
+
𝟏
𝒄
𝟏 𝒄𝒂
𝟏
𝒂
+
𝟏
𝒄
(Operate C2 →
𝟏
𝒂𝒃𝒄
C2. and
value is 0
(x)
𝒔𝒊𝒏𝜶 𝒄𝒐𝒔𝜶 𝐜𝐨𝐬(𝜶 + 𝜹)
𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜷 𝐜𝐨𝐬(𝜷 + 𝜹)
𝒔𝒊𝒏𝜸 𝒄𝒐𝒔𝜸 𝐜𝐨𝐬(𝜸 + 𝜹)
(Operating C3 → C3 – cos𝜹.C2+sin 𝜹 .C1 and value is 0)
Q. Prove that :
(a)
𝟏 + 𝒂² − 𝒃² 𝟐𝒂𝒃 −𝟐𝒃
𝟐𝒂𝒃 𝟏 − 𝒂² + 𝒃² 𝟐𝒂
𝟐𝒃 −𝟐𝒂 𝟏 − 𝒂² − 𝒃²
= (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 → (C2+aC3)
(b)
𝒄𝒐𝒔𝜶𝒄𝒐𝒔𝜷 𝒄𝒐𝒔𝜶𝒔𝒊𝒏𝜷 −𝒔𝒊𝒏𝜶
−𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜷 𝟎
𝒔𝒊𝒏𝜶𝒄𝒐𝒔𝜷 𝒔𝒊𝒏𝜶𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜶
= 1 (Apply R3 → sin𝛂R3 +cos𝛂R1)
(C)
( 𝒚 + 𝒛)² 𝒙𝒚 𝒛𝒙
𝒙𝒚 ( 𝒙 + 𝒛)² 𝒚𝒛
𝒙𝒛 𝒛𝒚 ( 𝒙 + 𝒚)²
= xyz(x+y+z)3 (Apply R1 → x R1,
R2 → y R2, R3 → z R3 and take x,y,z commonfrom C1,C2,C3 resp.)
(d
( 𝒃 + 𝒄)² 𝒂 𝟐
𝒃𝒄
𝒄 + 𝒂)² 𝒃 𝟐
𝒄𝒂
𝒂 + 𝒃)² 𝒄 𝟐
𝒂𝒃
=(a-b)(b-c)(c-a)(a+b+c)(a2+b2+c2 )
( Apply C1 → (C1+C2 – 2C3)
(e)
𝒂 + 𝒃 + 𝒄 −𝒄 −𝒃
−𝒄 𝒂 + 𝒃 + 𝒄 −𝒂
−𝒃 −𝒂 𝒂 + 𝒃 + 𝒄
= 2(a+b)(b+c)(c+a) (Apply C1
→(C1+C3 )and C2 →(C2+C3))
(f)
𝒂 𝒃 − 𝒄 𝒄 + 𝒃
𝒂 + 𝒄 𝒃 𝒄 − 𝒂
𝒂 − 𝒃 𝒂 + 𝒃 𝒄
= (a+b+c)(a2+b2+c2)
(g)
𝒃 + 𝒄 𝒄 + 𝒂 𝒂 + 𝒃
𝒒 + 𝒓 𝒓 + 𝒑 𝒑 + 𝒒
𝒚 + 𝒛 𝒛 + 𝒙 𝒙 + 𝒚
= 2
𝒂 𝒃 𝒄
𝒑 𝒒 𝒓
𝒙 𝒚 𝒛
(apply C1→C1-C2-C3, C2→C2-
C1,C3→C3-C1, C2↔C3)
(h)
𝟏 + 𝒂 𝟏 𝟏
𝟏 𝟏 + 𝒃 𝟏
𝟏 𝟏 𝟏 + 𝒄
= abc (
𝟏
𝒂
+
𝟏
𝒃
+
𝟏
𝒄
+1(ab+bc+ca+abc).
(Hint taking a,b,c commonfrom each row, apply R1→R1+R2+R3 then
expand along first row).
(i)
𝟏 + 𝒂² − 𝒃² 𝟐𝒂𝒃 −𝟐𝒃
𝟐𝒂𝒃 𝟏 − 𝒂² + 𝒃² 𝟐𝒂
𝟐𝒃 −𝟐𝒂 𝟏 − 𝒂² − 𝒃²
= (𝟏 + 𝒂² + 𝒃²)3
Apply C1→C1-bC3, C2→C2+aC3, we get
𝟏 + 𝒂² + 𝒃² 𝟎 −𝟐𝒃
𝟎 𝟏 + 𝒂² + 𝒃² 𝟐𝒂
𝒃(𝟏+ 𝒂 𝟐
+ 𝒃 𝟐
) −𝒂(𝟏+ 𝒂 𝟐
+ 𝒃 𝟐
) 𝟏 − 𝒂² − 𝒃²
= (𝟏 + 𝒂² + 𝒃²)²
𝟏 𝟎 −𝟐𝒃
𝟎 𝟏 𝟐𝒂
𝒃 −𝒂 𝟏 − 𝒂² − 𝒃²
expand along C1, We get (𝟏 +
𝒂² + 𝒃²)³ .
**(h) Evaluate
( 𝑿
𝟏
) ( 𝑿
𝟐
) ( 𝑿
𝟑
)
( 𝒀
𝟏
) ( 𝒀
𝟐
) ( 𝒀
𝟑
)
( 𝒁
𝟏
) ( 𝒁
𝟐
) ( 𝒁
𝟑
)
where ( 𝑿
𝟏
) =C(x,1) ( binomial
coefficient)
Solution:
𝒙
𝟏!
𝒙(𝒙−𝟏)
𝟐!
𝒙( 𝒙−𝟏)(𝒙−𝟐)
𝟑!
𝒚
𝟏!
𝒚(𝒚−𝟏)
𝟐!
𝒚( 𝒚−𝟏)(𝒚−𝟐)
𝟑!
𝒛
𝟏!
𝒛(𝒛−𝟏)
𝟐!
𝒛( 𝒛−𝟏)(𝒛−𝟐)
𝟑!
=
𝒙𝒚𝒛
𝟐!𝟑!
𝟏 𝒙 − 𝟏 ( 𝒙 − 𝟏)(𝒙 − 𝟐)
𝟏 𝒚 − 𝟏 (𝒚 − 𝟏)(𝒚 − 𝟐)
𝟏 𝒛 − 𝟏 ( 𝒛 − 𝟏)(𝒛 − 𝟐)
( taking x,y,z commonfrom
R1,R2,R3 resp.and
½!,1/3!
From C2,C3 resp.) ( by
formulaof C(n,r) =
𝒏!
( 𝒏−𝒓)!𝒏!
)
Apply C3→C3 + C2 and put a= x-1, b=y-1, c=z-1
𝒙𝒚𝒛
𝟏𝟐
𝟏 𝒂 𝒂²
𝟏 𝒃 𝒃²
𝟏 𝒄 𝒄²
=
𝒙𝒚𝒛
𝟏𝟐
(a-b)(b-c)(c-a) =
𝒙𝒚𝒛
𝟏𝟐
(x-y)(y-z)(z-x).
Question: If x,y,z are all different and if
𝒙 𝒙² 𝟏 + 𝒙³
𝒚 𝒚² 𝟏 + 𝒚³
𝒛 𝒛² 𝟏 + 𝒛³
= 0 , prove
that xyz = -1
Solution:
𝒙 𝒙² 𝟏 + 𝒙³
𝒚 𝒚² 𝟏 + 𝒚³
𝒛 𝒛² 𝟏 + 𝒛³
=
𝒙 𝒙² 𝟏
𝒚 𝒚² 𝟏
𝒛 𝒛² 𝟏
+
𝒙 𝒙² 𝒙³
𝒚 𝒚² 𝒚³
𝒛 𝒛² 𝒛³
=
𝒙 𝒙² 𝟏
𝒚 𝒚² 𝟏
𝒛 𝒛² 𝟏
+ xyz
𝟏 𝒙 𝒙²
𝟏 𝒚 𝒚²
𝟏 𝒛 𝒛²
= 0
𝟏 𝒙 𝒙²
𝟏 𝒚 𝒚²
𝟏 𝒛 𝒛²
(1+xyz) = 0 ⇨ (x-y)(y-z)(z-x)(1+xyz) =0 ⇨ xyz=-1 ∵
x ≠y≠ z.
Question: By using properties ofdeterminant,showthat
𝒂² + 𝟏 𝒂𝒃 𝒂𝒄
𝒂𝒃 𝒃² + 𝟏 𝒃𝒄
𝒄𝒂 𝒄𝒃 𝒄² + 𝟏
= 1+a2+b2+c2
[Hint: multiply and divide by a,b,c with R1,R2,R3respectively,taking
a,b,c commonfrom C1,C2,C3respectively
R1→R1+R2+R3]
Question: showthat
𝒂 − 𝒃 − 𝒄 𝟐𝒂 𝟐𝒂
𝟐𝒃 𝒃 − 𝒄 − 𝒂 𝟐𝒃
𝟐𝒄 𝟐𝒄 𝒄 − 𝒂 − 𝒃
=(a+b+c)3.[Hint:R1→R1+R2+R3]
Question(i) Using matrix method,solve the following system of
equations:
𝟐
𝒙
+
𝟑
𝒚
+
𝟏𝟎
𝒛
= 4,
𝟒
𝒙
-
𝟔
𝒚
+
𝟓
𝒛
= 1,
𝟔
𝒙
+
𝟗
𝒚
-
𝟐𝟎
𝒛
= 2; x, y, z ≠ 0.
[X=2,Y=3,Z=5,|A|=1200,adjA =[
𝟕𝟓 𝟏𝟓𝟎 𝟕𝟓
𝟏𝟏𝟎 −𝟏𝟎𝟎 𝟑𝟎
𝟕𝟐 𝟎 −𝟐𝟒
] ]
(ii)
𝟏
𝒙
-
𝟏
𝒚
+
𝟏
𝒛
=4,
𝟐
𝒙
+
𝟏
𝒚
-
𝟑
𝒛
= 0,
𝟏
𝒙
+
𝟏
𝒚
+
𝟏
𝒛
= 2
[ x=1/2,y=-1,z=1 adjA= [
𝟒 𝟐 𝟐
−𝟓 𝟎 𝟓
𝟏 −𝟐 𝟑
] |A| = 10]
(iii)
𝟐
𝒙
-
𝟑
𝒚
+
𝟑
𝒛
= 10,
𝟏
𝒙
+
𝟏
𝒚
+
𝟏
𝒛
= 10,
𝟑
𝒙
-
𝟏
𝒚
+
𝟐
𝒛
=13; X, Y, Z ≠ 0.
[X=1/2,Y=1/3,Z=1/5,|A|=-9, adjA = [
𝟑 𝟑 −𝟔
𝟏 −𝟓 𝟏
−𝟒 −𝟕 𝟓
] ]
Relations & functions
Q.1. Show that the relation R defined by (a, b) R (c, d) ⟹ a + d = b + c
on the set N×N is an equivalence relation.
Q.Show that ƒ: N N defined by ƒ (x) = , if n is odd
, if n is even is many-one onto function.
CompositionofFunction and Invertible Function.
Q.1. If f(x) = x + 7 and g(x) = x – 7, x ε R, find (fog)(7).
Solution:
We have, f(x) = x + 7 and g(x) = x – 7.
Then (fog)(x) =fo(g(x)) =f(x – 7) = (x – 7) + 7 = x.
Therefore,(fog)(7) =7. [Ans.]
Q.2. If f : R→ R and g : R→ R are defined respectively as f(x) = x2 + 3x + 1
and g(x) = 2x – 3, find (a) fog , (b) gof.
Solution: [ 4x2 – 6x + 1. [Ans.] , 2x2 + 6x – 1. [Ans.]
Q.3. If f : R→ R defined as f(x) = (2x – 7)/4 is an invertible function, find
f–1. [f –1(x) = (4x + 7)/2. [Ans.]
Q.4. If f : R→ R defined by f(x) = (3x + 5)/2 is an invertible function,
find f–1.Solution: [ f–1(x) = (2x – 5)/3. [Ans.]
1.4 Binary Operations.
Q.1.
i. Is the binary operation*, defined on set N, given by a*b= (a+
b)/2, for all a, b ε N, commutative ?
ii. Is the above binary operation* associative ?
Q.2. Let * be a binary operationdefined by a*b = 2a + b – 3. Find 3*4.
ASSIGNMENT( Relations & Functions)
Q. 1 Showthat the function f : N →N, defined by f(x) = x2 + x +1 is one-
one but not onto.
Q. 2. Let f: N be a function defined as f ( x ) = 4x2 + 12x + 15 .Show
that f: N where S is the range of f, is invertible .Find the inverse of
f .
Q. 3. Showthat the relationR on the set R ofall real numbers, defined
as R = { (a, b): a2 + b2 =1 } is neither reflexive nor transitive but
symmetric
Q. 4. Showthat ƒ : R –{0} R–{0} given by ƒ (x) = 3/x is invertible and it
is inverse of itself.
Q. 5 IfR1 & R2 are equivalence relations onset A. Show that R1 U R2 is
reflexive,symmetric but not transitive.
Q. 6. Let ƒ (x) = [x] and g(x) = |x|, find goƒ(-5/3) – ƒog (-5/3)
Q. 7. Showthat the function ƒ: R R defined by ƒ (x) = 3x3 + 5 for x R
is a bijection.
Q. 8. Showthat the relationR on the set R ofall real numbers, defined
as R = { (a, b): |a| ≤ b } is neither reflexive nor symmetric but
transitive.
Q. 9. Showthat the function ƒ : N → N given by ƒ (1) = ƒ (2) =1 and ƒ (x) =
x-1,for every x > 2 is onto but not one-one.
Q. 10. If * is a binary operationon R defined by a * b = a/4+ b/7for a,
b R, find the value of ( 2 * 5 ) * 7
Q. 11 Let R be a relationon NXN, defined by (a,b) R (c,d) ⇔ ad(b+c)) =
bc(a+d), ∀(a,b) ,(c,d) є NXN ,Show that R is an equivalence relationon
NXN.
Same for addition
Q. 12 Let R be a relationon NXN, defined by (a,b) R (c,d) ⇔ ad = bc,
∀(a,b) , (c,d) є NXN ,Showthat R is an equivalence relationon NXN.
Q.13 Let A = {1,2,3}. Find the number ofrelations on A containing (1,2)
& (2,3) which are reflexive, transitive but not symmetric.
Q.14 Showthat the function ƒ : N → N givenby f(x) =x – (-1)x, is
bijection.( same as ques.2 of misc.)
Q.15 Let f: [-1,∞) →[-1,∞) is given by f(x)= (x+1)2 – 1. Showthat f is
invertible and find the set = { x : f(x) = f-1 (x) }.
Relations & functions for class—XII Level—2
Q.1 If f: R→ R is givenby f(x) = (3 – x3)1/3 showthat fof =Ig where Igis the
identity mapon R.
Q.2 Show that the functionf: [-1, 1]→R definedby f(x) =
𝒙
𝟐+𝒙
is 1-1 . Find the
range of f. Alsofindthe inverse of the functionf: [-1, 1]→ range of f.
Q.3 Show that the functionf: R → R definedby f(x) = cos (5x+2) is neither 1-1
nor onto?
Q.4 If f: R → R be givenby f(x) = sin2
x +sin2
(x+π/3) +cosx .cos(x+π/3) ∀ x Є R,
and g: R → R be a function suchthat g(5/4) =1 , then prove that (gof) : R → R
is a constant function.
Q.5 Let R1=R – {-1} and an operation* is definedon R1 by a*b = a + b + ab ∀
a, b Є R1 .Findthe identity elementandinverse of an element.
ANSWERS OF Level—2
Ans.1 As f: R → R, fof exists andfof : R → R is givenby (fof) (x) = f(f(x)) = f(3 –
x3
)1/3
= (3 – ((3 – x3
)1/3
)3
)1/3
= (3 – (3 – x3
))1/3
=x ∀ x ЄR Ans.2 f is 1-1, as
consider any x1, x2 Є [-1, 1]such that f(x1) = f(x2) ⇨
𝒙₁
𝟐+𝑿₁
=
𝑿₂
𝟐+𝑿₂
⇨ x1x2+2x1 =
x1x2 +2x2 ⇨ x1 = x2 For the range of f
Let y = f(x) ⇨ y =
𝒙
𝟐+𝒙
⇨ xy +2y =x ⇨ (y – 1) x= -2y ⇨ x =
−𝟐𝒚
𝒚−𝟏
As x Є [-1, 1], so -1 ≤
−𝟐𝒚
𝒚−𝟏
≤1 , but (y – 1)2
>0 , y ≠ 1⇨ -(y – 1)2
≤
−𝟐𝒚
𝒚−𝟏
(y-1)2
≤(y
– 1)2
⇨ -(y2
– 2y +1) ≤ -2y2
+2y ≤ y2
– 2y +1 ,y≠ 1
⇨ Y2
– 1 ≤ 0 and 0 ≤ 3y2
– 4y +1 ⇨ y ε [-1,1]and (y – 1/3) (y – 1) ≥ 0 ,y ≠ 1
⇨y Є [-1,1] and y ε (-∞ ,1/3]U [1,∞) , y ≠ 1
⇨ y Є [-1,1] and y ε (-∞ ,1/3]U (1,∞)⇨ y Є[-1,1/3].
∴ its inverse existsas f is 1-1 and onto, to find f-1
𝒙
𝟐+𝒙
= y ⇨ xy +2y =x ⇨ 2y = x (1 – y) ⇨ x=
𝟐𝒚
𝟏−𝒚
f-1
(y) = x =
𝟐𝒚
𝟏−𝒚
.
Ans. 3 For f is not 1-1, 5x+2 = π/2 ⇨ x = (π – 4)/10
∴ 5x+2 =π/2, again 5x+2 = 3π/2 ⇨ x = (3π – 4)/10, Nowf ((π – 4)/10))
= cos[5((π – 4)/10) +2]= cosπ/2 =0
f ((3π – 4)/10) = cos[5((3π –4)/10) +2] = cos3π/2 = 0.
For f is not onto, as -1 ≤ cos (5x+2) ≤1, then -1≤ y ≤1, range of f = [-1, 1] = {y : -
1≤ y ≤1 } ≠ co-domain R.
Ans. 4 ½[ 2sin2
x +2sin2
(x+π/3) +2cosx cos(x+π/3)]
f (x)= ½[ 1 – cos2x +1 – cos (2x+2π/3)+cos (2x+π/3)+cosπ/3]
( As we know that 2sin2
x=1 – cos2x and 2cosA cosB= Cos(A+B) + cos(A-B).)
½[5/2 – {cos2x +cos(2x+2π/3)} +cos(2x+π/3) ⇨ ½[5/2 –2cos(2x+π/3) cos
π/3 + cos(2x+π/3)]=5/4 ∀ x ЄR
∴ for any x Є R , we have (gof)(x) = g(f(x)) = g(5/4)=1 ,soit is constant
function.
Ans. 5 * can be shown to be a binary operationon R1 as let a ≠ -1, b ≠ -1 .
a*b = a+b+ab Є R – {-1} ⇨ a+b+ab ≠ -1
⇨ a(1+b)+(b+1) ≠0 ⇨ (a+1) (1+b) ≠0 ⇨ a ≠ -1 and b ≠ -1 which is true.Nowif
e is the identity element, thena*e =a ⇨ a+e+ae =a ⇨ e (1+a) = 0 ⇨ e =0 or a
= -1 ⇨e =0 , 0 is the identity w.r.t. *
Let a’ be inverse of a, thena*a’ =0 ⇨ a+a’+aa’ = 0 ⇨ a’(1+a) = - a
∴ a’ = - a/(1+a) , is the inverse of a w.r.t. *.
Q.1 Let f(x) = 𝒙+𝟑,𝒊𝒇 𝒙<1𝟒𝒙−𝟐,𝒊𝒇 𝟏≤𝒙≤𝟒.𝒙²+𝟓,𝒊𝒇 𝒙>4 Find f(-1) ,f(4)
and f(5).
Q.2 If f(x) = x2 – 𝟏/𝒙², then find the value of f(x) + f ( 𝟏/𝒙²).
Q.3 Let Q be the setall rational numbers and relation on Q defined by R =
{(X, Y): 1+XY > 0}. Prove then R is reflexive and symmetric but not
transitive.
Q.4 Write the identity element for the binary operation*defined on setR
by a*b = 3ab/8 ∀ a, b ЄR.
Q.5 Show that the function f: R → R defined by f(x) = sin x is neither 1-1
nor onto.
Answers (Level—1)
Ans.1 f (-1) = 2, f (4) = 14, f(5)= 30. Ans.2 0. Ans.3 Considerany x, y Є Q,
since 1+x.x =1+x2 ≥ 1
⇨ (x,x)Є R ⇨ reflexive
Let (x,y) Є R ⇨ 1+xy > 0 ⇨ 1+yx > 0 ⇨ (y,x) ЄR ⇨ symmetric.
But not transitive . Since (-1, 0) and (0, 2) ЄR, because 1 > 0 by putting
values. But (-1, 2) ∉ R because -1<0. Ans.4 Let e be the identity element in
R. Then a *e =a =e*a ∀ aЄR ⇨ a*e =a ∀ a ЄR ⇨ e = 8/3 in R. Ans.5 f is not
1-1 because sin0 = 0 =sin π,so the different elements o, π have same images.
f is not onto because -1 ≤ sin x ≤ 1 for all x ЄR ∴ the range of f =[-1,1],
which is a proper subset of R.
Level
Inverse Trigonometric Functions
For suitable values of x and y
sin-1
x + sin-1
y= sin-1
(x√1-y2
+ y√1-x2
)
sin-1
x - sin-1
y=sin-1
(x√1-y2
- y√1-x2
)
cos-1
x + cos-1
y= cos-1
(xy- √1-x2
√1-y2
)
cos-1 x - cos-1 y= cos-1 (xy+√1-x2√1-y2)
tan-1
x + tan-1
y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀
𝟏−𝑿𝒀
; xy<1
tan-1
x – tan-1
y= 𝐭𝐚𝐧−𝟏 𝑿−𝒀
𝟏+𝑿𝒀
; xy>-1
2tan-1 x= 𝐭𝐚𝐧−𝟏 𝟐𝒙
𝟏−𝒙²
= 𝐬𝐢𝐧−𝟏 𝟐𝒙
𝟏+𝒙²
= 𝐜𝐨𝐬−𝟏 𝟏−𝒙²
𝟏+𝒙²
Q. 1. Find the value of : tan-1
(1) + cos -1
(-1/2) + sin-1
(-1/2).
Q.2 Prove that 𝐭𝐚𝐧−𝟏
𝒙 + 𝐭𝐚𝐧−𝟏 𝟐𝒙
𝟏−𝒙²
= 𝐭𝐚𝐧−𝟏 𝟑𝒙−𝒙³
𝟏−𝟑𝒙²
, x<
𝟏
√𝟑
Q. 3. Solve : tan-1
2x + tan-1
3x =
𝝅
𝟒
Q. 4. Prove :
Q. 5. Solve : sin-1
( 1 –x) – 2sin-1
x = .
Q. 6. Evaluate:tan-1 - sec-1
(-2) + cosec-1
.
Q. 7. Prove :
tan-1
=
Q. 8. Simplify :
sin-1
,
Q. 9. Prove: sec2 (tan-1
2) + cosec2 ( cot-1
3) = 15.
Q. 10. Simplify :
tan-1
Q. 11. Prove :
tan-1
=
Q. 12. If sin(sin-1
, then findthe value of x.
Q. 13. Prove that :
2tan-1
= cos-1
Q. 14. Find the principal value of Sec-1
Q. 15. Find value of
Sin
Q. 1. If 2 tan-1
(Cos q) = tan-1
(2 cosec q), find q.
Q.2
Question.3 if 𝐜𝐨𝐬−𝟏 𝒙
𝒂
+ 𝐜𝐨𝐬−𝟏 𝒚
𝒃
= 𝜽, then prove that
𝒙²
𝒂²
-
𝟐𝒙𝒚
𝒂𝒃
cos𝜽
+
𝒚²
𝒂²
= sin2
𝜽 [Hint: 𝐜𝐨𝐬−𝟏 𝒙
𝒂
+ 𝐜𝐨𝐬−𝟏 𝒚
𝒃
= 𝐜𝐨𝐬−𝟏
[
𝒙𝒚
𝒂𝒃
- √ 𝟏 −
𝒙²
𝒂²
√ 𝟏 −
𝒚²
𝒃²
] =
𝜽 ⇨ (
𝒙𝒚
𝒂𝒃
− cos𝜽 )2
= (√ 𝟏 −
𝒙²
𝒂²
√ 𝟏 −
𝒚²
𝒃²
)2
Simplify it]
Question.4 *(i) sin-1
x + sin-1
y + sin-1
z = π, then prove that
X4
+y4
+z4
+4x2
y2
z2
= 2(x2
y2
+y2
z2
+z2
x2
)
(ii) If 𝐭𝐚𝐧−𝟏
𝒙 + 𝐭𝐚𝐧−𝟏
𝒚 + 𝐭𝐚𝐧−𝟏
𝒛 = π/2 ; prove that xy+yz+xz = 1.
(iii) If 𝐭𝐚𝐧−𝟏
𝒙 + 𝐭𝐚𝐧−𝟏
𝒚 + 𝐭𝐚𝐧−𝟏
𝒛 = π , provethat x+y+z = xyz.
[Hint: for (i) sin-1
x + sin-1
y = π - sin-1
z ⇨ cos(sin-1
x + sin-1
y)
=cos( π - sin-1
z) Use cos(A-B) = cosAcosB – sinAsinB and cos(π –
𝛂)= -cos𝛂
It becomes √(𝟏 − 𝒙²)(𝟏− 𝒚²) - xy = - √ 𝟏 − 𝒛² and simply it.
[Hint: for (ii) tan-1
x + tan-1
y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀
𝟏−𝑿𝒀
]
Question.5 Write the following functions in the simplest
form:(i)𝐭𝐚𝐧−𝟏
(
𝒄𝒐𝒔𝒙
𝟏+𝒔𝒊𝒏𝒙
) (ii) 𝐭𝐚𝐧−𝟏
(
𝒄𝒐𝒔𝒙
𝟏−𝒔𝒊𝒏𝒙
) (iii) 𝐭𝐚𝐧−𝟏
√
𝒂−𝒙
𝒂+𝒙
, -a<x<a
[Hint: for (i) write cosx = cos2
x/2 – sin2
x/2 and 1+sinx =(cosx/2
+sinx/2)2
, then use tan(A-B), answer is π/4 – x/2 ]
[ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2 – x),
then use formula of 1-cos(π/2 – x)= 2sin2
(π/4 – x/2) and sin(π/2 – x) =
2 sin(π/4 – x/2) cos(π/4 – x/2)
Same method can be applied for (i) part also. Answer is π/4 +
x/2][ for (iii) put x=a cos𝛂, then answer will be ½ 𝐜𝐨𝐬−𝟏 𝒙
𝒂
]Question.6 If y = 𝐜𝐨𝐭−𝟏
(√ 𝒄𝒐𝒔𝒙) - 𝐭𝐚𝐧−𝟏
(√ 𝒄𝒐𝒔𝒙), prove that siny =
tan2
(x/2). [Hint: y =
𝝅
𝟐
- 2 𝐭𝐚𝐧−𝟏
(√ 𝒄𝒐𝒔𝒙) , use formula 2𝐭𝐚𝐧−𝟏
𝒙 =
𝐜𝐨𝐬−𝟏
(
𝟏−𝒙²
𝟏+𝒙²
)]
Question.7 (i) Prove that 𝐭𝐚𝐧−𝟏
𝟏 + 𝐭𝐚𝐧−𝟏
𝟐 + 𝐭𝐚𝐧−𝟏
𝟑 = π.
(ii) Prove that 𝐜𝐨𝐭−𝟏
(
𝒂𝒃+𝟏
𝒂−𝒃
) +𝐜𝐨𝐭−𝟏
(
𝒄𝒃+𝟏
𝒃−𝒄
) + 𝐜𝐨𝐭−𝟏
(
𝒂𝒄+𝟏
𝒄−𝒂
) = 0.
[Hint: for (i) 𝐭𝐚𝐧−𝟏
𝟐 =
𝝅
𝟐
- 𝐜𝐨𝐭−𝟏
𝟐 =
𝝅
𝟐
− 𝐭𝐚𝐧−𝟏 𝟏
𝟐
, then use
formula of tan-1
x + tan-1
y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀
𝟏−𝑿𝒀
]
(ii) [Hint: write 𝐜𝐨𝐭−𝟏
𝒙 = 𝐭𝐚𝐧−𝟏 𝟏
𝒙
]
Question.8 Solve the following equations:
(i) 𝐬𝐢𝐧−𝟏
𝒙 + 𝐬𝐢𝐧−𝟏
(𝟏 − 𝒙) = 𝐜𝐨𝐬−𝟏
𝒙 .
(ii) 𝐭𝐚𝐧−𝟏
√ 𝒙(𝒙 + 𝟏) + 𝐬𝐢𝐧−𝟏 √ 𝒙² + 𝒙 + 𝟏 =
𝝅
𝟐
.
(i) [Hint: write 𝐜𝐨𝐬−𝟏
𝒙 =
𝝅
𝟐
- 𝐬𝐢𝐧−𝟏
𝒙, put 𝐬𝐢𝐧−𝟏
𝒙 = y]
(ii) [Hint: use 𝐭𝐚𝐧−𝟏
𝒙 = 𝐜𝐨𝐬−𝟏 𝟏
√ 𝟏+𝒙²
] Indu thakur
Question.9 Using principal values, evaluate 𝐜𝐨𝐬−𝟏
(𝒄𝒐𝒔
𝟐𝝅
𝟑
) +
𝐬𝐢𝐧−𝟏
( 𝒔𝒊𝒏
𝟐𝝅
𝟑
). [answer is π]
Question.10Show thattan(
𝟏
𝟐
𝐬𝐢𝐧−𝟏 𝟑
𝟒
) =
𝟒− √𝟕
𝟑
and justify why the
other value is ignored?
[ Hint: put
𝟏
𝟐
𝐬𝐢𝐧−𝟏 𝟑
𝟒
=∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2
∅), find
tan∅]
ASSESSMENTOF PROBABILITY FOR CLASS –XII
Level—1
Q. 1 If the mean and variance of a binomial distributionare 4 and
4/3 respectively,find P(X≥1).
Q. 2 If P(A) = 3/8 , P(B) = ½ and P(A∩B) = ¼ , find P(𝑨/𝑩).
Q.3 A bag contains 4 white and 2 black balls.Another bag contains 3
white and 5 black balls.
If one ball is drawn from each bag, find the probability that
(i) Both are white balls.
(ii) One is white and one is black.
Q.4 IfA and B are independent events and P(A∩B) = 1/8, P( A’ ∩ B’)
=3/8 , find P(A) and P(B).
Q.5 The probabilities ofP, Q and R solving a problem are ½, 1/3 and
¼ respectively.Ifthe problem is attempted by
all simultaneously,find the probability of exactly one of them
solving it.
Answers of Level—1 Ans.1 np = 4, npq = 4/3 ⟹ q=1/3 ⟹ p = 1 -
1/3 = 2/3⟹ n=6 ⟹P(X≥1) =1– C(6,0) (2/3)0 (1/3)6 = 1 -
𝟏
𝟕𝟐𝟗
=
𝟕𝟐𝟖
𝟕𝟐𝟗
.
Ans.2 P(𝑨/𝑩) =P(𝑨∩ 𝑩 ) /P(B) =
𝟏−𝑷(𝑨𝑼𝑩)
𝟏−𝑷(𝑩)
=
𝟏−[
𝟑
𝟖
+
𝟏
𝟐
−
𝟏
𝟒
]
𝟏−𝟏/𝟐
= ¾.
Ans.3 (i) P(A∩ B) = P(A).P(B) = (2/3).(3/8) [A,Bare independent
events] (ii) P(A’∩ B) + P(A ∩ B’) = P(A’).P(B)+P(A).P(B’) =(
1/3).(3/8)+(2/3).(5/8)=13/24.[A’, B are indep. Events, B’ A are
indep. events], where A = drawing a white ball from first bag. B=
drawing a same ball from secondbag.A’ = drawing a blackball from
first bag and B’ =drawing from secondbag. Ans.4 P(A∩B) = P(A).P(B)
= 1/8 let x=P(A),y= P(B),P( A’ ∩ B’) =3/8 = P( A’) .P( B’) =(1- X)(1– Y)
⟹ X+Y – XY = 5/8 ⟹ X=1/2, Y= ¼.
Ans. 5 P(A’)=1/2,P(B’) = 1-1/3=2/3, P(C’)=3/4 ∴ Req. Prob.=
P(A)P(B’)P(C’) =P(A’)P(B)P(C’)+P(A’)P(B’)P(C)[A,B,C are indep.
events] = (1/2)(2/3)(3/4)+(1/2)(1/3)(3/4)+(1/2)(2/3)(1/4)=11/24.
Level---2
Q.1 If A ∩ B = ф, show that P(A/B) =0, where A and B are possible
events.
Q.2 A pair ofdice is thrownif the sum is even, find the probability that
at least one of the dice Shows three.
Q.3 Let X denotes the number of hours you study during a randomly
selectedschool day.The probability that X can take the value x, has the
following form,where k is some unknown constant P(X=0)=0.1and
P(X=x) = {
𝒌𝒙 𝒊𝒇 𝒙 = 𝟏𝒐𝒓𝟐
𝒌( 𝟓− 𝒙) 𝒊𝒇 𝒙 = 𝟑𝒐𝒓𝟒
𝒐, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
(i) Find k. (ii) What is the probability that you study at least two
hour? Exactly two hour? At most two hours?
Q.4 Six dice are thrown 729 times.Howmany do you expect at least
three dice to showa 5 or 6?
Q.5 In a class;5% of the boys and 10% of the girls have an I.Q. of more
than 150. In this class 60% ofthe students are boys. If a student is
selectedat random and is formedand is found To have an I.Q. ofmore
than 150, find the probability that the student is a boy.
Answers of Level—2
Ans.1 A and B are possible events ⟹A≠ ф⇨P(A)≠0 , P(B) ≠0 But A∩B
= ф ⟹ P(A∩B) =P(ф) = P(A/B) =
𝑷(𝑨∩𝑩)
𝑷(𝑩)
=0. Ans. 2 n(S)=36,
n(A)=18 Out of these 18, the cases which at least one die shows up 3
are (1, 3),(3,1),(3,3),(3,5),(5,3) Requiredprobability=5/18. Ans.3
X 0 1 2 3 4
P(X) 0.1 K 2K 2K K
(i) k=0.15 (ii) 0.75, 0.3, o.55.
Ans.4 P(success)=2/6=1/3 ∴ q=2/3
P(x success i.e., getting a 5 or 6)= C(6, x) Px q6-x P(at least three
successes insix trials) =P(x≥3)=1 – [p(0)+p(1)+p(2)]
By using above result we get 1 – (16/81)(31/9) =233/729 ∴
requiredanswer is 233/729x729=233. Ans.5 Let E1: The student
chosenis a boy. P(E1)=60/100 ∴ E2: ........................................girl. P(E2)
= 40/100 E1, E2 are mutually exclusive.A: a student has an I.Q. of
more than 150. P(A/ E1)= 5/100, P(A/ E2)= 10/100 By Baye’s
theorem P(E1/A) =3/7.
Probability
Q. 1. A die is throwntwice and the sum of the numbers appearing is
observedto be 6. What is the conditional probability that the number
4 has appeared at least once. [2/5]
Q. 2. Assume that each born childis equally likely to be a boy or a girl.
If a family has two children,what is the conditional probability that
both are girls given that at least one is a girl. [1/3]
Q. 3. If A and B are two independent events, showthat the probability
of occurrence ofat least one of A and B is givenby : 1 – P(A').P(B')
Q. 4. An urn contains 5 red and 5 blackballs. A ball is drawn at
random, its colour is noted and is returned to the urn. Moreover,2
additional balls of the colour drawnare put in the urn and then a ball
is drawn at random.What is the probability that the secondball is
red.[1/2]
Q. 5. Two cards are drawn simultaneously (or successively without
replacement) from a well shuffled pack of 52 cards. Find the mean
and standarddeviation of the number of kings.[34/221, 0.37]
Q. 6.
Q. 7. A doctor is to visit a patient .From the past experiences , it is
known that the probabilities that he will come by train , bus , scooter
or by other means of transport are respectively . The
probabilities that he will be late are , if he comes by train ,
bus, and scooter respectively ,but if he come by other means of
transport , then he will not be late .When he arrives he is late .What is
the probability that he comes by train? whichlife skill is the doctor
lacking?, ans. Lacks responsibility &dedicationtohis work.
13.1. Conditional Probability.
Q.1. A and B toss a coinalternately till one of them gets a head and
wins the game.If A starts first,find the probability that B will win the
game. [1/3.]
Q.2. A and B throwtwo dice simultaneously turn by turn. A will win if
he throws a total of 5, B will win if he throws a doublet.Find the
probability that B will win the game,though A startedit. [4/7]
Q.3. Two dice are rolledonce. Find the probability that :
i. the numbers on two dices are different.
ii. the total of numbers on the two dice is at least 4.
[11/12]
Q.4. Two unbiased dice are tossed simultaneously.Find the
probability that the sum of the numbers will be a multiple of 3 or 5.
[19/36]
Q.5. Two unbiased dice are thrown. Find the probability that the sum
of the numbers obtained on the two dice is neither a multiple of 3 nor
a multiple of 4.
[4/9]
13.2. MultiplicationTheorem on Probability/Independent Events.
Q.1. There are two bags.The first bag contains 4 white and 2 black
balls, while the second bag contains 3 white and 4 black balls. A
ball is picked up at random and a ball is drawn out. Find the
probability that it is a white ball.
[23/42]
Q.2. In a groupof 9 students, there are 5 boys and 4 girls.A team of
4 students is to be selectedfor a quiz competition.Find the
probability that there will be 2 boys and 2 girls in that
team.[10/21]
Q.3. 12 cards numbered 1 to 12, are placed in a box, mixed up
thoroughly and then a cardis drawn at random from the box. If it is
known that the number on the drawn card is more than 3, find the
probability that it is an even number. [5/9]
13.3. Bayes’ Theorem.
Q.1. In a bolt factory,machines A, B, C manufacture 25%, 35% and
40% respectively of the total bolts.Of their output 5%, 4% and 2%
respectively are defective bolts. A bolt is drawn at random and is
found to be defective. Find the probability that it is manufactured
by machine B.
28/69.[Ans.]
Q.2. In a bulb factory,machines A, B and C manufacture 60%, 30%
and 10% bulbs respectively.1%,2% and 3% of the bulbs produced
respectively by A, B and C are found to be defective. A bulb is
picked up at random from the total productionand found to be
defective. Find the probability that this bulb was produced by the
machine A.
Ans. = 2/5.]
Q.3. A class consists of50 students out of which there are 10 girls.
In the class 2 girls and 5 boys are rank holders in an examination.If
a student is selectedat random from the class and is found to be a
rank holder, what is the probability that the student selectedis a
girl ?
2/7 . [Ans.]
Q.4. A man is known to speak the truth 3 out of 4 times.He throws a
die and reports that it is a six. Find the probability that it is actually a
six. How is truthfulness helpfulinlife?----- Ans.--- truthfulness is essential life
skill. Everybody trusts atruthful person.
3/8. [Ans.]
Q.5. A company has two plants whichmanufacture scooters.Plant I
manufactures 80% ofthe scooters while Plant II manufactures
20% of the scooters.At Plant I, 85 out of 100 scooters are ratedas
being of standardquality, while at Plant II only 65 out of 100
scooters are ratedas being of standard quality. If a scooter is of
standard quality , what is the probability that it come from Plant I.
0.84 .[Ans.]
Q.6. A manufacturing firm produces steel pipes in three plants A, B
and C with daily productionof 500, 1000 and 2000 units
respectively.The fractions ofdefective steel pipes output produced
by the plant A, B and C are respectively 0.005, 0.008 and 0.010. Ifa
pipe is selectedfrom a day’s total productionand found to be
defective, find out the probability that it came from the first plant.
5/61[Ans.]
Q.7. An insurance company insured 6000 scooter drivers,3000 car
drivers and 9000 truckdrivers.The probability of an accident
involving a scooter,a car and a truck is 0.02, 0.06 and 0.30
respectively.One of the insuredpersons meets with an accident.
Find the probability that he is a car driver.
0.06.[Ans.]
Q.8. An insurance company insured 4000 doctors,8000 teachers
and 12000 engineers.The probabilities ofa doctor,a teacher and
an engineer dying before the age of 58 years are 0.01, 0.03 and 0.05
respectively.Ifone of the insured persondies before the age of58
years,find the probability that he is a doctor.
Solution: Do yourself.[Ans. = 1/22]
Q.9. A car manufacturing factory has two plants X and Y. Plant X
manufactures 70% ofthe cars and plant Y manufactures 30%.At
plant X, 80% ofthe cars are rated of standardquality and at plant
Y, 90% are ratedof standard quality. A car is picked up at random
and is found to be of standard quality. Find the probability that it
has come from plant X.
56/83. [Ans.]
Q. 10 in an examination, an examinee either guesses or copies or
knows the answer of MCQs with four choices.The prob.That he makes
a guess is 1/3and the prob. That he copies answer is 1/6, the prob.
That his answer is correct, giventhat he copied it , is 1/8. Find the
prob. That he copies the answer, given that he correctly answeredit.if
a student copies an answer, what value is he violating?
[hint P(A)=1/3, P(B)=1/6 , P(C) = 1 – 1/3– 1/6 =1/2, P(E/A)=1/4,
P(E/B)=1/8,P(E/C)=1& P(B/E)=1/29]
13.4. Probability Distribution.
Q.1. Two cards are drawn successively withreplacement from a well
shuffled pack of 52 cards. Find the probability distributionof
number of jacks.
Hence required probability distributionis :
x 0 1 2
P(X) 144/169 24/169 1/169
Q.2. Two cards are drawn successively withreplacement from a well
shuffled deck of 52 cards. Find the probability distributionof number
of aces.
Solution:Do yourself.[Ans. of Q.1. above]
Q.3. A pair of dice is tossedtwice. Ifthe random variable X is defined
as the number of doublets, find the probability distributionof X.
Hence, the requiredprobability distributionis
x 0 1 2
P(X) 25/36 10/36 1/36
Q.4. Find the probability distributionofthe number of successes in
two tosses of a die, where a success is defined as a number less than 3.
Also find the mean and the variance ofthe distribution.
Mean = Σpixi = 4/9×0+ 4/9×1+ 1/9×2= 6/9= 2/3.[Ans.]
Variance = Σpixi2 – (Mean)2= (4/9×0+ 4/9×12 + 1/9×22) – (2/3)2
= 8/9 – 4/9 = 4/9.[Ans.]
Q.5. An urn contains 5 white and 3 red balls. Find the probability
distributionof the number of red balls, withreplacements,in three
draws.
Hence the required probability distributionis
x 0 1 2 3
P(X) 125/512 225/512 135/512 27/512
Q.6. A pair of dice is thrown 4 times.If getting a doublet is considered
a success,find the probability distributionof number ofsuccess.
Therefore,required probability distributionis
x 0 1 2 3 4
P(X) 625/1296 500/1296 150/1296 20/1296 1/1296
13.5. Binomial Distribution.
Q.1. The mean and variance of a binomial distributions are 4 and 4/3
respectively.Find the distributionand P(X ≥ 1).
728/729.[Ans.]
Q. 2 A drunk man takes a stepforward withprob. 0.4 &backwardwith
prob. 0.6. find the prob. That at the end of11 steps,he is just one step
away from the starting point. Is drinking alcohol a goodhabit?
[ hint required prob.= P(X=5) +P(X=6) = 462(0.24)5 ]
ASSIGNMENT(3-DIMENTIAL GEOMETRY)
Direction Ratio & Direction Cosines of a Line.
Q.1. The equation of a line is given by (4 – x)/2= (y + 3)/3 = (z + 2)/6.
Write the directioncosines of a line parallel to the above line.
l = – 2/7, m = 3/7, n = 6/7 [Ans.]
Q.2. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the
direction cosines of a line parallel to this line. [Ans. = 2/7, 3/7, – 6/7]
11.2. Equation of a Line in Space.
Q.1. Find the foot of the perpendicular drawn from the point P(1, 6, 3)
on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P.
PQ = √13units. [Ans.]
Q.3. Find the point on the line : (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a
distance 3√2 from the point (1, 2, 3).
P(56/17,43/17, 111/17).
Q.4. Find the equationof the perpendicular drawn from the point (2,
4, – 1) to the line (x + 5)/1= (y + 3)/4 = (z – 6)/–9.
(x – 2)/6= (y – 4)/3= (z + 1)/2[Ans.]
Q.2. Find the lengthand the foot of the perpendicular drawn from the
point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11.
[Ans. = Point (1, 2, 3); distance = √14 units.]
11.3. Angle Between Two Lines.
Q.1. Find the angle between the pair of lines given by
r→ = (2i – 5j + k) + λ(3i + 2j + 6k) and r→ = 7i – 6j – 6k + μ(i + 2j + 2k).
θ = cos –1(19/21).[Ans.]
11.4. Shortest Distance Between Two Lines.
Q.1. The vector equations of two lines are :
r→ = i + 2j + 3k + λ(i – 3j + 2k ) and
r→ = 4i + 5j + 6k + μ(2i + 3j + k).
Find the shortest distance betweenthe above lines.
3/√19units. [Ans.]
Q.2. Find the shortest distance betweenthe following lines :
(x – 3)/1= (y – 5)/–2=(z – 7)/1and (x + 1)/7= (y + 1)/–6= (z + 1)/1.
2√29. [Ans.]
Equation of a Plane.
Q.1. Find the equation of the plane passing through the points (1, 2, 3)
and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3.
6x – 3y + z = 3. [Ans.]
Q.2. Find the equation of the plane passing through the points (0, – 1, –
1), (4, 5, 1) and (3, 9, 4). 5x – 7y + 11z + 4 = 0. [Ans.]
Q.3. Find the equationof the plane passing throughthe point (–1,–1,
2) and perpendicular to each of the following planes :
2x + 3y – 3z = 2 and 5x – 4y + z = 6.
11x + 17y + 23z – 18 = 0 [Ans.]
Q.4.Find the equation of the plane passing through the points (3, 4, 1)
and (0, 1, 0) and parallel to the line (x + 3)/2= (y – 3)/7 = (z – 2)/5.
8x – 13y + 15z + 13 = 0. [Ans.]
11.6. Distance of a Point from a Plane.
Q.1. Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.
(3, 6, 11). [Ans.]
Q.2. Find the co-ordinates ofthe image of the point (1, 3, 4) in the
plane 2x – y + z + 3 = 0. [Ans. = (–3, 5, 2) ]
Q.3. From the point P(1, 2, 4), a perpendicular is drawn on the plane
2x + y – 2z + 3 = 0. Find the equation, the lengthand the co-ordinates
of the foot of the perpendicular. (11/9, 19/9, 34/9) [Ans.]
Length of perpendicular from (1, 2, 4) is
PM = |{2(1) + 2 – 2(4) + 3}/√(4+ 1 + 4)| = 1/3 unit. [Ans.]
Q.4. Find the distance between the point P(6,5, 9) and the plane
determinedby the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).
6/√(34).[Ans.]
Q.5. Find the co-ordinates ofthe point where the line (x + 1)/2= (y +
2)/3= (z + 3)/4 meets the plane x + y + 4z = 6. P(1, 1, 1) [Ans.]
Q.6. Find the distance of the point (– 2, 3 – 4) from the line
(x + 2)/3= (2y + 3)/4= (3z + 4)/5
measuredparallel to the plane 4x + 12y – 3z + 1 = 0. 17/2.[Ans.]
Class – XII Subject – Mathematics (Three Dimensional Geometry)
1. Find the d.c’s of X, Y and Z-axis.[example 1,0,0(x-axis)0,1,0(y-axis]
2. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6.
Write the direction cosines of a line parallel to the above
line. [Ans: – 2/7, 3/7, 6/7]
3. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the
direction cosines of a line parallel to this line. [ Ans : 2/7, 3/7, – 6/7]
4. Find coordinates of the foot of the per. drawn from the origin to
the plane 2x – 3y + 4z -6 = 0. [example ans. 12/29, -18/29, 24/29]
5. Find the angle between the line (x + 1)/2 = y/3 = (z-3)/6 and the
plane 10x + 2y – 11z = 3.[example ans. sinφ = |
𝒃⃗⃗ .𝒏⃗⃗
| 𝒃⃗⃗ ||𝒏⃗⃗ |
| = 8/21]
*6. Find the image of the point (1, 6, 3) in the line x = (y – 1)/2 = (z –
2)/3. [ Hint: it will be found that foot of per. from P to the line is N
(1,3,5). If Q(𝜶, 𝜷, 𝜸) is the image of P then N is the mid point of PQ
⇨
𝜶+𝟏
𝟐
= 𝟏,
𝜷+𝟔
𝟐
= 𝟑,
𝜸+𝟑
𝟐
= 𝟓 ⇨ Q(1, 0, 7)]
7. Find the foot of the perpendicular drawn from the point P(1, 6, 3)
on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from
P. [ Ans : √13 units., eqn. Of line is (x-1)/0=(y-6)/-3=(z+1)/2]
8. Find the length and the foot of the perpendicular drawn from the
point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –
11. [ Ans. = Point (1, 2, 3); distance = √14 units.]
*9. Show that the angles between the diagonals of a cube is cos-
1
(1/3).
Ans. (same as example. 26 of misc.) Let the length of cube be a , in
fig. Of example : A,B,C are on x-axis, y-axis, z-axis resp.& F,G,D are
in yz, xz ,xy –axis , O is origin , d.r’s of OE (E (a,a,a)) & AF are a,a,a &
-a,a,a resp. ∴ cosѲ =(-a2
+a2
+a2
)/√3a.√3a = 1/3(angle b/w two lines)
10. Find the length of the perpendicular drawn from the point (2, 3,
7) to the plane 3x – y – z = 7 . Also find the coordinates of the foot
of the perpendicular. [ same as Q. 8 ]
*11. Find the point on the line (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a
distance 3√2 from the point (1, 2, 3).
Ans. (x + 2)/3 = (y + 1)/2 = (z – 3)/2 =k , any point on the line A(3k-2,
2k-1, 2k+3) , if its distance from B(1,2,3) is 3√2, then AB² = (3√2)²
⇨ K= 0 OR 30/17 , ∴ point A(-2,-1,3) OR A(56/17,43/17,111/17) [by
putting value of k in A]
12. Find the equation of the perpendicular drawn from the point (2,
4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 .
[ Ans : (x – 2)/6 = (y – 4)/3 = (z + 1)/2]
*13. Find the equation of the line passing through the point (-1, 3, -
2) and perpendicular to the lines x = y/2 = z/3 and (x + 2)/(-3) = (y –
1)/2 = (z + 1)/5.
Ans. eqn. Of line (x+1)/a=(y-3)/b=(z+2)/c ........(i) ,where a,b,c are
d.r’s , since (i) is per. to lines ∴ a+2b+3c=0 & -3a+2b+5c=0, after
solving , we get a/4=b/-14=c/8 =k(say) , so eqn. Of line (by putting
the values of a=4k, b=-14k,c=8k) is (x+1)/4 = (y-3)/-14 = (z+2)/8
14. Find the foot of the per. drawn from the point (0, 2, 3) on the
line (x + 3)/5 = (y – 1)/2 = (z + 4)/3 Also, find the length of the
perpendicular. [ans..(2,3,-1), use distance formula, length= √21]
*15. Find the equation of the plane passing through the line
intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and
parallel to the line with direction ratios (1, 2, 1) Also, find the
perpendicular distance of the point P(3, 1, 2) from this plane.
Ans. equation of the plane passing through the line intersection of
the planes x – 2y + z -1 +k(2x + y + z – 8)=0 or (1+2k)x+(-
2+k)y+(1+k)z-1-8k=0 ........(i) is parallel to line ∴ (1+2k).1+(-
2+k).2+(1+k).1=0 ⇨ k= 2/5(normal to the plane must be per. to the
line), put k in (i) ⇨ 9x-8y+7z-21=0, per. dis. From P Is 22/√(194)
*19. Find the shortest distance between the following lines :
(x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and
(x + 1)/7 = (y + 1)/–6 = (z + 1)/1.
[ncert Ans : 2√29 , use formula ]
20. Find the equation of the plane passing through the points (1, 2,
3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3.
[Ans :same as Q. 26 , 6x – 3y + z = 3]
21. Find the shortest distance between the following pairs of lines
whose cartesian equations are : (x -1)/2 = (y + 1)/3 = z and (x + 1)/3
= (y – 2) , z = 2. [ SAME as Q. 19 ]
*22. Find the distance of the point P(2, 3, 4) from the plane 3x + 2y
+2 z + 5=0 measured parallel to the line (x+3)/3 = (y-2)/6 = z/2
Ans. Any point on line (x+3)/3 = (y-2)/6 = z/2=k is
Q(3k+2,6k+3,2k+4) It lies on plane ⇨ k=-1 ∴ point Q( -1,-3,2),PQ=7
23. Find the equation of the plane passing through the points (0, –
1, – 1), (4, 5, 1) and (3, 9, 4). [ Ans : 5x – 7y + 11z + 4 = 0]
*24. Find the equation of the plane passing through the point (–1, –
1, 2) and perpendicular to each of the following planes :
2x + 3y – 3z = 2 and 5x – 4y + z = 6.
[ Ans : 9x + 17y + 23z – 20 = 0, same as example of misc. Eqn. Of
plane a(x+1)+b(y+1)+c(z-2)=0........(i) By condition of
perpendicularity to the plane (i) with the planes 2a+3b-3c=0 & 5a-
4b+c =0 , after solving , we get a=9c/23 & b=17c/23, put in (i) ]
25. Find the equation of the plane passing through the point (1, 1, -
1) and perpendicular to the planes x + 2y + 3z – 7=0 and 2x – 3y +
4z = 0. [ SAME as Q. 24]
26. Find the equation of the plane passing through the points (3, 4,
1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z –
2)/5.
Ans : 8x – 13y + 15z + 13 = 0, eqn. Of plane a(x-3)+b(y-4)+c(z-
1)=0.......(i),It passes through (0,1,0) ,then -3a-3b-c=0 // to line ∴
2a+7b+5c=0, after solving a=8c/15, b=-13c/15, put in (i)
*27. Find the image of the point P(1,2,3) in the plane x+2y+4Z=38
[ Ans :Let M is foot of per. (mid point of PQ) , if Q (𝜶, 𝜷, 𝜸 ) , d.r’s of
a normal to the plane are 1,2,4, eqn. Of line PM is (x-1)/1=(y-
2)/2=(z-3)/4 = k ⇨ any point on a line Q(1+K,2+2K,3+4K) ∴
M{(2+k)/2,(4+2k)/2,(6+4k)/2} put in plane ⇨ k=2 ∴ Q(3,6,11) ]
28. Find the co-ordinates of the image of the point (1, 3, 4) in the
plane 2x – y + z + 3 = 0. [Ans : (–3, 5, 2)]
*29. Find the distance between the point P(6, 5, 9) and the plane
determined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).
Ans. [example28 of misc. Ans: 6/√(34),we can find eqn. Of plane
through A,B,C and find distance of P from plane or can be found as
PD = 𝑨𝑷⃗⃗⃗⃗⃗⃗ . 𝑨𝑩⃗⃗⃗⃗⃗⃗ × 𝑨𝑪⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ( proj. Of 𝑨𝑷⃗⃗⃗⃗⃗⃗ on 𝑨𝑩⃗⃗⃗⃗⃗⃗ × 𝑨𝑪⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ , D is foot of per. from
P) or find eqn. of plane passes through A,B,C , then find dist. ]
30. Find the co-ordinates of the point where the line (x + 1)/2 = (y +
2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6. [Ans: P(1, 1, 1)]
*31. Find the distance of the point A(– 2, 3 – 4) from the line
(x + 2)/3 = (2y + 3)/4 = (3z + 4)/5
measured parallel to the plane 4x + 12y – 3z + 1 = 0.
[Ans: |AP|= 17/2, eqn. Can be written as (x + 2)/3 = (y + 3/2)/2 =
(z + 4/3)/5/3 = t ⇨ P{ -2+3t, (-3/2)+2t, (-4/3)+5t/3} be any point on
line D.R’S of AP {3t, 2t-(9/2), (5t/3)+(8/3)} , since AP is parallel to
plane 4.3t+ (12).[ 2t-(9/2)]+ (-3). (5t/3)+(8/3) = 0 ⇨ t=2 ∴ P
(4,5/2,2) ]
Ques. 32 Find the distance of the point P(-2, 3, -4) from the line
(x+2)/3 = (2y+3)/4 = (3z+4)/6 measured // to the plane 4x + 12y -3 z
+ 1=0.
Ans. same as Q. 22, point on lineQ(3k-2,2k-(3/2),(5k/3)-(4/3)), d.r’s
of PQ are (3k,2k-(9/2),(5k/3)+(8/3)) , since PQ is // to plane ∴ PQ is
per. to normal to the plane ⇨ (3k).4+{2k-(9/2)}.12+{(5k/3)+(8/3)}.(-
3)=0 ∴ k= 2 ⇨ Q(4,5/2,2), PQ = 17/2
Question 33 find whether the lines 𝒓⃗ = (i-j-k) +λ (2i+j) and 𝒓⃗ =
(2i-j) +μ (i+j-k) intersect or not . if intersecting, find their point of
intersection.
Ans. 𝒂 𝟐⃗⃗⃗⃗ -𝒂 𝟏⃗⃗⃗⃗ = I +k , 𝒃 𝟏
⃗⃗⃗⃗ × 𝒃 𝟐
⃗⃗⃗⃗ =
𝒊 𝒋 𝒌
𝟐 𝟏 𝟎
𝟏 𝟏 −𝟏
= -i+2j+k , its
magnitude is √6
Shortest distance = = 0/√6 = 0
Hence lines intersect. Point of intersection is given by
(i-j-k) +λ (2i+j) = (2i-j) +μ (i+j-k) ⇨ (1+2λ) = 2+μ , -1+λ - -1+μ
and -1 = -μ ⇨ λ = 1 and μ =1 satisfy μ = 2λ -1 , put the values of
λ and μ in given lines , we obtain the position vector of point of
intersection of the two given lines as 3i- k i.e., the point of
intersection is (3,o,-1).
Ques. 34 Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (-
4,4,4) are coplanar. Also find the eqn. of the plane containing them.
Ans. Eqn. of plane passes through (0,-1,-1) is a(x-0)+b(y+1)+(z+1)=0
It passes through (4,5,1), (3,9,4), we get 4a+6b+2c=0, 3a+10b+5c=0
⇨ eqn. of plane is 5x-7y+11z+4=0 and (-4,4,4) will satisff the eqn. of
plane to be coplanar.
Q. 35 Show that lines (x+3)/-3 = (y-1)/1 = (z-5)/5 & (x+1)/-1=(y-2)/2
=(z-5)/5 are coplanar . find the eqn. of plane.
Ans.(as example 21), The given lines are coplanar if
(𝒙𝟐 − 𝒙𝟏) (𝒚𝟐 − 𝒚𝟏) (𝒛𝟐 − 𝒛𝟏)
𝒍𝟏 𝒎𝟏 𝒏𝟏
𝒍𝟐 𝒎𝟐 𝒏𝟐
=
(−𝟏 + 𝟑) (𝟐 − 𝟏) 𝟎
−𝟑 𝟏 𝟓
−𝟏 𝟐 𝟓
= 0 , so lines are coplanar. Eqn of plane
=
(𝒙 − 𝒙𝟏) (𝒚 − 𝒚𝟏) (𝒛 − 𝒛𝟏)
𝒍𝟏 𝒎𝟏 𝒏𝟏
𝒍𝟐 𝒎𝟐 𝒏𝟐
=
(𝒙 + 𝟑) (𝒚 − 𝟏) (𝒛 − 𝟓)
−𝟑 𝟏 𝟓
−𝟏 𝟐 𝟓
= x-2y+z=0.
Q.36 Find the shortest distance and eqn. Of shortest distance b/w
the lines (x-6)/3 = (y-7)/-1 = (z-4)/1 & x/-3 = (y+9)/2 = (z-2)/4 .
S (3p+6,-p+7,p+4) ......(i)
D (-3q,2q-9,4q+2) .......(ii)
Ans. let SD be the shortest dist. , then co-ordinates of eqn.(i) are (
3p+6, -p+7, p+4) & eqn.(ii) are (-3q, 2q-9, 4q+2)
d.c’s of SD are proportional to -3q-3p-6, 2q+p-16, 4q-p-2 and if SD is
at right angles to both lines , we get 3.( -3q-3p-6)+(-1).( 2q+p-16)+ 1.
(4q-p-2)=0 and -3.(-3q-3p-6)+2.( 2q+p-16)+4.( 4q-p-2) =0 ⇨
7q+11p+4=0 & 29q+7p-22=0 ⇨ p=-1 , q=1 and SD=3√(30) , eqn. Will
be (x-3)/-6 = (y-8)/-15 = (z-3)/3
Q.37 Find the shortest distance and eqn. b/w the lines 𝒓⃗ = (8+3λ)i-
(9+16λ)j+(10+7λ)k & 𝒓⃗ = (15i+29j+5k)+μ(3i+8j-5k).
Ans. let SD be the shortest dist. b/w two lines , let position vector
of S be (8i-9j+10k)+λ(3i-16j+7k) and position vector of D is
(15i+29j+5k+μ(3i+8j-5k), 𝑺𝑫⃗⃗⃗⃗⃗ = 𝑫⃗⃗ − 𝑺⃗⃗ = (3μ-3λ+7)i+(8μ+16λ+38)j+(-
7λ-5μ-5)k since SD is perpendicular on line (i) & (ii) then
[(3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k]. (3i-16j+7k)=0 ⇨
157λ+77μ+311=0...(iii) & 77λ+49μ+175=0...(iv), by solving λ = -1,μ=-
2 ∴ 𝑺𝑫⃗⃗⃗⃗⃗ = 𝑫⃗⃗ − 𝑺⃗⃗ = 4i+6j+12k ⇨| 𝑺𝑫⃗⃗⃗⃗⃗ |= 14 & eqn. Is 𝒓⃗ =
(5i+7j+3k)+t[(9i+13j+15k)-(5i+7j+3k)] = (5i+7j+3k)+t(4i+6j+12k)
where 𝑺⃗⃗ =5i+7j+3k, 𝑫⃗⃗ =9i+13j+15k & use 𝒓⃗ =𝒂⃗⃗ +t(𝒃⃗⃗ − 𝒂⃗⃗ ).
Vector Algebra
Q. 1. Find a vector in the direction of vector that has
magnitude 7 units.
Q. 2. Show that the points A, B and C with position
vectors, respectively, form the
vertices of a right angled triangle.
Q. 3. Find
, if two vectors are such that .
Q. 4. Find the area of the parallelogram whose adjacent sides are
determined by the vectors
Q. 5. If a unit vector makes angles with , and acute
angle θ with , then find θ and hence the components of .
Q. 6. Let , and be three vectors such that and
each one of them being perpendicular to the sum of other two,
Find
Q. 7. Find the value of
Q. 8. The scalar product of the vector with a unit vector along
the sum of vectors Find the value
of .
Q. 9. If the sum of two unit vectors is a unit vector, Prove that the
magnitude of their difference is .
Q. 10. If are position vectors of points A and B respectively,
then find the position vector of points of trisection of AB.
Q. 11. Prove that the line segment joining the mid-points of two sides
of a triangle is parallel to the third side and equal to half of it.
Q. 12. ABCD is a parallelogram. If the coordinates of A, B, C are (-2, -1),
(3, 0) and (1, -2) respectively, Find the co-ordinate of D.
Q. 13. Show that the points A, B, C with position
vectors
Q. 14. If a vector makes with OX, OY and OZ respectively, prove
that
Q. 15. If inclined at an angle , then prove that
sin = .
Q. 16. If .
Q. 17. If .
Q. 18. Consider two points P and Q with position vectors
. Find the position vector of a point R which
divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii)
externally.
Q. 19. Find a unit vector perpendicular to each of the
vectors
Q. 20. Show that the vectors 2 i – j + k , i– 3 j – 5 k and 3 i – 4 j – 4 k
form the vertices of a right angled triangle
Q. 21.
ASSIGNMENT(LINEARPROGRAMMING)
Q.1. David wants to invest at most Rs12,000in Bonds A and B.
According to the rule, he has to invest at least Rs2,000in Bond A and
at least Rs4,000 in Bond B. If the rate of interest in bonds A and B
respectively are 8% and 10% per annum, formulate the problem as
L.P.P. and solve it graphically for maximum interest.Also determine
the maximum interest receivedin a year. Do you think your
investment in bond A &B help in the well being of the nation?Do you
think that a person shouldstart saving at an early age for his
retirement?
[ANS:Therefore,the interest I is maximum at C(2000, 10000) and the
maximum interest is Rs1160.]
Q.2. A farmer has a supply of chemical fertilizer oftype A which
contains 10% of nitrogenand 6% of phosphoric acidand of type B
which contains 5% ofnitrogen and 10% of phosphoric acid. After soil
testing it is found that at least 7 kg of nitrogenand the same quantity
of phosphoric acid is requiredfor a goodcrop. The fertilizer oftype A
costs Rs5.00per kg and the type B costs Rs8.00per kg. Using linear
programming findhow many kgs ofeach type of the fertilizer should
be bought to meet the requirement and the cost be minimum.Solve
the problem graphically.
[ANS: Therefore,cost is minimum at P(50,40). That is 50 kg of type A
and 40 kg oftype B is mixed to meet the requirement.
Q.3. A dietician wishes to mix two types of foods in such a way that
vitamincontents ofthe mixture contain at least 8 units of vitamin A
and 10 units ofvitamin C. Food I contains 2 units/kg of vitaminA and
1 unit/kg ofvitamin C. Food II contains 1 unit/kg of vitaminA and 2
units/kg of vitaminC. It costs Rs50per kg to purchase Food I and Rs70
per kg to purchase FoodII. Formulate this problem as a linear
programming problem to minimize the cost of such a mixture.
[ANS:Hence,the minimum value of Z is Rs380.]
Q.4. Two tailors A and B are paid Rs150and Rs200per day
respectively.A can stich6 shirts and 4 paints while B can stich10
shirts and 4 paints per day. Form a linear programming problem to
minimize the labour cost to produce at least 60 shirts and 32 paints.
Solve the problem graphically. These shirts & paints have been
specially designed for a ‘VAN MAHOTSAVA’ programme.Wouldyou
like to participate in this programme? Why?
[Ans. = Tailor A works for 5 days and B for 3 days; Minimum cost =
Rs1,350]
Q.5. A cooperative society offarmers has 50 hectare of land to grow
two crops X and Y. The profit from crops X and Y per hectare are
estimatedas Rs10,500 and Rs9,000respectively.To control weeds,a
liquid herbicide has to be used for crops X and Y at rates of20 litres
and 10 litres per hectare. Further, not more than 800 litres of
herbicide should be used in order to protect fishand wild life using a
pond which collects drainage from this land. Howmuch land should
be allocatedto eachcrop so as to maximize the total profit of the
society ? In what ways does a cooperative society helpthe farmers?
[ANS:Hence , society will get the maximum profit of Rs4,95,000 by
allocating 30 hectare for cropX and 20 hectare for cropY.]
Q.6. Kellogg is a new cereal formedof a mixture of bran and rice that
contains at least 88 grams ofproteinand at least 36 milligrams of
iron. Knowing that bran contains 80 grams ofprotein and 40
milligrams ofironper kilogram,and that rice contains 100 grams of
proteinand 30 milligrams ofironper kilogram,find the minimum
cost of producing this new cereal if bran costs Rs5per kilogram and
rice costs Rs4per kilogram.
[ANS: Therefore,minimum cost of producing this cereal is Rs4.60 per
kg.]
Q.7. A new cereal, formedof a mixture of bran and rice, contains at
least 88 grams of proteinand at least 36 milligrams ofiron.Knowing
that bran contains 80 grams of proteinand 40 milligrams ofironper
kilogram,and that rice contains 100 grams ofprotein and 30
milligrams ofironper kilogram,find the minimum cost of producing a
kilogram ofthis new cereal if bran costs Rs.28 per kilogram and rice
costs Rs.25 per kilogram.
Solution: – Do yourself.[Ans. = Rs.26.8]
Q.8. A dealer wishes to purchase a number of fans and CFLs. He has
only Rs5,760to invest and has space for at most 20 items.A fan cost
him Rs360and a CFL Rs240. He expects to sell a fan at a profit of Rs22
and a CFL at a profit of Rs18. Assuming that he can sell all the atoms
that he buys, how should he invest his money to maximize the profit ?
Solve graphically and find the maximum profit.What is the full form of
CFL? Which one iswmore energy efficient: CFL or ordinary
incandescent bulb?
[ANS: Therefore,for maximum profit he shouldbuy and sell 8 fans and
12 CFLs. His maximum profit is Rs392.]
Q.9. A man has Rs1,500 for purchasing rice and wheat. A bag ofrice
and a bag ofwheat cost Rs180and Rs120respectively.He has the
storage capacity of at most 10 bags.He earns a profit ofRs11 and Rs9
per bag ofrice and wheat respectively.Formulate the above problem
as an LPP to maximize the profit and solve it graphically.
Solution:
Do yourself.
[Ans. Max. Profit = Rs100, No.of rice bags = 5 = No.of wheat bags]
Q.10. A factory owner purchases two types ofmachines, A and B for
his factory.The requirements and the limitations for the machines are
as follows:
Machine
Area
occupied
Labour
force
Daily
output
(in
units)
A 1000 m2
12
men
60
B 1200 m2 8 men 40
He has maximum area of 9000 m2 available,and 72 skilledlabourers
who can operate both the machines.How many machines of eachtype
should he buy to maximize the daily output?
Solution: [ANS: The maximum output is at B and C. But the number of
machines cannot be a fraction. Hence no. ofmachines of type A = 6 and
no. of machines of type B = 0 ]
Q.11 Ifa 19 years oldgirl drives her car at 25 km/hr,she has to spend
Rs.2/km on petrol.If she drives it at a faster speedof 40 km/hr, the
petrol cost increases to Rs.5/km.She has Rs.100 to spend on petrol
and wishes to find the max. Distance she can travel within one hour .
express it as a L.P.P. and then solve it . Is the girl eligible for a driving
licence? What is the benefit of driving at economic speed?
Solution:[ans: max distance is 30 km.at (50/3,40/3) .yes ∵ minimum
age require for licence is 18 years.Reduces fuel consumptionand
causes less pollution.]
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Class xii practice questions

  • 1. CLASS XII (ASSIGNMENTS) Matrices and Determinants Properties of determinants :- 1. The value of a determinant remains unchanged if its rows and columns are interchanged. 2. The sign of value of a determinant is changed if its any two rows or columns are interchanged. 3. The value of a determinant is zero if its any two rows or columns are identical. 4. If a determinant is multiplied by a scalar (number) , its only one row or column gets multiplied by that constant. 5. If any two row or column of a determinant are proportional, its value becomes zero. 6. If all elements of a row or column are expressed as sum of two or more elements ,the whole of the determinant can be expressed in sum of two or more determinants. 7. If some multiple of one row or column is added or subtracted to another row or column (elementwise), its value remains unchanged. Tips to solve properties based problems:- 1. If a determinant is of order ,we can apply only n-1 propertis at a time to it. 2. The format of applicationof properties is :-Row affected Row affected n (Rowused) Ex. 3. The format for interchanging Rows or columns :- 4. You can never multiply a number to Row affected, it is always multiplied to Row used. 5. First always try to make elements of any one row or column identical so that you could take out common from that row or column. It makes all the elements of that Row or column unity(1) and then you make at the most two elements of that row or columns zero (0).Now expand that the determinant by that row or column.
  • 2. EXAMPLE: We shall apply and Taking out common b from Expand by = Important Questions (ASSIGNMENT) Q. 1. Construct a 3X4 matrix, whose elements are given by aij= Q. 2. Construct a 2x3 matrix A = [aij] whose elements are given by aij
  • 3. = , i ≠ j = |i + 2j|, i = j Q.3. Construct a 3 × 3 matrix whose elements aij are givenby Q.4. Howmany number of matrices are possible of order 3 × 3 with each entry 0 or 1. Q. 5. Express the matrix A= as the sum of symmetric and skew-symmetric matrix Q. 6. Obtain the inverse of the matrix A = using elementary transformations. Q. 7. If f(x)= Prove that f(x). f(y) = f(x + y) 8. Find x,y,z if [x 3 2] =[0 0 1] Q. 9. If A and B are invertible matrices of the same order, then prove that (AB)-1 = B-1A-1 10. Express as a sum of a symmetric and a skew – symmetric matrix.
  • 4. Q. 11. If A = Show that A2 -5A + 7I = 0, Use this to find A-1. Q. 12. Express the matrix A = as the sum of a symmetric and a skew-symmetric matrix. Q. 13. Find the values of x, y, z if the matrix A = satisfy the equation A'A = I3. Q. 14. Show that : = Q. 15. Find the inverse of , if a2 + b2 + c2 + d2 = 1. Q. 16.(i) Find x such that:
  • 5. (ii) find x, p, y &q (iii) Find the value of x, if Q. 17. For what value of k the matrix A = has no inverse. Q. 18. Prove that the product of matrices and is the null matrix, when and differ by an odd multiple of . Q. 19. A matrix X has a + b rows and a + 2 columns while the matrix Y has b + 1 rows and a + 3 columns. Both matrices XY and YX exist. Find a and b. Can you say XY and YX are of the same type? Are they equal. Q. 20. Find the matrix A satisfying the matrix equation A = 21. If A = , B = , c = Find a matrix D such that CD - AB = 0. 22. Let A = , Verify that.
  • 6. 23. If A = ,find k so that. 24.Find B if . 25. Find A = , find a and b such that such that where I is unit matrix of order 2. Q. 26.Prove that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric. Hence represent as above. 27. If , prove that ,n N 28. If A = ,Using principle of mathematically induction prove that 29. If A = , B = , c = Find a matrix D such that CD - AB = 0. 30. Let A = , Verify that. Q. 31.(i)Prove that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric. Hence represent
  • 7. as above. (ii) Express the matrix as the sum of symmetric and skew symmetric matrix. Q.32(i).Using elementary row transformation find the inverse of the matrix (ii) Using the method of reduction (i.e elementary row transformations),find the inverse of (iii)IfA = ; showthat A2 = A-1 . (Without using elementary transformations) (iv) Using elementary transformation , find the inverse of the matrix.
  • 8. Q. 33.(i) Showthat the matrix A = satisfies the equation x2 – 4x –5=0. Hence find A-1 (ii) Solve the equations by matrix method (iii) Solve using matrices : (iv) Solve by matrix method: 2x + y + z = 1 x - 2y – z = 3/2 3y - 5z = 9 34.(i) If A = and B = find the product AB and use this result to solve the following system of equations: (ii) If find A-1 and use it solve the system of equations: (iii) Determine the product
  • 9. and use it solve the system ofequations : x-y+z=4 x-2y-2z=9 2x+y+3z=1 (iv) If A = , find A-1, using A solve the following system of linear equations. 2x – y + z + 3 = 0 3x – z + 8 = 0 2x + 6y – 2 = 0 35.(i) For what value of a and b ,the following system of equations is consistent? (ii) Find whether the following system ofequations is consistent or not, find the solutionofthe system also.3x-y+2z=3, x-2y-z = 1, 2x+y + 3z = 5. (iii) Using matrix methed, solve the following system ofequations: (iv) Solve the following system oflinear equation (v)The sum of three numbers is 6. If we multiply third number by 3 and add secondnumber to it, we get 11. By adding first and third numbers, we get double of the secondnumber. Represent it algebraically and find the numbers using matrix method.
  • 10. (vi) The management committee of a residential colony decided to award some of its members for honesty, some for helping others and some others for supervising the workerstokeepthe colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervisionaddedtotwo times the number of awardees for honesty is 33. If the sum of the number awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from above values, suggest one more value which the management of the colony must include for awards. (vii) Showthat the following system ofequations is consistent 2x – y + 3z = 5, 3x + 2y – z = 7, 4x + 5y – 5z = 9. Also,find the solution. (viii)Using matrix method,solve the following system of equations for x, y and z : Determinants Q. 1. (i)Find the equation of the line joining A(1,3) and B(0,0) using determinants and find if D (K, 0) is a point such that area of a triangle ABD is 3 square units. (ii) if | 𝒙 + 𝟏 𝒙 − 𝟏 𝒙 − 𝟑 𝒙 + 𝟐 | = | 𝟒 −𝟏 𝟏 𝟑 |, then write the value of x. Q. 2. Using determinants,find the area ofthe triangle whose vertices are (1, 4), (2, 3), (-5, 3). Are the givenpoints collinear. Q. 3. If the points (a1, b1), (a2, b2) and (a1 + a2,(b1 + b2) are collinear, Showthat a1b2 = a2b1. Q. 4. If a, b and c are real numbers,and
  • 11. , show that either a + b + c = 0 or a = b = c. Q.5. Using properties ofdeterminants, prove that = 0. Q. 6. Without expanding showthat Q. 7. Prove that : Q. 8. Solve : Q. 9. If a, b, c are all positive and are pth , qth and rth terms ofG.P., then showthat Q. 10. If = 0, then Prove that a, b, c are in G.P or x, y, z are in G.P Q. 11. If x, y, z are different and
  • 12. Q. 12. Showthat points A (a, b + c), B (b,c + a),C (c, a + b) are collinear. Q.13. Using properties of determinants,prove that: Q.14. Using properties ,prove that: Q. 15. Using properties of determinants solve for x: Q16. Using properties prove that : Q.17. Prove using properties of determinants. Q. 18. Showthat :
  • 13. Q. 19. Prove that : Q. 20. ASSIGNMENT ASSIGNMENT(matrices) Qoestion.1 Using matrices, solvethe following systemof equations (i) x+2y+z = 1 , 2x – y+z = 5 , 3x+y – z = 0. [Hint use AX= B ⇨ X = A-1 B, |A|=15≠0 meansA is invertible.Adj(A)= [ 𝟎 𝟑 𝟑 𝟓 −𝟒 𝟏 𝟓 𝟓 −𝟓 ] ,A-1 = 𝒂𝒅𝒋(𝑨) |𝑨| Ans. x =1, y=-1, z=2] (ii) 2x+y – 3z = 13, x + y – z = 6 , 2x – y+4z = -12. [ Ans. |A| = 9, adj(A) = [ 𝟑 −𝟏 𝟐 −𝟔 𝟏𝟒 −𝟏 −𝟑 𝟒 𝟏 ] , x=1, y=2, z=-3.] (iii) 2x+y+z = 1 , x – 2y – z = 3/2 , 3y – 5z = 9. [Hint |A| = 34, adj (A) = [ 𝟏𝟑 𝟖 𝟏 𝟓 −𝟏𝟎 𝟑 𝟑 −𝟔 −𝟓 ], x=1, y= 1/2., z=-3/2.]
  • 14. Question.2 Use the product [ −𝟒 𝟒 𝟒 −𝟕 𝟏 𝟑 𝟓 −𝟑 −𝟏 ] [ 𝟏 −𝟏 𝟏 𝟏 −𝟐 −𝟐 𝟐 𝟏 𝟑 ] to solve the equations x – y+z = 4, x – 2y – 2z = 9, 2x+y+3z = 1. [Hint take product of above two matrices,we get identity matrix,then use AB=BA = I means B is the inverse of A Or A is the inverse of B. ⇨ [ −𝟒 𝟒 𝟒 −𝟕 𝟏 𝟑 𝟓 −𝟑 −𝟏 ] [ 𝟏 −𝟏 𝟏 𝟏 −𝟐 −𝟐 𝟐 𝟏 𝟑 ] = 8I3 , according to above equation let A let B (1/8) B is the inverse of A. Ans. x=3, y=-2, z=-1.] Question.3 Solve the following system ofhomogenous equations: 2x+3y – z = 0, x – y – 2z = 0, 3x+y+3z = 0. Solution: system of homogenous equations can be writtenas AX = O , A = [ 𝟐 𝟑 −𝟏 𝟏 −𝟏 −𝟐 𝟑 𝟏 𝟑 ] ,|A| = -33 So, the system has only the trivial solutiongivenby x=y=z=0. If|A| = 0 then system has non-trivial solution.] Question.4 Showthat system ofequations x+y – z = 0, x – 2y+z = 0, 3x+6y – 5z = 0 has non-trivial solution.Find sol.Answer: |A| = 0, it has infinitely many solutions ∴ let z = k a arbitrary x+y = z = k , x – 2y = -z = -k , 3y = 2k i.e, y = 2k/3 ⇨ x = k/3from first equation by putting the values of x, y & z in thirdequation, we get 0 which is true. The requiredsolutionis z = k, y = 2k/3, x = k/3 where k is arbitrary. Question.5 Showthat system ofequations3x+2y +7 z = 0, 4x – 3y - 2z = 0, 5x+9y +23z = 0 has non-trivial solution.Find the solution. [Hint x = -k, y = -2k, z = k] Question.6 The system ofequations 2x+3y = 7 , 14x+21y = 49 has (a) only one solution (b) finitely many solution (c) no solution (d) infinitely many solution . [give reason]
  • 15. Question.6 Find the inverse (using elementary transformations) of following matrices: (i) A = [ 𝟎 𝟏 𝟐 𝟏 𝟐 𝟑 𝟑 𝟏 𝟏 ] (ii) A = [ 𝟑 𝟒 𝟐 𝟎 𝟐 −𝟑 𝟏 −𝟐 𝟔 ] [Hint: A-1 = [ 𝟑 −𝟏𝟒 −𝟖 −𝟑/𝟐 𝟖 𝟗/𝟐 −𝟏 𝟓 𝟑 ] ,R1↔R3,R3→R3 – 3R1,R1→R1+R2,R2→1/2R2,R3→R3 – 10R2, R3→-R3,R1→R1– 3R3,R2→R2+3/2R3] (iii) A = [ 𝟏 𝟑 −𝟐 −𝟑 𝟎 −𝟓 𝟐 𝟓 𝟎 ] [Hint: A-1 = [ 𝟏 −𝟐/𝟓 −𝟑/𝟓 −𝟐/𝟓 𝟒/𝟓 𝟏𝟏/𝟐𝟓 −𝟑/𝟓 𝟏/𝟐𝟓 𝟗/𝟐𝟓 ], Question. IfA is singular matrix then under what conditionset of equations AX = B may be consistent. [answer if (adjA)B = O ,then eqns. Will have infinitly many sols.Hence consistent.] Question. If A is a square matrix of order 3 such that |adjA| = 289, find |A|. [ |A| = ±17 ∵ |adjA| = |A|n-1.] Q. The management committee of a residential colony decided to award some of its members for honesty, some for helping others and some others for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervisionadded to two times the number of awardees for honesty is 33. If the sum of the number awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from above values, suggest one more value which the management of the colony must include for awards. [HINT: x+y+z = 12, 3(y+z) + 2x = 33, x – 2y +z = 0, |A|= 3, Adj A = 𝟗 −𝟑 𝟎 𝟏 𝟎 −𝟏 −𝟕 𝟑 𝟏 X=3, Y=4 &Z=5. Regularity, sincerity.] ASSIGNMENT( WITH HINTS)(determinant)
  • 16. Question: (i) Let ∆ = 𝑨𝒙 𝒙² 𝟏 𝑩𝒚 𝒚² 𝟏 𝑪𝒛 𝒛² 𝟏 and ∆₁ = 𝑨 𝑩 𝑪 𝒙 𝒚 𝒛 𝒛𝒚 𝒛𝒙 𝒙𝒚 , then ∆ - ∆′₁ = 0 [Hint ∆₁ = 𝒙𝒚𝒛 𝒙𝒚𝒛 𝑨𝒙 𝑩𝒚 𝑪𝒛 𝒙² 𝒚² 𝒛² 𝟏 𝟏 𝟏 ] (ii) If f(x) = 𝟎 𝒙 − 𝒂 𝒙 − 𝒃 𝒙 + 𝒂 𝟎 𝒙 − 𝒄 𝒙 + 𝒃 𝒙 + 𝒄 𝟎 , then which is correctf(a)=0, f(b)=0, f(0)=0 and f(1)=0 [ Hint f(0)=0 det.(skewsymm.matrix)=0]. **(iii) Let f(t) = 𝒄𝒐𝒔𝒕 𝒕 𝟏 𝟐𝒔𝒊𝒏𝒕 𝒕 𝟐𝒕 𝒔𝒊𝒏𝒕 𝒕 𝒕 , then 𝐥𝐢𝐦 𝒕→𝟎 𝒇(𝒕) 𝒕² is equal to 0,1,2,3. [Hint 0, 𝒇(𝒕) 𝒕² = 𝒄𝒐𝒔𝒕 𝒕 𝟏 𝟐𝒔𝒊𝒏𝒕 𝒕 𝟏 𝟐 𝒔𝒊𝒏𝒕 𝒕 𝟏 𝟏 → 𝟏 𝟎 𝟏 𝟐 𝟏 𝟐 𝟏 𝟏 𝟏 as t→∞]. (iv) There are two values of a which makes determinant ∆ = 𝟏 −𝟐 𝟓 𝟐 𝒂 −𝟏 𝟎 𝟒 𝟐𝒂 = 86, then sum of these numbers is 4,5,-4,9. [Hint a=- 4, operate R2 – 2R1] Question.1 Prove that the points P (a, b+c), Q(b, c+a), R(c, a+b) are collinear. Answer : If P,Q and R are collinear then 𝒂 𝒃 + 𝒄 𝟏 𝒃 𝒄 + 𝒂 𝟏 𝑪 𝒂 + 𝒃 𝟏 = 0 By applying C2 → C2+C1 𝒂 𝒂 + 𝒃 + 𝒄 𝟏 𝒃 𝒂 + 𝒃 + 𝒄 𝟏 𝒄 𝒂 + 𝒃 + 𝒄 𝟏 = (a+b+c) 𝒂 𝟏 𝟏 𝒃 𝟏 𝟏 𝒄 𝟏 𝟏 =0 ( ∵ C2, C3 are identical) Question.2 Find the value of k if the areaof the triangle with vertices (-2,0),(0,4) and (0,k) is 4 square units. Answer: Area of ∆ = ½ −𝟐 𝟎 𝟏 𝟎 𝟒 𝟏 𝟎 𝒌 𝟏 = 4
  • 17. ⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨ the absolute value of (k – 4) = 4 ⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8, 0. Question.3 Without expanding, showthat (i) 𝒃 − 𝒄 𝒄 − 𝒂 𝒂 − 𝒃 𝒄 − 𝒂 𝒂 − 𝒃 𝒃 − 𝒄 𝒂 − 𝒃 𝒃 − 𝒄 𝒄 − 𝒂 = 0 Operating C1 → C1+C2+C3, we get 𝟎 𝒄 − 𝒂 𝒂 − 𝒃 𝟎 𝒂 − 𝒃 𝒃 − 𝒄 𝟎 𝒃 − 𝒄 𝒄 − 𝒂 = 0. (ii) 𝟎 𝒂 −𝒃 −𝒂 𝟎 −𝒄 𝒃 𝒄 𝒐 = 0 Taking out (-1) from C1,C2 and C3, we get, ∆ = (- 1)(-1)(-1) 𝟎 −𝒂 𝒃 𝒂 𝟎 𝒄 −𝒃 −𝒄 𝒐 = -1 𝟎 𝒂 −𝒃 −𝒂 𝟎 −𝒄 𝒃 𝒄 𝒐 = - ∆ (by interchanging rows and columns) 2∆ = 0 ⇨ ∆ = 0 (iii) 𝒃²𝒄² 𝒃𝒄 𝒃 + 𝒄 𝒄²𝒂² 𝒄𝒂 𝒄 + 𝒂 𝒂²𝒃² 𝒂𝒃 𝒂 + 𝒃 = 0 ⇨ 𝒂𝒃𝒄 𝒂𝒃𝒄 𝒃²𝒄² 𝒃𝒄 𝒃 + 𝒄 𝒄²𝒂² 𝒄𝒂 𝒄 + 𝒂 𝒂²𝒃² 𝒂𝒃 𝒂 + 𝒃 = 𝟏 𝒂𝒃𝒄 𝒂𝒃²𝒄² 𝒂𝒃𝒄 𝒂𝒃 + 𝒂𝒄 𝒃𝒄²𝒂² 𝒃𝒄𝒂 𝒃𝒄 + 𝒃𝒂 𝒄𝒂²𝒃² 𝒄𝒂𝒃 𝒄𝒂 + 𝒄𝒃 𝒂𝒃𝒄.𝒂𝒃𝒄 𝒂𝒃𝒄 𝒃𝒄 𝟏 𝒂𝒃 + 𝒂𝒄 𝒄𝒂 𝟏 𝒃𝒄 + 𝒃𝒂 𝒂𝒃 𝟏 𝒄𝒂 + 𝒄𝒃 = abc 𝒃𝒄 𝟏 𝒂𝒃 + 𝒃𝒄+ 𝒂𝒄 𝒄𝒂 𝟏 𝒂𝒄 + 𝒃𝒄 + 𝒃𝒂 𝒂𝒃 𝟏 𝒂𝒃 + 𝒄𝒂 + 𝒄𝒃 ( Operating C3 → C3+C1) abc(ab+bc+ac) x 0 = 0 ( two cols.Are identical) (iv) 𝟏 𝒂 𝒂² − 𝒃𝒄 𝟏 𝒃 𝒃² − 𝒄𝒂 𝟏 𝒄 𝒄² − 𝒂𝒃 = 0 ⇨ 𝟏 𝒂 𝒂² 𝟏 𝒃 𝒃² 𝟏 𝒄 𝒄² - 𝟏 𝒂 𝒃𝒄 𝟏 𝒃 𝒄𝒂 𝟏 𝒄 𝒂𝒃 = 𝟏 𝒂 𝒂² 𝟏 𝒃 𝒃² 𝟏 𝒄 𝒄² - 𝟏 𝒂𝒃𝒄 𝒂 𝒂² 𝒂𝒃𝒄 𝒃 𝒃² 𝒃𝒄𝒂 𝒄 𝒄𝟐 𝒄𝒂𝒃 𝟏 𝒂 𝒂² 𝟏 𝒃 𝒃² 𝟏 𝒄 𝒄² - 𝒂 𝒂² 𝟏 𝒃 𝒃² 𝟏 𝒄 𝒄𝟐 𝟏 = 0 (PassC3overthefirsttwocolumns.)
  • 18. (v) 𝟏 𝒂 𝒂𝟐 𝟏 𝒃 𝒃𝟐 𝟏 𝒄 𝒄𝟐 = 𝟏 𝒃𝒄 𝒃 + 𝒄 𝟏 𝒄𝒂 𝒄 + 𝒂 𝟏 𝒂𝒃 𝒂 + 𝒃 R.H.S. 𝟏 𝒂𝒃𝒄 𝒂 𝒂𝒃𝒄 𝒂𝒃 + 𝒂𝒄 𝒃 𝒂𝒃𝒄 𝒃𝒄+ 𝒃𝒂 𝒄 𝒂𝒃𝒄 𝒄𝒂 + 𝒄𝒃 = 𝟏 𝒂 𝒂𝒃 + 𝒂𝒄 𝟏 𝒃 𝒃𝒄 + 𝒃𝒂 𝟏 𝒄 𝒄𝒂 + 𝒄𝒃 ( applying C1 ↔ C2) = - 𝟏 𝒂 −𝒃𝒄 𝟏 𝒃 −𝒄𝒂 𝟏 𝒄 −𝒂𝒃 (apply C3 → C3 – (ab+bc+ca))C1) = 𝟏 𝒂𝒃𝒄 𝒂 𝒂² 𝒂𝒃𝒄 𝒃 𝒃² 𝒂𝒃𝒄 𝒄 𝒄𝟐 𝒂𝒃𝒄 = 𝟏 𝒂 𝒂𝟐 𝟏 𝒃 𝒃𝟐 𝟏 𝒄 𝒄𝟐 (apply C2 ↔ C3 and C1 ↔ C2) Q. If a,b,c are +ve and are the pth,qth and rth terms resp.Of a G.P.,show without expanding that ** (vi) 𝐥𝐨𝐠 𝒂 𝒑 𝟏 𝐥𝐨𝐠 𝐛 𝒒 𝟏 𝐥𝐨𝐠 𝐜 𝒓 𝟏 = 0 (put a=xyp-1 ,b=xyq-1 ,c=xyr-1 , apply C1 →C1-logx.C3 ,C1→C1+C3) (vii) 𝟏 𝒂 𝒃𝒄 𝟏 𝒃 𝒄𝒂 𝟏 𝒄 𝒂𝒃 = 𝟏 𝒂 𝒂² 𝟏 𝒃 𝒃² 𝟏 𝒄 𝒄² (same methodas given below) (viii) 𝒂 𝒂² 𝒃𝒄 𝒃 𝒃² 𝒄𝒂 𝒄 𝒄𝟐 𝒂𝒃 = 𝟏 𝒂² 𝒂³ 𝟏 𝒃² 𝒃³ 𝟏 𝒄² 𝒄³ ( Multiply by abc as R1with a,R2 by b and R3by c then divide with abc ) Find the values of: (ix) 𝟏 𝒂𝒃 𝟏 𝒂 + 𝟏 𝒃 𝟏 𝒃𝒄 𝟏 𝒃 + 𝟏 𝒄 𝟏 𝒄𝒂 𝟏 𝒂 + 𝟏 𝒄 (Operate C2 → 𝟏 𝒂𝒃𝒄 C2. and value is 0 (x) 𝒔𝒊𝒏𝜶 𝒄𝒐𝒔𝜶 𝐜𝐨𝐬(𝜶 + 𝜹) 𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜷 𝐜𝐨𝐬(𝜷 + 𝜹) 𝒔𝒊𝒏𝜸 𝒄𝒐𝒔𝜸 𝐜𝐨𝐬(𝜸 + 𝜹) (Operating C3 → C3 – cos𝜹.C2+sin 𝜹 .C1 and value is 0) Q. Prove that :
  • 19. (a) 𝟏 + 𝒂² − 𝒃² 𝟐𝒂𝒃 −𝟐𝒃 𝟐𝒂𝒃 𝟏 − 𝒂² + 𝒃² 𝟐𝒂 𝟐𝒃 −𝟐𝒂 𝟏 − 𝒂² − 𝒃² = (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 → (C2+aC3) (b) 𝒄𝒐𝒔𝜶𝒄𝒐𝒔𝜷 𝒄𝒐𝒔𝜶𝒔𝒊𝒏𝜷 −𝒔𝒊𝒏𝜶 −𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜷 𝟎 𝒔𝒊𝒏𝜶𝒄𝒐𝒔𝜷 𝒔𝒊𝒏𝜶𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜶 = 1 (Apply R3 → sin𝛂R3 +cos𝛂R1) (C) ( 𝒚 + 𝒛)² 𝒙𝒚 𝒛𝒙 𝒙𝒚 ( 𝒙 + 𝒛)² 𝒚𝒛 𝒙𝒛 𝒛𝒚 ( 𝒙 + 𝒚)² = xyz(x+y+z)3 (Apply R1 → x R1, R2 → y R2, R3 → z R3 and take x,y,z commonfrom C1,C2,C3 resp.) (d ( 𝒃 + 𝒄)² 𝒂 𝟐 𝒃𝒄 𝒄 + 𝒂)² 𝒃 𝟐 𝒄𝒂 𝒂 + 𝒃)² 𝒄 𝟐 𝒂𝒃 =(a-b)(b-c)(c-a)(a+b+c)(a2+b2+c2 ) ( Apply C1 → (C1+C2 – 2C3) (e) 𝒂 + 𝒃 + 𝒄 −𝒄 −𝒃 −𝒄 𝒂 + 𝒃 + 𝒄 −𝒂 −𝒃 −𝒂 𝒂 + 𝒃 + 𝒄 = 2(a+b)(b+c)(c+a) (Apply C1 →(C1+C3 )and C2 →(C2+C3)) (f) 𝒂 𝒃 − 𝒄 𝒄 + 𝒃 𝒂 + 𝒄 𝒃 𝒄 − 𝒂 𝒂 − 𝒃 𝒂 + 𝒃 𝒄 = (a+b+c)(a2+b2+c2) (g) 𝒃 + 𝒄 𝒄 + 𝒂 𝒂 + 𝒃 𝒒 + 𝒓 𝒓 + 𝒑 𝒑 + 𝒒 𝒚 + 𝒛 𝒛 + 𝒙 𝒙 + 𝒚 = 2 𝒂 𝒃 𝒄 𝒑 𝒒 𝒓 𝒙 𝒚 𝒛 (apply C1→C1-C2-C3, C2→C2- C1,C3→C3-C1, C2↔C3) (h) 𝟏 + 𝒂 𝟏 𝟏 𝟏 𝟏 + 𝒃 𝟏 𝟏 𝟏 𝟏 + 𝒄 = abc ( 𝟏 𝒂 + 𝟏 𝒃 + 𝟏 𝒄 +1(ab+bc+ca+abc). (Hint taking a,b,c commonfrom each row, apply R1→R1+R2+R3 then expand along first row). (i) 𝟏 + 𝒂² − 𝒃² 𝟐𝒂𝒃 −𝟐𝒃 𝟐𝒂𝒃 𝟏 − 𝒂² + 𝒃² 𝟐𝒂 𝟐𝒃 −𝟐𝒂 𝟏 − 𝒂² − 𝒃² = (𝟏 + 𝒂² + 𝒃²)3
  • 20. Apply C1→C1-bC3, C2→C2+aC3, we get 𝟏 + 𝒂² + 𝒃² 𝟎 −𝟐𝒃 𝟎 𝟏 + 𝒂² + 𝒃² 𝟐𝒂 𝒃(𝟏+ 𝒂 𝟐 + 𝒃 𝟐 ) −𝒂(𝟏+ 𝒂 𝟐 + 𝒃 𝟐 ) 𝟏 − 𝒂² − 𝒃² = (𝟏 + 𝒂² + 𝒃²)² 𝟏 𝟎 −𝟐𝒃 𝟎 𝟏 𝟐𝒂 𝒃 −𝒂 𝟏 − 𝒂² − 𝒃² expand along C1, We get (𝟏 + 𝒂² + 𝒃²)³ . **(h) Evaluate ( 𝑿 𝟏 ) ( 𝑿 𝟐 ) ( 𝑿 𝟑 ) ( 𝒀 𝟏 ) ( 𝒀 𝟐 ) ( 𝒀 𝟑 ) ( 𝒁 𝟏 ) ( 𝒁 𝟐 ) ( 𝒁 𝟑 ) where ( 𝑿 𝟏 ) =C(x,1) ( binomial coefficient) Solution: 𝒙 𝟏! 𝒙(𝒙−𝟏) 𝟐! 𝒙( 𝒙−𝟏)(𝒙−𝟐) 𝟑! 𝒚 𝟏! 𝒚(𝒚−𝟏) 𝟐! 𝒚( 𝒚−𝟏)(𝒚−𝟐) 𝟑! 𝒛 𝟏! 𝒛(𝒛−𝟏) 𝟐! 𝒛( 𝒛−𝟏)(𝒛−𝟐) 𝟑! = 𝒙𝒚𝒛 𝟐!𝟑! 𝟏 𝒙 − 𝟏 ( 𝒙 − 𝟏)(𝒙 − 𝟐) 𝟏 𝒚 − 𝟏 (𝒚 − 𝟏)(𝒚 − 𝟐) 𝟏 𝒛 − 𝟏 ( 𝒛 − 𝟏)(𝒛 − 𝟐) ( taking x,y,z commonfrom R1,R2,R3 resp.and ½!,1/3! From C2,C3 resp.) ( by formulaof C(n,r) = 𝒏! ( 𝒏−𝒓)!𝒏! ) Apply C3→C3 + C2 and put a= x-1, b=y-1, c=z-1 𝒙𝒚𝒛 𝟏𝟐 𝟏 𝒂 𝒂² 𝟏 𝒃 𝒃² 𝟏 𝒄 𝒄² = 𝒙𝒚𝒛 𝟏𝟐 (a-b)(b-c)(c-a) = 𝒙𝒚𝒛 𝟏𝟐 (x-y)(y-z)(z-x). Question: If x,y,z are all different and if 𝒙 𝒙² 𝟏 + 𝒙³ 𝒚 𝒚² 𝟏 + 𝒚³ 𝒛 𝒛² 𝟏 + 𝒛³ = 0 , prove that xyz = -1
  • 21. Solution: 𝒙 𝒙² 𝟏 + 𝒙³ 𝒚 𝒚² 𝟏 + 𝒚³ 𝒛 𝒛² 𝟏 + 𝒛³ = 𝒙 𝒙² 𝟏 𝒚 𝒚² 𝟏 𝒛 𝒛² 𝟏 + 𝒙 𝒙² 𝒙³ 𝒚 𝒚² 𝒚³ 𝒛 𝒛² 𝒛³ = 𝒙 𝒙² 𝟏 𝒚 𝒚² 𝟏 𝒛 𝒛² 𝟏 + xyz 𝟏 𝒙 𝒙² 𝟏 𝒚 𝒚² 𝟏 𝒛 𝒛² = 0 𝟏 𝒙 𝒙² 𝟏 𝒚 𝒚² 𝟏 𝒛 𝒛² (1+xyz) = 0 ⇨ (x-y)(y-z)(z-x)(1+xyz) =0 ⇨ xyz=-1 ∵ x ≠y≠ z. Question: By using properties ofdeterminant,showthat 𝒂² + 𝟏 𝒂𝒃 𝒂𝒄 𝒂𝒃 𝒃² + 𝟏 𝒃𝒄 𝒄𝒂 𝒄𝒃 𝒄² + 𝟏 = 1+a2+b2+c2 [Hint: multiply and divide by a,b,c with R1,R2,R3respectively,taking a,b,c commonfrom C1,C2,C3respectively R1→R1+R2+R3] Question: showthat 𝒂 − 𝒃 − 𝒄 𝟐𝒂 𝟐𝒂 𝟐𝒃 𝒃 − 𝒄 − 𝒂 𝟐𝒃 𝟐𝒄 𝟐𝒄 𝒄 − 𝒂 − 𝒃 =(a+b+c)3.[Hint:R1→R1+R2+R3] Question(i) Using matrix method,solve the following system of equations: 𝟐 𝒙 + 𝟑 𝒚 + 𝟏𝟎 𝒛 = 4, 𝟒 𝒙 - 𝟔 𝒚 + 𝟓 𝒛 = 1, 𝟔 𝒙 + 𝟗 𝒚 - 𝟐𝟎 𝒛 = 2; x, y, z ≠ 0. [X=2,Y=3,Z=5,|A|=1200,adjA =[ 𝟕𝟓 𝟏𝟓𝟎 𝟕𝟓 𝟏𝟏𝟎 −𝟏𝟎𝟎 𝟑𝟎 𝟕𝟐 𝟎 −𝟐𝟒 ] ] (ii) 𝟏 𝒙 - 𝟏 𝒚 + 𝟏 𝒛 =4, 𝟐 𝒙 + 𝟏 𝒚 - 𝟑 𝒛 = 0, 𝟏 𝒙 + 𝟏 𝒚 + 𝟏 𝒛 = 2 [ x=1/2,y=-1,z=1 adjA= [ 𝟒 𝟐 𝟐 −𝟓 𝟎 𝟓 𝟏 −𝟐 𝟑 ] |A| = 10] (iii) 𝟐 𝒙 - 𝟑 𝒚 + 𝟑 𝒛 = 10, 𝟏 𝒙 + 𝟏 𝒚 + 𝟏 𝒛 = 10, 𝟑 𝒙 - 𝟏 𝒚 + 𝟐 𝒛 =13; X, Y, Z ≠ 0.
  • 22. [X=1/2,Y=1/3,Z=1/5,|A|=-9, adjA = [ 𝟑 𝟑 −𝟔 𝟏 −𝟓 𝟏 −𝟒 −𝟕 𝟓 ] ] Relations & functions Q.1. Show that the relation R defined by (a, b) R (c, d) ⟹ a + d = b + c on the set N×N is an equivalence relation. Q.Show that ƒ: N N defined by ƒ (x) = , if n is odd , if n is even is many-one onto function. CompositionofFunction and Invertible Function. Q.1. If f(x) = x + 7 and g(x) = x – 7, x ε R, find (fog)(7). Solution: We have, f(x) = x + 7 and g(x) = x – 7. Then (fog)(x) =fo(g(x)) =f(x – 7) = (x – 7) + 7 = x. Therefore,(fog)(7) =7. [Ans.] Q.2. If f : R→ R and g : R→ R are defined respectively as f(x) = x2 + 3x + 1 and g(x) = 2x – 3, find (a) fog , (b) gof. Solution: [ 4x2 – 6x + 1. [Ans.] , 2x2 + 6x – 1. [Ans.] Q.3. If f : R→ R defined as f(x) = (2x – 7)/4 is an invertible function, find f–1. [f –1(x) = (4x + 7)/2. [Ans.] Q.4. If f : R→ R defined by f(x) = (3x + 5)/2 is an invertible function, find f–1.Solution: [ f–1(x) = (2x – 5)/3. [Ans.] 1.4 Binary Operations. Q.1.
  • 23. i. Is the binary operation*, defined on set N, given by a*b= (a+ b)/2, for all a, b ε N, commutative ? ii. Is the above binary operation* associative ? Q.2. Let * be a binary operationdefined by a*b = 2a + b – 3. Find 3*4. ASSIGNMENT( Relations & Functions) Q. 1 Showthat the function f : N →N, defined by f(x) = x2 + x +1 is one- one but not onto. Q. 2. Let f: N be a function defined as f ( x ) = 4x2 + 12x + 15 .Show that f: N where S is the range of f, is invertible .Find the inverse of f . Q. 3. Showthat the relationR on the set R ofall real numbers, defined as R = { (a, b): a2 + b2 =1 } is neither reflexive nor transitive but symmetric Q. 4. Showthat ƒ : R –{0} R–{0} given by ƒ (x) = 3/x is invertible and it is inverse of itself. Q. 5 IfR1 & R2 are equivalence relations onset A. Show that R1 U R2 is reflexive,symmetric but not transitive. Q. 6. Let ƒ (x) = [x] and g(x) = |x|, find goƒ(-5/3) – ƒog (-5/3) Q. 7. Showthat the function ƒ: R R defined by ƒ (x) = 3x3 + 5 for x R is a bijection. Q. 8. Showthat the relationR on the set R ofall real numbers, defined as R = { (a, b): |a| ≤ b } is neither reflexive nor symmetric but transitive. Q. 9. Showthat the function ƒ : N → N given by ƒ (1) = ƒ (2) =1 and ƒ (x) = x-1,for every x > 2 is onto but not one-one. Q. 10. If * is a binary operationon R defined by a * b = a/4+ b/7for a, b R, find the value of ( 2 * 5 ) * 7 Q. 11 Let R be a relationon NXN, defined by (a,b) R (c,d) ⇔ ad(b+c)) = bc(a+d), ∀(a,b) ,(c,d) є NXN ,Show that R is an equivalence relationon NXN.
  • 24. Same for addition Q. 12 Let R be a relationon NXN, defined by (a,b) R (c,d) ⇔ ad = bc, ∀(a,b) , (c,d) є NXN ,Showthat R is an equivalence relationon NXN. Q.13 Let A = {1,2,3}. Find the number ofrelations on A containing (1,2) & (2,3) which are reflexive, transitive but not symmetric. Q.14 Showthat the function ƒ : N → N givenby f(x) =x – (-1)x, is bijection.( same as ques.2 of misc.) Q.15 Let f: [-1,∞) →[-1,∞) is given by f(x)= (x+1)2 – 1. Showthat f is invertible and find the set = { x : f(x) = f-1 (x) }. Relations & functions for class—XII Level—2 Q.1 If f: R→ R is givenby f(x) = (3 – x3)1/3 showthat fof =Ig where Igis the identity mapon R. Q.2 Show that the functionf: [-1, 1]→R definedby f(x) = 𝒙 𝟐+𝒙 is 1-1 . Find the range of f. Alsofindthe inverse of the functionf: [-1, 1]→ range of f. Q.3 Show that the functionf: R → R definedby f(x) = cos (5x+2) is neither 1-1 nor onto? Q.4 If f: R → R be givenby f(x) = sin2 x +sin2 (x+π/3) +cosx .cos(x+π/3) ∀ x Є R, and g: R → R be a function suchthat g(5/4) =1 , then prove that (gof) : R → R is a constant function. Q.5 Let R1=R – {-1} and an operation* is definedon R1 by a*b = a + b + ab ∀ a, b Є R1 .Findthe identity elementandinverse of an element. ANSWERS OF Level—2 Ans.1 As f: R → R, fof exists andfof : R → R is givenby (fof) (x) = f(f(x)) = f(3 – x3 )1/3 = (3 – ((3 – x3 )1/3 )3 )1/3 = (3 – (3 – x3 ))1/3 =x ∀ x ЄR Ans.2 f is 1-1, as consider any x1, x2 Є [-1, 1]such that f(x1) = f(x2) ⇨ 𝒙₁ 𝟐+𝑿₁ = 𝑿₂ 𝟐+𝑿₂ ⇨ x1x2+2x1 = x1x2 +2x2 ⇨ x1 = x2 For the range of f Let y = f(x) ⇨ y = 𝒙 𝟐+𝒙 ⇨ xy +2y =x ⇨ (y – 1) x= -2y ⇨ x = −𝟐𝒚 𝒚−𝟏
  • 25. As x Є [-1, 1], so -1 ≤ −𝟐𝒚 𝒚−𝟏 ≤1 , but (y – 1)2 >0 , y ≠ 1⇨ -(y – 1)2 ≤ −𝟐𝒚 𝒚−𝟏 (y-1)2 ≤(y – 1)2 ⇨ -(y2 – 2y +1) ≤ -2y2 +2y ≤ y2 – 2y +1 ,y≠ 1 ⇨ Y2 – 1 ≤ 0 and 0 ≤ 3y2 – 4y +1 ⇨ y ε [-1,1]and (y – 1/3) (y – 1) ≥ 0 ,y ≠ 1 ⇨y Є [-1,1] and y ε (-∞ ,1/3]U [1,∞) , y ≠ 1 ⇨ y Є [-1,1] and y ε (-∞ ,1/3]U (1,∞)⇨ y Є[-1,1/3]. ∴ its inverse existsas f is 1-1 and onto, to find f-1 𝒙 𝟐+𝒙 = y ⇨ xy +2y =x ⇨ 2y = x (1 – y) ⇨ x= 𝟐𝒚 𝟏−𝒚 f-1 (y) = x = 𝟐𝒚 𝟏−𝒚 . Ans. 3 For f is not 1-1, 5x+2 = π/2 ⇨ x = (π – 4)/10 ∴ 5x+2 =π/2, again 5x+2 = 3π/2 ⇨ x = (3π – 4)/10, Nowf ((π – 4)/10)) = cos[5((π – 4)/10) +2]= cosπ/2 =0 f ((3π – 4)/10) = cos[5((3π –4)/10) +2] = cos3π/2 = 0. For f is not onto, as -1 ≤ cos (5x+2) ≤1, then -1≤ y ≤1, range of f = [-1, 1] = {y : - 1≤ y ≤1 } ≠ co-domain R. Ans. 4 ½[ 2sin2 x +2sin2 (x+π/3) +2cosx cos(x+π/3)] f (x)= ½[ 1 – cos2x +1 – cos (2x+2π/3)+cos (2x+π/3)+cosπ/3] ( As we know that 2sin2 x=1 – cos2x and 2cosA cosB= Cos(A+B) + cos(A-B).) ½[5/2 – {cos2x +cos(2x+2π/3)} +cos(2x+π/3) ⇨ ½[5/2 –2cos(2x+π/3) cos π/3 + cos(2x+π/3)]=5/4 ∀ x ЄR ∴ for any x Є R , we have (gof)(x) = g(f(x)) = g(5/4)=1 ,soit is constant function. Ans. 5 * can be shown to be a binary operationon R1 as let a ≠ -1, b ≠ -1 . a*b = a+b+ab Є R – {-1} ⇨ a+b+ab ≠ -1 ⇨ a(1+b)+(b+1) ≠0 ⇨ (a+1) (1+b) ≠0 ⇨ a ≠ -1 and b ≠ -1 which is true.Nowif e is the identity element, thena*e =a ⇨ a+e+ae =a ⇨ e (1+a) = 0 ⇨ e =0 or a = -1 ⇨e =0 , 0 is the identity w.r.t. * Let a’ be inverse of a, thena*a’ =0 ⇨ a+a’+aa’ = 0 ⇨ a’(1+a) = - a ∴ a’ = - a/(1+a) , is the inverse of a w.r.t. *.
  • 26. Q.1 Let f(x) = 𝒙+𝟑,𝒊𝒇 𝒙<1𝟒𝒙−𝟐,𝒊𝒇 𝟏≤𝒙≤𝟒.𝒙²+𝟓,𝒊𝒇 𝒙>4 Find f(-1) ,f(4) and f(5). Q.2 If f(x) = x2 – 𝟏/𝒙², then find the value of f(x) + f ( 𝟏/𝒙²). Q.3 Let Q be the setall rational numbers and relation on Q defined by R = {(X, Y): 1+XY > 0}. Prove then R is reflexive and symmetric but not transitive. Q.4 Write the identity element for the binary operation*defined on setR by a*b = 3ab/8 ∀ a, b ЄR. Q.5 Show that the function f: R → R defined by f(x) = sin x is neither 1-1 nor onto. Answers (Level—1) Ans.1 f (-1) = 2, f (4) = 14, f(5)= 30. Ans.2 0. Ans.3 Considerany x, y Є Q, since 1+x.x =1+x2 ≥ 1 ⇨ (x,x)Є R ⇨ reflexive Let (x,y) Є R ⇨ 1+xy > 0 ⇨ 1+yx > 0 ⇨ (y,x) ЄR ⇨ symmetric. But not transitive . Since (-1, 0) and (0, 2) ЄR, because 1 > 0 by putting values. But (-1, 2) ∉ R because -1<0. Ans.4 Let e be the identity element in R. Then a *e =a =e*a ∀ aЄR ⇨ a*e =a ∀ a ЄR ⇨ e = 8/3 in R. Ans.5 f is not 1-1 because sin0 = 0 =sin π,so the different elements o, π have same images. f is not onto because -1 ≤ sin x ≤ 1 for all x ЄR ∴ the range of f =[-1,1], which is a proper subset of R. Level Inverse Trigonometric Functions For suitable values of x and y sin-1 x + sin-1 y= sin-1 (x√1-y2 + y√1-x2 ) sin-1 x - sin-1 y=sin-1 (x√1-y2 - y√1-x2 ) cos-1 x + cos-1 y= cos-1 (xy- √1-x2 √1-y2 ) cos-1 x - cos-1 y= cos-1 (xy+√1-x2√1-y2) tan-1 x + tan-1 y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀 𝟏−𝑿𝒀 ; xy<1 tan-1 x – tan-1 y= 𝐭𝐚𝐧−𝟏 𝑿−𝒀 𝟏+𝑿𝒀 ; xy>-1
  • 27. 2tan-1 x= 𝐭𝐚𝐧−𝟏 𝟐𝒙 𝟏−𝒙² = 𝐬𝐢𝐧−𝟏 𝟐𝒙 𝟏+𝒙² = 𝐜𝐨𝐬−𝟏 𝟏−𝒙² 𝟏+𝒙² Q. 1. Find the value of : tan-1 (1) + cos -1 (-1/2) + sin-1 (-1/2). Q.2 Prove that 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝟐𝒙 𝟏−𝒙² = 𝐭𝐚𝐧−𝟏 𝟑𝒙−𝒙³ 𝟏−𝟑𝒙² , x< 𝟏 √𝟑 Q. 3. Solve : tan-1 2x + tan-1 3x = 𝝅 𝟒 Q. 4. Prove : Q. 5. Solve : sin-1 ( 1 –x) – 2sin-1 x = . Q. 6. Evaluate:tan-1 - sec-1 (-2) + cosec-1 . Q. 7. Prove : tan-1 = Q. 8. Simplify : sin-1 , Q. 9. Prove: sec2 (tan-1 2) + cosec2 ( cot-1 3) = 15. Q. 10. Simplify : tan-1 Q. 11. Prove : tan-1 = Q. 12. If sin(sin-1 , then findthe value of x.
  • 28. Q. 13. Prove that : 2tan-1 = cos-1 Q. 14. Find the principal value of Sec-1 Q. 15. Find value of Sin Q. 1. If 2 tan-1 (Cos q) = tan-1 (2 cosec q), find q. Q.2 Question.3 if 𝐜𝐨𝐬−𝟏 𝒙 𝒂 + 𝐜𝐨𝐬−𝟏 𝒚 𝒃 = 𝜽, then prove that 𝒙² 𝒂² - 𝟐𝒙𝒚 𝒂𝒃 cos𝜽 + 𝒚² 𝒂² = sin2 𝜽 [Hint: 𝐜𝐨𝐬−𝟏 𝒙 𝒂 + 𝐜𝐨𝐬−𝟏 𝒚 𝒃 = 𝐜𝐨𝐬−𝟏 [ 𝒙𝒚 𝒂𝒃 - √ 𝟏 − 𝒙² 𝒂² √ 𝟏 − 𝒚² 𝒃² ] = 𝜽 ⇨ ( 𝒙𝒚 𝒂𝒃 − cos𝜽 )2 = (√ 𝟏 − 𝒙² 𝒂² √ 𝟏 − 𝒚² 𝒃² )2 Simplify it] Question.4 *(i) sin-1 x + sin-1 y + sin-1 z = π, then prove that X4 +y4 +z4 +4x2 y2 z2 = 2(x2 y2 +y2 z2 +z2 x2 ) (ii) If 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 = π/2 ; prove that xy+yz+xz = 1. (iii) If 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 = π , provethat x+y+z = xyz. [Hint: for (i) sin-1 x + sin-1 y = π - sin-1 z ⇨ cos(sin-1 x + sin-1 y) =cos( π - sin-1 z) Use cos(A-B) = cosAcosB – sinAsinB and cos(π – 𝛂)= -cos𝛂 It becomes √(𝟏 − 𝒙²)(𝟏− 𝒚²) - xy = - √ 𝟏 − 𝒛² and simply it. [Hint: for (ii) tan-1 x + tan-1 y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀 𝟏−𝑿𝒀 ]
  • 29. Question.5 Write the following functions in the simplest form:(i)𝐭𝐚𝐧−𝟏 ( 𝒄𝒐𝒔𝒙 𝟏+𝒔𝒊𝒏𝒙 ) (ii) 𝐭𝐚𝐧−𝟏 ( 𝒄𝒐𝒔𝒙 𝟏−𝒔𝒊𝒏𝒙 ) (iii) 𝐭𝐚𝐧−𝟏 √ 𝒂−𝒙 𝒂+𝒙 , -a<x<a [Hint: for (i) write cosx = cos2 x/2 – sin2 x/2 and 1+sinx =(cosx/2 +sinx/2)2 , then use tan(A-B), answer is π/4 – x/2 ] [ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2 – x), then use formula of 1-cos(π/2 – x)= 2sin2 (π/4 – x/2) and sin(π/2 – x) = 2 sin(π/4 – x/2) cos(π/4 – x/2) Same method can be applied for (i) part also. Answer is π/4 + x/2][ for (iii) put x=a cos𝛂, then answer will be ½ 𝐜𝐨𝐬−𝟏 𝒙 𝒂 ]Question.6 If y = 𝐜𝐨𝐭−𝟏 (√ 𝒄𝒐𝒔𝒙) - 𝐭𝐚𝐧−𝟏 (√ 𝒄𝒐𝒔𝒙), prove that siny = tan2 (x/2). [Hint: y = 𝝅 𝟐 - 2 𝐭𝐚𝐧−𝟏 (√ 𝒄𝒐𝒔𝒙) , use formula 2𝐭𝐚𝐧−𝟏 𝒙 = 𝐜𝐨𝐬−𝟏 ( 𝟏−𝒙² 𝟏+𝒙² )] Question.7 (i) Prove that 𝐭𝐚𝐧−𝟏 𝟏 + 𝐭𝐚𝐧−𝟏 𝟐 + 𝐭𝐚𝐧−𝟏 𝟑 = π. (ii) Prove that 𝐜𝐨𝐭−𝟏 ( 𝒂𝒃+𝟏 𝒂−𝒃 ) +𝐜𝐨𝐭−𝟏 ( 𝒄𝒃+𝟏 𝒃−𝒄 ) + 𝐜𝐨𝐭−𝟏 ( 𝒂𝒄+𝟏 𝒄−𝒂 ) = 0. [Hint: for (i) 𝐭𝐚𝐧−𝟏 𝟐 = 𝝅 𝟐 - 𝐜𝐨𝐭−𝟏 𝟐 = 𝝅 𝟐 − 𝐭𝐚𝐧−𝟏 𝟏 𝟐 , then use formula of tan-1 x + tan-1 y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀 𝟏−𝑿𝒀 ] (ii) [Hint: write 𝐜𝐨𝐭−𝟏 𝒙 = 𝐭𝐚𝐧−𝟏 𝟏 𝒙 ] Question.8 Solve the following equations: (i) 𝐬𝐢𝐧−𝟏 𝒙 + 𝐬𝐢𝐧−𝟏 (𝟏 − 𝒙) = 𝐜𝐨𝐬−𝟏 𝒙 . (ii) 𝐭𝐚𝐧−𝟏 √ 𝒙(𝒙 + 𝟏) + 𝐬𝐢𝐧−𝟏 √ 𝒙² + 𝒙 + 𝟏 = 𝝅 𝟐 . (i) [Hint: write 𝐜𝐨𝐬−𝟏 𝒙 = 𝝅 𝟐 - 𝐬𝐢𝐧−𝟏 𝒙, put 𝐬𝐢𝐧−𝟏 𝒙 = y] (ii) [Hint: use 𝐭𝐚𝐧−𝟏 𝒙 = 𝐜𝐨𝐬−𝟏 𝟏 √ 𝟏+𝒙² ] Indu thakur Question.9 Using principal values, evaluate 𝐜𝐨𝐬−𝟏 (𝒄𝒐𝒔 𝟐𝝅 𝟑 ) + 𝐬𝐢𝐧−𝟏 ( 𝒔𝒊𝒏 𝟐𝝅 𝟑 ). [answer is π]
  • 30. Question.10Show thattan( 𝟏 𝟐 𝐬𝐢𝐧−𝟏 𝟑 𝟒 ) = 𝟒− √𝟕 𝟑 and justify why the other value is ignored? [ Hint: put 𝟏 𝟐 𝐬𝐢𝐧−𝟏 𝟑 𝟒 =∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2 ∅), find tan∅] ASSESSMENTOF PROBABILITY FOR CLASS –XII Level—1 Q. 1 If the mean and variance of a binomial distributionare 4 and 4/3 respectively,find P(X≥1). Q. 2 If P(A) = 3/8 , P(B) = ½ and P(A∩B) = ¼ , find P(𝑨/𝑩). Q.3 A bag contains 4 white and 2 black balls.Another bag contains 3 white and 5 black balls. If one ball is drawn from each bag, find the probability that (i) Both are white balls. (ii) One is white and one is black. Q.4 IfA and B are independent events and P(A∩B) = 1/8, P( A’ ∩ B’) =3/8 , find P(A) and P(B). Q.5 The probabilities ofP, Q and R solving a problem are ½, 1/3 and ¼ respectively.Ifthe problem is attempted by all simultaneously,find the probability of exactly one of them solving it. Answers of Level—1 Ans.1 np = 4, npq = 4/3 ⟹ q=1/3 ⟹ p = 1 - 1/3 = 2/3⟹ n=6 ⟹P(X≥1) =1– C(6,0) (2/3)0 (1/3)6 = 1 - 𝟏 𝟕𝟐𝟗 = 𝟕𝟐𝟖 𝟕𝟐𝟗 . Ans.2 P(𝑨/𝑩) =P(𝑨∩ 𝑩 ) /P(B) = 𝟏−𝑷(𝑨𝑼𝑩) 𝟏−𝑷(𝑩) = 𝟏−[ 𝟑 𝟖 + 𝟏 𝟐 − 𝟏 𝟒 ] 𝟏−𝟏/𝟐 = ¾. Ans.3 (i) P(A∩ B) = P(A).P(B) = (2/3).(3/8) [A,Bare independent events] (ii) P(A’∩ B) + P(A ∩ B’) = P(A’).P(B)+P(A).P(B’) =( 1/3).(3/8)+(2/3).(5/8)=13/24.[A’, B are indep. Events, B’ A are indep. events], where A = drawing a white ball from first bag. B= drawing a same ball from secondbag.A’ = drawing a blackball from first bag and B’ =drawing from secondbag. Ans.4 P(A∩B) = P(A).P(B) = 1/8 let x=P(A),y= P(B),P( A’ ∩ B’) =3/8 = P( A’) .P( B’) =(1- X)(1– Y) ⟹ X+Y – XY = 5/8 ⟹ X=1/2, Y= ¼.
  • 31. Ans. 5 P(A’)=1/2,P(B’) = 1-1/3=2/3, P(C’)=3/4 ∴ Req. Prob.= P(A)P(B’)P(C’) =P(A’)P(B)P(C’)+P(A’)P(B’)P(C)[A,B,C are indep. events] = (1/2)(2/3)(3/4)+(1/2)(1/3)(3/4)+(1/2)(2/3)(1/4)=11/24. Level---2 Q.1 If A ∩ B = ф, show that P(A/B) =0, where A and B are possible events. Q.2 A pair ofdice is thrownif the sum is even, find the probability that at least one of the dice Shows three. Q.3 Let X denotes the number of hours you study during a randomly selectedschool day.The probability that X can take the value x, has the following form,where k is some unknown constant P(X=0)=0.1and P(X=x) = { 𝒌𝒙 𝒊𝒇 𝒙 = 𝟏𝒐𝒓𝟐 𝒌( 𝟓− 𝒙) 𝒊𝒇 𝒙 = 𝟑𝒐𝒓𝟒 𝒐, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 (i) Find k. (ii) What is the probability that you study at least two hour? Exactly two hour? At most two hours? Q.4 Six dice are thrown 729 times.Howmany do you expect at least three dice to showa 5 or 6? Q.5 In a class;5% of the boys and 10% of the girls have an I.Q. of more than 150. In this class 60% ofthe students are boys. If a student is selectedat random and is formedand is found To have an I.Q. ofmore than 150, find the probability that the student is a boy. Answers of Level—2 Ans.1 A and B are possible events ⟹A≠ ф⇨P(A)≠0 , P(B) ≠0 But A∩B = ф ⟹ P(A∩B) =P(ф) = P(A/B) = 𝑷(𝑨∩𝑩) 𝑷(𝑩) =0. Ans. 2 n(S)=36, n(A)=18 Out of these 18, the cases which at least one die shows up 3 are (1, 3),(3,1),(3,3),(3,5),(5,3) Requiredprobability=5/18. Ans.3 X 0 1 2 3 4 P(X) 0.1 K 2K 2K K (i) k=0.15 (ii) 0.75, 0.3, o.55. Ans.4 P(success)=2/6=1/3 ∴ q=2/3
  • 32. P(x success i.e., getting a 5 or 6)= C(6, x) Px q6-x P(at least three successes insix trials) =P(x≥3)=1 – [p(0)+p(1)+p(2)] By using above result we get 1 – (16/81)(31/9) =233/729 ∴ requiredanswer is 233/729x729=233. Ans.5 Let E1: The student chosenis a boy. P(E1)=60/100 ∴ E2: ........................................girl. P(E2) = 40/100 E1, E2 are mutually exclusive.A: a student has an I.Q. of more than 150. P(A/ E1)= 5/100, P(A/ E2)= 10/100 By Baye’s theorem P(E1/A) =3/7. Probability Q. 1. A die is throwntwice and the sum of the numbers appearing is observedto be 6. What is the conditional probability that the number 4 has appeared at least once. [2/5] Q. 2. Assume that each born childis equally likely to be a boy or a girl. If a family has two children,what is the conditional probability that both are girls given that at least one is a girl. [1/3] Q. 3. If A and B are two independent events, showthat the probability of occurrence ofat least one of A and B is givenby : 1 – P(A').P(B') Q. 4. An urn contains 5 red and 5 blackballs. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover,2 additional balls of the colour drawnare put in the urn and then a ball is drawn at random.What is the probability that the secondball is red.[1/2] Q. 5. Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and standarddeviation of the number of kings.[34/221, 0.37] Q. 6. Q. 7. A doctor is to visit a patient .From the past experiences , it is known that the probabilities that he will come by train , bus , scooter or by other means of transport are respectively . The
  • 33. probabilities that he will be late are , if he comes by train , bus, and scooter respectively ,but if he come by other means of transport , then he will not be late .When he arrives he is late .What is the probability that he comes by train? whichlife skill is the doctor lacking?, ans. Lacks responsibility &dedicationtohis work. 13.1. Conditional Probability. Q.1. A and B toss a coinalternately till one of them gets a head and wins the game.If A starts first,find the probability that B will win the game. [1/3.] Q.2. A and B throwtwo dice simultaneously turn by turn. A will win if he throws a total of 5, B will win if he throws a doublet.Find the probability that B will win the game,though A startedit. [4/7] Q.3. Two dice are rolledonce. Find the probability that : i. the numbers on two dices are different. ii. the total of numbers on the two dice is at least 4. [11/12] Q.4. Two unbiased dice are tossed simultaneously.Find the probability that the sum of the numbers will be a multiple of 3 or 5. [19/36] Q.5. Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4. [4/9] 13.2. MultiplicationTheorem on Probability/Independent Events. Q.1. There are two bags.The first bag contains 4 white and 2 black balls, while the second bag contains 3 white and 4 black balls. A ball is picked up at random and a ball is drawn out. Find the probability that it is a white ball. [23/42]
  • 34. Q.2. In a groupof 9 students, there are 5 boys and 4 girls.A team of 4 students is to be selectedfor a quiz competition.Find the probability that there will be 2 boys and 2 girls in that team.[10/21] Q.3. 12 cards numbered 1 to 12, are placed in a box, mixed up thoroughly and then a cardis drawn at random from the box. If it is known that the number on the drawn card is more than 3, find the probability that it is an even number. [5/9] 13.3. Bayes’ Theorem. Q.1. In a bolt factory,machines A, B, C manufacture 25%, 35% and 40% respectively of the total bolts.Of their output 5%, 4% and 2% respectively are defective bolts. A bolt is drawn at random and is found to be defective. Find the probability that it is manufactured by machine B. 28/69.[Ans.] Q.2. In a bulb factory,machines A, B and C manufacture 60%, 30% and 10% bulbs respectively.1%,2% and 3% of the bulbs produced respectively by A, B and C are found to be defective. A bulb is picked up at random from the total productionand found to be defective. Find the probability that this bulb was produced by the machine A. Ans. = 2/5.] Q.3. A class consists of50 students out of which there are 10 girls. In the class 2 girls and 5 boys are rank holders in an examination.If a student is selectedat random from the class and is found to be a rank holder, what is the probability that the student selectedis a girl ? 2/7 . [Ans.] Q.4. A man is known to speak the truth 3 out of 4 times.He throws a die and reports that it is a six. Find the probability that it is actually a six. How is truthfulness helpfulinlife?----- Ans.--- truthfulness is essential life skill. Everybody trusts atruthful person. 3/8. [Ans.]
  • 35. Q.5. A company has two plants whichmanufacture scooters.Plant I manufactures 80% ofthe scooters while Plant II manufactures 20% of the scooters.At Plant I, 85 out of 100 scooters are ratedas being of standardquality, while at Plant II only 65 out of 100 scooters are ratedas being of standard quality. If a scooter is of standard quality , what is the probability that it come from Plant I. 0.84 .[Ans.] Q.6. A manufacturing firm produces steel pipes in three plants A, B and C with daily productionof 500, 1000 and 2000 units respectively.The fractions ofdefective steel pipes output produced by the plant A, B and C are respectively 0.005, 0.008 and 0.010. Ifa pipe is selectedfrom a day’s total productionand found to be defective, find out the probability that it came from the first plant. 5/61[Ans.] Q.7. An insurance company insured 6000 scooter drivers,3000 car drivers and 9000 truckdrivers.The probability of an accident involving a scooter,a car and a truck is 0.02, 0.06 and 0.30 respectively.One of the insuredpersons meets with an accident. Find the probability that he is a car driver. 0.06.[Ans.] Q.8. An insurance company insured 4000 doctors,8000 teachers and 12000 engineers.The probabilities ofa doctor,a teacher and an engineer dying before the age of 58 years are 0.01, 0.03 and 0.05 respectively.Ifone of the insured persondies before the age of58 years,find the probability that he is a doctor. Solution: Do yourself.[Ans. = 1/22] Q.9. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% ofthe cars and plant Y manufactures 30%.At plant X, 80% ofthe cars are rated of standardquality and at plant Y, 90% are ratedof standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant X. 56/83. [Ans.]
  • 36. Q. 10 in an examination, an examinee either guesses or copies or knows the answer of MCQs with four choices.The prob.That he makes a guess is 1/3and the prob. That he copies answer is 1/6, the prob. That his answer is correct, giventhat he copied it , is 1/8. Find the prob. That he copies the answer, given that he correctly answeredit.if a student copies an answer, what value is he violating? [hint P(A)=1/3, P(B)=1/6 , P(C) = 1 – 1/3– 1/6 =1/2, P(E/A)=1/4, P(E/B)=1/8,P(E/C)=1& P(B/E)=1/29] 13.4. Probability Distribution. Q.1. Two cards are drawn successively withreplacement from a well shuffled pack of 52 cards. Find the probability distributionof number of jacks. Hence required probability distributionis : x 0 1 2 P(X) 144/169 24/169 1/169 Q.2. Two cards are drawn successively withreplacement from a well shuffled deck of 52 cards. Find the probability distributionof number of aces. Solution:Do yourself.[Ans. of Q.1. above] Q.3. A pair of dice is tossedtwice. Ifthe random variable X is defined as the number of doublets, find the probability distributionof X. Hence, the requiredprobability distributionis x 0 1 2 P(X) 25/36 10/36 1/36 Q.4. Find the probability distributionofthe number of successes in two tosses of a die, where a success is defined as a number less than 3. Also find the mean and the variance ofthe distribution. Mean = Σpixi = 4/9×0+ 4/9×1+ 1/9×2= 6/9= 2/3.[Ans.] Variance = Σpixi2 – (Mean)2= (4/9×0+ 4/9×12 + 1/9×22) – (2/3)2 = 8/9 – 4/9 = 4/9.[Ans.]
  • 37. Q.5. An urn contains 5 white and 3 red balls. Find the probability distributionof the number of red balls, withreplacements,in three draws. Hence the required probability distributionis x 0 1 2 3 P(X) 125/512 225/512 135/512 27/512 Q.6. A pair of dice is thrown 4 times.If getting a doublet is considered a success,find the probability distributionof number ofsuccess. Therefore,required probability distributionis x 0 1 2 3 4 P(X) 625/1296 500/1296 150/1296 20/1296 1/1296 13.5. Binomial Distribution. Q.1. The mean and variance of a binomial distributions are 4 and 4/3 respectively.Find the distributionand P(X ≥ 1). 728/729.[Ans.] Q. 2 A drunk man takes a stepforward withprob. 0.4 &backwardwith prob. 0.6. find the prob. That at the end of11 steps,he is just one step away from the starting point. Is drinking alcohol a goodhabit? [ hint required prob.= P(X=5) +P(X=6) = 462(0.24)5 ] ASSIGNMENT(3-DIMENTIAL GEOMETRY) Direction Ratio & Direction Cosines of a Line. Q.1. The equation of a line is given by (4 – x)/2= (y + 3)/3 = (z + 2)/6. Write the directioncosines of a line parallel to the above line. l = – 2/7, m = 3/7, n = 6/7 [Ans.] Q.2. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the direction cosines of a line parallel to this line. [Ans. = 2/7, 3/7, – 6/7] 11.2. Equation of a Line in Space.
  • 38. Q.1. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P. PQ = √13units. [Ans.] Q.3. Find the point on the line : (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a distance 3√2 from the point (1, 2, 3). P(56/17,43/17, 111/17). Q.4. Find the equationof the perpendicular drawn from the point (2, 4, – 1) to the line (x + 5)/1= (y + 3)/4 = (z – 6)/–9. (x – 2)/6= (y – 4)/3= (z + 1)/2[Ans.] Q.2. Find the lengthand the foot of the perpendicular drawn from the point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11. [Ans. = Point (1, 2, 3); distance = √14 units.] 11.3. Angle Between Two Lines. Q.1. Find the angle between the pair of lines given by r→ = (2i – 5j + k) + λ(3i + 2j + 6k) and r→ = 7i – 6j – 6k + μ(i + 2j + 2k). θ = cos –1(19/21).[Ans.] 11.4. Shortest Distance Between Two Lines. Q.1. The vector equations of two lines are : r→ = i + 2j + 3k + λ(i – 3j + 2k ) and r→ = 4i + 5j + 6k + μ(2i + 3j + k). Find the shortest distance betweenthe above lines. 3/√19units. [Ans.] Q.2. Find the shortest distance betweenthe following lines : (x – 3)/1= (y – 5)/–2=(z – 7)/1and (x + 1)/7= (y + 1)/–6= (z + 1)/1. 2√29. [Ans.] Equation of a Plane. Q.1. Find the equation of the plane passing through the points (1, 2, 3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3.
  • 39. 6x – 3y + z = 3. [Ans.] Q.2. Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4). 5x – 7y + 11z + 4 = 0. [Ans.] Q.3. Find the equationof the plane passing throughthe point (–1,–1, 2) and perpendicular to each of the following planes : 2x + 3y – 3z = 2 and 5x – 4y + z = 6. 11x + 17y + 23z – 18 = 0 [Ans.] Q.4.Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line (x + 3)/2= (y – 3)/7 = (z – 2)/5. 8x – 13y + 15z + 13 = 0. [Ans.] 11.6. Distance of a Point from a Plane. Q.1. Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38. (3, 6, 11). [Ans.] Q.2. Find the co-ordinates ofthe image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0. [Ans. = (–3, 5, 2) ] Q.3. From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the lengthand the co-ordinates of the foot of the perpendicular. (11/9, 19/9, 34/9) [Ans.] Length of perpendicular from (1, 2, 4) is PM = |{2(1) + 2 – 2(4) + 3}/√(4+ 1 + 4)| = 1/3 unit. [Ans.] Q.4. Find the distance between the point P(6,5, 9) and the plane determinedby the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6). 6/√(34).[Ans.] Q.5. Find the co-ordinates ofthe point where the line (x + 1)/2= (y + 2)/3= (z + 3)/4 meets the plane x + y + 4z = 6. P(1, 1, 1) [Ans.] Q.6. Find the distance of the point (– 2, 3 – 4) from the line (x + 2)/3= (2y + 3)/4= (3z + 4)/5 measuredparallel to the plane 4x + 12y – 3z + 1 = 0. 17/2.[Ans.] Class – XII Subject – Mathematics (Three Dimensional Geometry)
  • 40. 1. Find the d.c’s of X, Y and Z-axis.[example 1,0,0(x-axis)0,1,0(y-axis] 2. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6. Write the direction cosines of a line parallel to the above line. [Ans: – 2/7, 3/7, 6/7] 3. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the direction cosines of a line parallel to this line. [ Ans : 2/7, 3/7, – 6/7] 4. Find coordinates of the foot of the per. drawn from the origin to the plane 2x – 3y + 4z -6 = 0. [example ans. 12/29, -18/29, 24/29] 5. Find the angle between the line (x + 1)/2 = y/3 = (z-3)/6 and the plane 10x + 2y – 11z = 3.[example ans. sinφ = | 𝒃⃗⃗ .𝒏⃗⃗ | 𝒃⃗⃗ ||𝒏⃗⃗ | | = 8/21] *6. Find the image of the point (1, 6, 3) in the line x = (y – 1)/2 = (z – 2)/3. [ Hint: it will be found that foot of per. from P to the line is N (1,3,5). If Q(𝜶, 𝜷, 𝜸) is the image of P then N is the mid point of PQ ⇨ 𝜶+𝟏 𝟐 = 𝟏, 𝜷+𝟔 𝟐 = 𝟑, 𝜸+𝟑 𝟐 = 𝟓 ⇨ Q(1, 0, 7)] 7. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P. [ Ans : √13 units., eqn. Of line is (x-1)/0=(y-6)/-3=(z+1)/2] 8. Find the length and the foot of the perpendicular drawn from the point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ – 11. [ Ans. = Point (1, 2, 3); distance = √14 units.] *9. Show that the angles between the diagonals of a cube is cos- 1 (1/3). Ans. (same as example. 26 of misc.) Let the length of cube be a , in fig. Of example : A,B,C are on x-axis, y-axis, z-axis resp.& F,G,D are in yz, xz ,xy –axis , O is origin , d.r’s of OE (E (a,a,a)) & AF are a,a,a & -a,a,a resp. ∴ cosѲ =(-a2 +a2 +a2 )/√3a.√3a = 1/3(angle b/w two lines)
  • 41. 10. Find the length of the perpendicular drawn from the point (2, 3, 7) to the plane 3x – y – z = 7 . Also find the coordinates of the foot of the perpendicular. [ same as Q. 8 ] *11. Find the point on the line (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a distance 3√2 from the point (1, 2, 3). Ans. (x + 2)/3 = (y + 1)/2 = (z – 3)/2 =k , any point on the line A(3k-2, 2k-1, 2k+3) , if its distance from B(1,2,3) is 3√2, then AB² = (3√2)² ⇨ K= 0 OR 30/17 , ∴ point A(-2,-1,3) OR A(56/17,43/17,111/17) [by putting value of k in A] 12. Find the equation of the perpendicular drawn from the point (2, 4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 . [ Ans : (x – 2)/6 = (y – 4)/3 = (z + 1)/2] *13. Find the equation of the line passing through the point (-1, 3, - 2) and perpendicular to the lines x = y/2 = z/3 and (x + 2)/(-3) = (y – 1)/2 = (z + 1)/5. Ans. eqn. Of line (x+1)/a=(y-3)/b=(z+2)/c ........(i) ,where a,b,c are d.r’s , since (i) is per. to lines ∴ a+2b+3c=0 & -3a+2b+5c=0, after solving , we get a/4=b/-14=c/8 =k(say) , so eqn. Of line (by putting the values of a=4k, b=-14k,c=8k) is (x+1)/4 = (y-3)/-14 = (z+2)/8 14. Find the foot of the per. drawn from the point (0, 2, 3) on the line (x + 3)/5 = (y – 1)/2 = (z + 4)/3 Also, find the length of the perpendicular. [ans..(2,3,-1), use distance formula, length= √21] *15. Find the equation of the plane passing through the line intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios (1, 2, 1) Also, find the perpendicular distance of the point P(3, 1, 2) from this plane.
  • 42. Ans. equation of the plane passing through the line intersection of the planes x – 2y + z -1 +k(2x + y + z – 8)=0 or (1+2k)x+(- 2+k)y+(1+k)z-1-8k=0 ........(i) is parallel to line ∴ (1+2k).1+(- 2+k).2+(1+k).1=0 ⇨ k= 2/5(normal to the plane must be per. to the line), put k in (i) ⇨ 9x-8y+7z-21=0, per. dis. From P Is 22/√(194) *19. Find the shortest distance between the following lines : (x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and (x + 1)/7 = (y + 1)/–6 = (z + 1)/1. [ncert Ans : 2√29 , use formula ] 20. Find the equation of the plane passing through the points (1, 2, 3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3. [Ans :same as Q. 26 , 6x – 3y + z = 3] 21. Find the shortest distance between the following pairs of lines whose cartesian equations are : (x -1)/2 = (y + 1)/3 = z and (x + 1)/3 = (y – 2) , z = 2. [ SAME as Q. 19 ] *22. Find the distance of the point P(2, 3, 4) from the plane 3x + 2y +2 z + 5=0 measured parallel to the line (x+3)/3 = (y-2)/6 = z/2 Ans. Any point on line (x+3)/3 = (y-2)/6 = z/2=k is Q(3k+2,6k+3,2k+4) It lies on plane ⇨ k=-1 ∴ point Q( -1,-3,2),PQ=7 23. Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4). [ Ans : 5x – 7y + 11z + 4 = 0] *24. Find the equation of the plane passing through the point (–1, – 1, 2) and perpendicular to each of the following planes : 2x + 3y – 3z = 2 and 5x – 4y + z = 6. [ Ans : 9x + 17y + 23z – 20 = 0, same as example of misc. Eqn. Of plane a(x+1)+b(y+1)+c(z-2)=0........(i) By condition of
  • 43. perpendicularity to the plane (i) with the planes 2a+3b-3c=0 & 5a- 4b+c =0 , after solving , we get a=9c/23 & b=17c/23, put in (i) ] 25. Find the equation of the plane passing through the point (1, 1, - 1) and perpendicular to the planes x + 2y + 3z – 7=0 and 2x – 3y + 4z = 0. [ SAME as Q. 24] 26. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z – 2)/5. Ans : 8x – 13y + 15z + 13 = 0, eqn. Of plane a(x-3)+b(y-4)+c(z- 1)=0.......(i),It passes through (0,1,0) ,then -3a-3b-c=0 // to line ∴ 2a+7b+5c=0, after solving a=8c/15, b=-13c/15, put in (i) *27. Find the image of the point P(1,2,3) in the plane x+2y+4Z=38 [ Ans :Let M is foot of per. (mid point of PQ) , if Q (𝜶, 𝜷, 𝜸 ) , d.r’s of a normal to the plane are 1,2,4, eqn. Of line PM is (x-1)/1=(y- 2)/2=(z-3)/4 = k ⇨ any point on a line Q(1+K,2+2K,3+4K) ∴ M{(2+k)/2,(4+2k)/2,(6+4k)/2} put in plane ⇨ k=2 ∴ Q(3,6,11) ] 28. Find the co-ordinates of the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0. [Ans : (–3, 5, 2)] *29. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6). Ans. [example28 of misc. Ans: 6/√(34),we can find eqn. Of plane through A,B,C and find distance of P from plane or can be found as PD = 𝑨𝑷⃗⃗⃗⃗⃗⃗ . 𝑨𝑩⃗⃗⃗⃗⃗⃗ × 𝑨𝑪⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ( proj. Of 𝑨𝑷⃗⃗⃗⃗⃗⃗ on 𝑨𝑩⃗⃗⃗⃗⃗⃗ × 𝑨𝑪⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ , D is foot of per. from P) or find eqn. of plane passes through A,B,C , then find dist. ] 30. Find the co-ordinates of the point where the line (x + 1)/2 = (y + 2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6. [Ans: P(1, 1, 1)]
  • 44. *31. Find the distance of the point A(– 2, 3 – 4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y – 3z + 1 = 0. [Ans: |AP|= 17/2, eqn. Can be written as (x + 2)/3 = (y + 3/2)/2 = (z + 4/3)/5/3 = t ⇨ P{ -2+3t, (-3/2)+2t, (-4/3)+5t/3} be any point on line D.R’S of AP {3t, 2t-(9/2), (5t/3)+(8/3)} , since AP is parallel to plane 4.3t+ (12).[ 2t-(9/2)]+ (-3). (5t/3)+(8/3) = 0 ⇨ t=2 ∴ P (4,5/2,2) ] Ques. 32 Find the distance of the point P(-2, 3, -4) from the line (x+2)/3 = (2y+3)/4 = (3z+4)/6 measured // to the plane 4x + 12y -3 z + 1=0. Ans. same as Q. 22, point on lineQ(3k-2,2k-(3/2),(5k/3)-(4/3)), d.r’s of PQ are (3k,2k-(9/2),(5k/3)+(8/3)) , since PQ is // to plane ∴ PQ is per. to normal to the plane ⇨ (3k).4+{2k-(9/2)}.12+{(5k/3)+(8/3)}.(- 3)=0 ∴ k= 2 ⇨ Q(4,5/2,2), PQ = 17/2 Question 33 find whether the lines 𝒓⃗ = (i-j-k) +λ (2i+j) and 𝒓⃗ = (2i-j) +μ (i+j-k) intersect or not . if intersecting, find their point of intersection. Ans. 𝒂 𝟐⃗⃗⃗⃗ -𝒂 𝟏⃗⃗⃗⃗ = I +k , 𝒃 𝟏 ⃗⃗⃗⃗ × 𝒃 𝟐 ⃗⃗⃗⃗ = 𝒊 𝒋 𝒌 𝟐 𝟏 𝟎 𝟏 𝟏 −𝟏 = -i+2j+k , its magnitude is √6 Shortest distance = = 0/√6 = 0 Hence lines intersect. Point of intersection is given by (i-j-k) +λ (2i+j) = (2i-j) +μ (i+j-k) ⇨ (1+2λ) = 2+μ , -1+λ - -1+μ and -1 = -μ ⇨ λ = 1 and μ =1 satisfy μ = 2λ -1 , put the values of λ and μ in given lines , we obtain the position vector of point of intersection of the two given lines as 3i- k i.e., the point of intersection is (3,o,-1).
  • 45. Ques. 34 Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (- 4,4,4) are coplanar. Also find the eqn. of the plane containing them. Ans. Eqn. of plane passes through (0,-1,-1) is a(x-0)+b(y+1)+(z+1)=0 It passes through (4,5,1), (3,9,4), we get 4a+6b+2c=0, 3a+10b+5c=0 ⇨ eqn. of plane is 5x-7y+11z+4=0 and (-4,4,4) will satisff the eqn. of plane to be coplanar. Q. 35 Show that lines (x+3)/-3 = (y-1)/1 = (z-5)/5 & (x+1)/-1=(y-2)/2 =(z-5)/5 are coplanar . find the eqn. of plane. Ans.(as example 21), The given lines are coplanar if (𝒙𝟐 − 𝒙𝟏) (𝒚𝟐 − 𝒚𝟏) (𝒛𝟐 − 𝒛𝟏) 𝒍𝟏 𝒎𝟏 𝒏𝟏 𝒍𝟐 𝒎𝟐 𝒏𝟐 = (−𝟏 + 𝟑) (𝟐 − 𝟏) 𝟎 −𝟑 𝟏 𝟓 −𝟏 𝟐 𝟓 = 0 , so lines are coplanar. Eqn of plane = (𝒙 − 𝒙𝟏) (𝒚 − 𝒚𝟏) (𝒛 − 𝒛𝟏) 𝒍𝟏 𝒎𝟏 𝒏𝟏 𝒍𝟐 𝒎𝟐 𝒏𝟐 = (𝒙 + 𝟑) (𝒚 − 𝟏) (𝒛 − 𝟓) −𝟑 𝟏 𝟓 −𝟏 𝟐 𝟓 = x-2y+z=0. Q.36 Find the shortest distance and eqn. Of shortest distance b/w the lines (x-6)/3 = (y-7)/-1 = (z-4)/1 & x/-3 = (y+9)/2 = (z-2)/4 . S (3p+6,-p+7,p+4) ......(i) D (-3q,2q-9,4q+2) .......(ii) Ans. let SD be the shortest dist. , then co-ordinates of eqn.(i) are ( 3p+6, -p+7, p+4) & eqn.(ii) are (-3q, 2q-9, 4q+2)
  • 46. d.c’s of SD are proportional to -3q-3p-6, 2q+p-16, 4q-p-2 and if SD is at right angles to both lines , we get 3.( -3q-3p-6)+(-1).( 2q+p-16)+ 1. (4q-p-2)=0 and -3.(-3q-3p-6)+2.( 2q+p-16)+4.( 4q-p-2) =0 ⇨ 7q+11p+4=0 & 29q+7p-22=0 ⇨ p=-1 , q=1 and SD=3√(30) , eqn. Will be (x-3)/-6 = (y-8)/-15 = (z-3)/3 Q.37 Find the shortest distance and eqn. b/w the lines 𝒓⃗ = (8+3λ)i- (9+16λ)j+(10+7λ)k & 𝒓⃗ = (15i+29j+5k)+μ(3i+8j-5k). Ans. let SD be the shortest dist. b/w two lines , let position vector of S be (8i-9j+10k)+λ(3i-16j+7k) and position vector of D is (15i+29j+5k+μ(3i+8j-5k), 𝑺𝑫⃗⃗⃗⃗⃗ = 𝑫⃗⃗ − 𝑺⃗⃗ = (3μ-3λ+7)i+(8μ+16λ+38)j+(- 7λ-5μ-5)k since SD is perpendicular on line (i) & (ii) then [(3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k]. (3i-16j+7k)=0 ⇨ 157λ+77μ+311=0...(iii) & 77λ+49μ+175=0...(iv), by solving λ = -1,μ=- 2 ∴ 𝑺𝑫⃗⃗⃗⃗⃗ = 𝑫⃗⃗ − 𝑺⃗⃗ = 4i+6j+12k ⇨| 𝑺𝑫⃗⃗⃗⃗⃗ |= 14 & eqn. Is 𝒓⃗ = (5i+7j+3k)+t[(9i+13j+15k)-(5i+7j+3k)] = (5i+7j+3k)+t(4i+6j+12k) where 𝑺⃗⃗ =5i+7j+3k, 𝑫⃗⃗ =9i+13j+15k & use 𝒓⃗ =𝒂⃗⃗ +t(𝒃⃗⃗ − 𝒂⃗⃗ ). Vector Algebra Q. 1. Find a vector in the direction of vector that has magnitude 7 units. Q. 2. Show that the points A, B and C with position vectors, respectively, form the vertices of a right angled triangle. Q. 3. Find , if two vectors are such that . Q. 4. Find the area of the parallelogram whose adjacent sides are determined by the vectors
  • 47. Q. 5. If a unit vector makes angles with , and acute angle θ with , then find θ and hence the components of . Q. 6. Let , and be three vectors such that and each one of them being perpendicular to the sum of other two, Find Q. 7. Find the value of Q. 8. The scalar product of the vector with a unit vector along the sum of vectors Find the value of . Q. 9. If the sum of two unit vectors is a unit vector, Prove that the magnitude of their difference is . Q. 10. If are position vectors of points A and B respectively, then find the position vector of points of trisection of AB. Q. 11. Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it. Q. 12. ABCD is a parallelogram. If the coordinates of A, B, C are (-2, -1), (3, 0) and (1, -2) respectively, Find the co-ordinate of D. Q. 13. Show that the points A, B, C with position vectors Q. 14. If a vector makes with OX, OY and OZ respectively, prove that Q. 15. If inclined at an angle , then prove that sin = .
  • 48. Q. 16. If . Q. 17. If . Q. 18. Consider two points P and Q with position vectors . Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii) externally. Q. 19. Find a unit vector perpendicular to each of the vectors Q. 20. Show that the vectors 2 i – j + k , i– 3 j – 5 k and 3 i – 4 j – 4 k form the vertices of a right angled triangle Q. 21. ASSIGNMENT(LINEARPROGRAMMING) Q.1. David wants to invest at most Rs12,000in Bonds A and B. According to the rule, he has to invest at least Rs2,000in Bond A and at least Rs4,000 in Bond B. If the rate of interest in bonds A and B respectively are 8% and 10% per annum, formulate the problem as L.P.P. and solve it graphically for maximum interest.Also determine the maximum interest receivedin a year. Do you think your investment in bond A &B help in the well being of the nation?Do you think that a person shouldstart saving at an early age for his retirement? [ANS:Therefore,the interest I is maximum at C(2000, 10000) and the maximum interest is Rs1160.] Q.2. A farmer has a supply of chemical fertilizer oftype A which contains 10% of nitrogenand 6% of phosphoric acidand of type B which contains 5% ofnitrogen and 10% of phosphoric acid. After soil testing it is found that at least 7 kg of nitrogenand the same quantity of phosphoric acid is requiredfor a goodcrop. The fertilizer oftype A
  • 49. costs Rs5.00per kg and the type B costs Rs8.00per kg. Using linear programming findhow many kgs ofeach type of the fertilizer should be bought to meet the requirement and the cost be minimum.Solve the problem graphically. [ANS: Therefore,cost is minimum at P(50,40). That is 50 kg of type A and 40 kg oftype B is mixed to meet the requirement. Q.3. A dietician wishes to mix two types of foods in such a way that vitamincontents ofthe mixture contain at least 8 units of vitamin A and 10 units ofvitamin C. Food I contains 2 units/kg of vitaminA and 1 unit/kg ofvitamin C. Food II contains 1 unit/kg of vitaminA and 2 units/kg of vitaminC. It costs Rs50per kg to purchase Food I and Rs70 per kg to purchase FoodII. Formulate this problem as a linear programming problem to minimize the cost of such a mixture. [ANS:Hence,the minimum value of Z is Rs380.] Q.4. Two tailors A and B are paid Rs150and Rs200per day respectively.A can stich6 shirts and 4 paints while B can stich10 shirts and 4 paints per day. Form a linear programming problem to minimize the labour cost to produce at least 60 shirts and 32 paints. Solve the problem graphically. These shirts & paints have been specially designed for a ‘VAN MAHOTSAVA’ programme.Wouldyou like to participate in this programme? Why? [Ans. = Tailor A works for 5 days and B for 3 days; Minimum cost = Rs1,350] Q.5. A cooperative society offarmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimatedas Rs10,500 and Rs9,000respectively.To control weeds,a liquid herbicide has to be used for crops X and Y at rates of20 litres and 10 litres per hectare. Further, not more than 800 litres of herbicide should be used in order to protect fishand wild life using a pond which collects drainage from this land. Howmuch land should be allocatedto eachcrop so as to maximize the total profit of the society ? In what ways does a cooperative society helpthe farmers? [ANS:Hence , society will get the maximum profit of Rs4,95,000 by allocating 30 hectare for cropX and 20 hectare for cropY.]
  • 50. Q.6. Kellogg is a new cereal formedof a mixture of bran and rice that contains at least 88 grams ofproteinand at least 36 milligrams of iron. Knowing that bran contains 80 grams ofprotein and 40 milligrams ofironper kilogram,and that rice contains 100 grams of proteinand 30 milligrams ofironper kilogram,find the minimum cost of producing this new cereal if bran costs Rs5per kilogram and rice costs Rs4per kilogram. [ANS: Therefore,minimum cost of producing this cereal is Rs4.60 per kg.] Q.7. A new cereal, formedof a mixture of bran and rice, contains at least 88 grams of proteinand at least 36 milligrams ofiron.Knowing that bran contains 80 grams of proteinand 40 milligrams ofironper kilogram,and that rice contains 100 grams ofprotein and 30 milligrams ofironper kilogram,find the minimum cost of producing a kilogram ofthis new cereal if bran costs Rs.28 per kilogram and rice costs Rs.25 per kilogram. Solution: – Do yourself.[Ans. = Rs.26.8] Q.8. A dealer wishes to purchase a number of fans and CFLs. He has only Rs5,760to invest and has space for at most 20 items.A fan cost him Rs360and a CFL Rs240. He expects to sell a fan at a profit of Rs22 and a CFL at a profit of Rs18. Assuming that he can sell all the atoms that he buys, how should he invest his money to maximize the profit ? Solve graphically and find the maximum profit.What is the full form of CFL? Which one iswmore energy efficient: CFL or ordinary incandescent bulb? [ANS: Therefore,for maximum profit he shouldbuy and sell 8 fans and 12 CFLs. His maximum profit is Rs392.] Q.9. A man has Rs1,500 for purchasing rice and wheat. A bag ofrice and a bag ofwheat cost Rs180and Rs120respectively.He has the storage capacity of at most 10 bags.He earns a profit ofRs11 and Rs9 per bag ofrice and wheat respectively.Formulate the above problem as an LPP to maximize the profit and solve it graphically. Solution: Do yourself. [Ans. Max. Profit = Rs100, No.of rice bags = 5 = No.of wheat bags]
  • 51. Q.10. A factory owner purchases two types ofmachines, A and B for his factory.The requirements and the limitations for the machines are as follows: Machine Area occupied Labour force Daily output (in units) A 1000 m2 12 men 60 B 1200 m2 8 men 40 He has maximum area of 9000 m2 available,and 72 skilledlabourers who can operate both the machines.How many machines of eachtype should he buy to maximize the daily output? Solution: [ANS: The maximum output is at B and C. But the number of machines cannot be a fraction. Hence no. ofmachines of type A = 6 and no. of machines of type B = 0 ] Q.11 Ifa 19 years oldgirl drives her car at 25 km/hr,she has to spend Rs.2/km on petrol.If she drives it at a faster speedof 40 km/hr, the petrol cost increases to Rs.5/km.She has Rs.100 to spend on petrol and wishes to find the max. Distance she can travel within one hour . express it as a L.P.P. and then solve it . Is the girl eligible for a driving licence? What is the benefit of driving at economic speed? Solution:[ans: max distance is 30 km.at (50/3,40/3) .yes ∵ minimum age require for licence is 18 years.Reduces fuel consumptionand causes less pollution.]