CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-13
Fill in the blanks.
Q.1827/log The expression
 
1
.
0
20
log
)
05
.
0
( is a perfect square of the natural number ______.
(where 0 1
. denotes 0.111111 ..... ) [Ans. 9]
[Sol.











 9
1
log
2 20
20
1
=
2
20 3
log
2
20


=
4
20 3
log
20 = 81  9 ]
Q.2808/log The expression, xlogy  logz · ylogz  logx · zlogx  logy when simplified reduces to ______. [Ans. 1]
[Sol. P = xlogy  logz · ylogz  logx · zlogx  logy
log P = (log y– log z) log x + (log z – log x) log y+ (log x – log y) log z
log P = 0  P = 1 ]
Select the correct alternative : (Only one is correct)
Q.3log If logx(logyz) = 0 and logy(logzx) = 0, where x, y, z > 1, then 2z – x – y equals
(A*) 0 (B) 1 (C) xy (D)yz
[Hint: x = y = z ]
Q.4 Whichofthefollowingisthelargest?
(A)
6
log5
2 (B) 5
log6
3 (C*) 6
log5
3 (D) 3
Q.5805/QE If sin  and cos  are the roots of the equation ax2 – bx + c = 0, then
(A) a2 – b2 = 2ac (B) a2 + b2 = 2ac (C) a2 + b2 + 2ac = 0 (D*) b2 – a2 = 2ac
Q.642/ph-1 Let a = cos x + cos 




 

3
2
x + cos 




 

3
4
x and b = sin x + sin 




 

3
2
x + sin 




 

3
4
x then
whichoneof thefollowingdoesnot hold good?
(A) a = 2b (B) b = 2a (C) a + b = 0 (D*) a  b
[Sol. Using cos(A + B) = cosA cosB – sinA sinB
a = cos x + 2 cos (x + )cos
3

(using C-Drelation)
= cos x – cos x = 0
|||ly b = sin x + 2 sin x( + x)cos
3

= sin x – sin x = 0
= sin x – sin x = 0
Hence a = b = 0 ]
Q.7log Suppose that
log10(x – 2) + log10y = 0 and y
x
2
y
x 



Then the value of (x + y), is
(A) 2 (B) 2
2 (C*) 2 + 2
2 (D) 4 + 2
2

[Sol. y(x – 2) = 1
 x =
y
1
+ 2 .....(1)
x + 2
y  = y
x 
x + y – 2 + 2 )
2
y
(
x  = x + y
1 = )
2
y
(
x 
x(y – 2) = 1 ....(2)
using(1)in(2)








 2
y
1
(y – 2) = 1
1 –
y
2
+ 2y – 4 = 1
y2 – 2y – 1 = 0
y =
2
4
4
2 

y = 1 ± 2 ....(3)
rejecting y = 1 – 2
y = 1 + 2 using(3)in(1)
x = 1 + 2
 x + y = 2 + 2 2 Ans. ]
Q.829/log Let x, y, z are real numbers greater than 1 and 'w' is a positive real number. If logxw = 24, logyw = 40
and logxyzw =12 then logwz has the value equal to
(A)
120
1
(B*)
120
2
(C)
120
3
(D)
120
5
[Sol. Given logwx =
24
1
....(1); logwy=
40
1
....(2), Let logwz = k ....(3)
(1) + (2) + (3)  logw(x y z) =
24
1
+
40
1
+ k

12
1
=
24
1
+
40
1
+ k
 k =
12
1
–
24
1
–
40
1
=
120
3
5
10 

=
120
2
,
hence logwz =
120
2
Ans. ]
Q.949/log The value of the expression, log4 







4
x2
 2 log4 (4 x4) when x =  2 is :
(A*)  6 (B)  5 (C)  4 (D) meaningless
[Hint: (log4x2 – 1) –2(1 + log4x4)
 when x = – 2 the equation becomes
log44 – 3 – 2 log416 = 1 – 7 = – 6 Ans ]

Q.10 The solution set of the equation, 3 log10 x + 2 log10
1
x
= 2 is :
(A) {10, 102} (B) {10, 103} (C*) {10, 104} (D) {10, 102, 104}
[Sol. 3 log10 x – 2 log10
x = 2
log10 x = y
3y – 2y2
= 2
2y2
– 3y + 2 = 0
2y2
– 2y – y + 2 = 0
2y(y – 1) – (y – 1) = 0
(y – 1)(2y – 1) = 0
y = 1; y = 1/2
if log10 x = 1  x = 10
log10 x = 1/2  x = 104]
Subjective :
Q.1113/6 If  and  are the roots of the quadratic equation (sin 2a)x2 – 2(sin a +cos a)x + 2 = 0, findthem and
hence prove that 2 + 2 = 2 · 2. [Ans. sec a , cosec a] [5]
[Sol. x =
a
2
sin
2
a
2
sin
8
)
a
cos
a
(sin
4
)
a
cos
a
(sin
2 2




=
a
2
sin
a
2
sin
2
)
a
cos
a
(sin
)
a
cos
a
(sin 2




=
a
2
sin
a
2
sin
2
a
2
sin
1
)
a
cos
a
(sin 



=
a
2
sin
)
a
2
sin
1
(
)
a
cos
a
(sin 


= a
cos
·
a
sin
2
)
a
cos
a
(sin
)
a
cos
a
(sin 


with+vesign a
cos
·
a
sin
2
a
sin
2
= sec a
with–vesign a
cos
·
a
sin
2
a
cos
2
= cosec a
now 2 + 2 = sec2a + cosec2a
=
a
cos
1
2 +
a
sin
1
2 =
a
cos
·
a
sin
1
2
2 = sec2a · cosec2a = 2 · 2 hence proved]
Q.12 Findallintegral solutionof theequation,      
3
x
2
2
x
4
2
x x
log
3
x
log
2
x
log
4 
 . [5]
[Ans. x = 1, x = 4]
[Sol.
 
2
x
log
x
log
4
2
2
+
)
x
4
(
log
)
x
(
log
2
2
2
2
=
)
x
2
(
log
)
x
(
log
3
2
3
2
1
x
log
)
x
(
log
2
1
·
4
2
2

+ )
x
(
log
2
)
x
(
log
4
2
2
 = )
x
(
log
1
)
x
(
log
9
2
2


let log2x = t
1
t
t
2

+
2
t
t
4

=
1
t
t
9


1
t
2

+
2
t
4

=
1
t
9

)
2
t
)(
1
t
(
4
t
4
4
t
2





=
1
t
9

 6t (t + 1) = 9(t2 + t – 2)  6t2 + 6t = 9t2 + 9t – 18  3t2 + 3t – 18 = 0
 t2 + t – 6 = 0  (t + 3)(t – 2) = 0
t = 0, t = 2 & t = – 3,
x = 1, x = 4, x =
8
1
(rejected  it is not integral value)]

Select the correct alternative : (Only one is correct)
Q.1log Let N =
2
2
12
22
2
12
22 




 

 then log2N equals
(A) 2 (B) 3 (C*) 4 (D) none
[Sol. N =
2
2
2
)
2
2
3
(
)
2
2
3
( 




 

 = 1
 2
)
2
2
3
(
)
2
2
3
( 

 = 16
hence log2N = log216 = 4 Ans. ]
Q.2log The sum of all valuesof x so that )
2
x
3
x
(
)
1
x
3
x
( 2
2
8
16 



 ,is
(A) 0 (B) 3 (C*) – 3 (D) – 5
[Sol. )
2
x
3
x
(
)
1
x
3
x
( 2
2
8
16 



  )
2
x
3
x
(
3
)
1
x
3
x
(
4 2
2
2
2 



  6
x
9
x
3
)
4
x
12
x
4 2
2
2
2 




4x2 + 12x – 4 = 3x2 + 9x + 6  x2 + 3x – 10 = 0  (x + 5)(x – 2) = 0
x = – 5 or x = 2
 the sum of values of x is (– 5) + 2 = – 3 Ans. ]
Q.3 The equation, | sin x | = sin x + 3 in [0, 2] has
(A*) no root (B) onlyone root (C) two roots (D) more than two roots
Q.420/log Given log2(log8x) = log8(log2x) then (log2x)2 has the value equal to
(A) 9 (B) 12 (C*) 27 (D) 3
3
[Sol. Given log2






x
log
3
1
2 =
3
1
log2(log2x)
let log2x = 3y
 log2y=
3
1
log23y  3log2y= log23 + log2y  2 log2y = log23
 y2 = 3 
9
1
(log2x)2 = 3  (log2x)2 = 27 Ans. ]
Q.530/log The reals x and ysatisfy
log8x + log4(y2) = 5 and log8y+ log4(x2) = 7
thenthe valueofxyis
(A) 1024 (B*) 512 (C) 256 (D) 81
[Sol. equation(1) 
3
1
log2x + log2y= 5 ....(3) and
equation(2) 
3
1
log2y+ log2x = 7 .....(4)
(3) + (4) 
3
1
log2(xy) + log2(x y) = 12 
3
4
log2(xy) = 12
log2(xy) = 9  xy = 29 = 512 Ans.]
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-14

Q.6 If sin 2x =
2025
2024
, where
4
5
< x <
4
9
, the value of the sin x – cos x is equal to
(A*) –
45
1
(B)
45
1
(C) ±
2025
1
(D) none
[Sol. (sin x – cos x)2 = 1 – sin 2x
= 1 –
2025
2024
=
2025
1
sin x – cos x =
45
1
or –
45
1
in
4
5
to
4
9
, sin x – cos x < 0
Hence answer is B ]
Q.7log The equation ln 







  )
1
k
(
1
k
1
)
1
k
(
k
= F(k) · 













 k
n
k
1
1
k
1
1
n l
l is true for all k wherever defined.
F(100) has the value equal to
(A) 100 (B*)
101
1
(C) 5050 (D)
100
1
[Sol. L.H.S. = k
n
k
1
l –
1
k
1

ln(k + 1) =
)
1
k
(
k
)
1
k
(
n
k
k
n
)
1
k
(



 l
l
R.H.S. =
k
)
k
(
F
[k ln k – k ln(k + 1) + ln k]
=
k
)
k
(
F
[(k + 1) ln k – k ln(k + 1)]
 L.H.S. = R.H.S.

)
1
k
(
k
)
1
k
(
n
k
k
n
)
1
k
(



 l
l
=
k
)
k
(
F
[(k + 1) ln k – k ln(k + 1)]
 F (k) =
1
k
1

F(100) =
101
1
Ans. ]
Q.848/log If x1 and x2 are two solutions of the equation log3 2x  7 = 1 where x1 < x2 , then the number of
integer(s) between x1 and x2 is/are :
(A*) 2 (B) 3 (C) 4 (D) 5
[Hint: log3| 2x – 7| = 1  |2x – 7| = 3
Case 1 : x > 7/2  2x – 7 = 3  x = 5
Case 2 : x < 7/2  – 2x + 7 = 3  x = 2
number of integers between 2 and 5 are 3 and 4 = 2 ]

Select the correct alternative : (More than one are correct)
Q.9522/log The equation 1 27
 logx
log3 x + 1 = 0 has :
(A*) no integral solution (B) oneirrationalsolution
(C) two real solutions (D*) no primesolution
[ Hint : x
log
x
log
2
3
1 3
3








 + 1 = 0
let log3x = y
y
y
2
3
1 







 = – 1  2
y
1
y
2
3
1 








  2
y
1
y
2
3
y
2


2y2 + 3y – 2 = 0  2y2 + 4y – y – 2 = 0  ( y +2) (2y – 1) = 0
y = 1/2 or y = – 2  x = 31/2 (rejected) or x = 1/9 ]
Subjective :
Q.1012/1 Find the set of values of ‘x’ satisfying the equation x
64 – x 3
x
3
2  + 12 = 0. [Ans: x = 3 ]
[Hint: Here x  N, therefore the 2nd value x = 6
log
3
2
has to be rejected
given equation reduced to  2
x
3
2 – 8 · x
3
2 + 12 = 0 ; put x
3
2 = y and proceed. ]
Q.11 Let aandbaretworealnumbers suchthat, sina+sinb=
2
2
andcos a+cosb =
2
6
.Findthevalueof
(i) cos(a – b) and (ii) sin(a + b). [3]
[Sol. squaringandadding [Ans. (i) 0, (ii)
2
3
]
2 + 2 cos(a – b) =
2
1
+
2
3
= 2
cos(a – b) = 0
again 2 sin
2
b
a 
cos
2
b
a 
=
2
1
2cos
2
b
a 
cos
2
b
a 
=
2
3
 tan
2
b
a 
=
3
1
 sin(a + b) =
2
b
a
tan
1
2
b
a
tan
2
2 


=
3
1
1
3
1
2


=
4
3
3
2
 =
2
3
]
Q.12 For any 3 angle ,  and , prove that
sin  + sin  + sin  = sin( +  + ) + 4 sin 




 


2
·sin 




 


2
·sin 




 


2
. [5]

[Sol. TPT sin  + sin  + sin  – sin( +  + ) = 4 sin 




 


2
·sin 




 


2
·sin 




 


2
LHS = 




 







 










 







 


2
sin
2
cos
2
2
cos
2
sin
2
= 










 










 







 


2
2
cos
2
cos
2
sin
2
= 










 







 







 


2
sin
2
sin
2
2
sin
2
= 4 sin 




 


2
·sin 




 


2
·sin 




 


2
= RHS hence proved. ]

CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-15
Fill in the blanks :
Q.1803/log Solution set of the equation, x
log2
10
+ log10x2 = 2
log2
10
 1 is ______ . [Ans.
1
20
,
1
5
]
[Sol. (log10x)2 + 2log10x + 1 = 2
log2
10
= (log10x + 1)2 = 2
log2
10
log10x + 1 = + log102 log10x + 1 = – log102
log10x + 1 = log102 log10(2x) = – 1
log10  
x
2 = 1 2x = 10–1
 
x
2 = 10 = 5
1 x = 20
1 ]
Select the correct alternative : (Only one is correct)
Q.2 Let m denotesthe number of digits in 264 and n denotes the number of zeroes between decimal point
and the first significant digit in 2–64, then theordered pair (m, n) is (you mayuse log102=0.3)
(A) (20, 21) (B) (20, 20) (C) (19, 19) (D*) (20, 19)
[Sol. x = 264 y = 2–64
log10x = 64 log10x log10y = – 64 log102
= 64 × 0.3 = – 64 × 0.3
log10x = 19.2 log10y = – 19.2
m = 19 + 1 log10y = 8
.
20
m = 20 n = 19
 (20, 19) ]
Q.326/log The ratio
1
a
7
a
2
3
2
a
log
4
log
a
log
49
3
2
4
/
1 )
( 1
a
27
2





simplifies to :
(A) a2  a  1 (B) a2 + a  1 (C) a2  a + 1 (D*) a2 + a + 1
[Hint:
N
D
r
r =
a a a
a a
4 2
2
2 1
1
  
 
 D ]
Q.4 Whichoneof the followingdenotesthegreatest positive properfraction?
(A)
6
log 2
4
1






(B)
5
log3
3
1






(C*)
2
log3
3

(D)








 2
log
1
3
8
[Hint: A: a = 6
log
2
2
2
1
= 36
log
2
2
1
=
1
36
; B: b = 5
log
3
3
1
=
1
5
C : c = 2
1
3
1
2
log
3
 ; D: d =
3
log
2
8

=
27
log
2
2

= 27
1
2
1
27
log
2
 ]
Select the correct alternative : (More than one are correct)
Q.5152/log The solution set of the system of equations, log12x
1
2
2
log
log
x
y






 = log2 x and
log2 x . (log3 (x + y)) = 3 log3 x is:
(A*) x = 6; y = 2 (B) x = 4; y = 3 (C*) x = 2; y = 6 (D) x = 3; y = 4
[Hint: log12x(log2xy) = log2x  log2xy = log212 xy= 12 ....(1)
log3( x + y) = 3. log32  x + y = 8 ....(2)
from (1) and (2) x = 6 , y = 2 or y = 6 , x = 2 ]

Subjective :
Q.6830/log If log18 36 = a & log2472 = b, then find the value of 4 (a +b)  5 ab. [Ans. 2]
[Hint: a =
y
2
y
2
2
2
3
log
2
3
log
18
log
36
log
2
3
2
2
3





 where log32 = y
b =
1
y
3
2
y
3
3
2
log
3
2
log
24
log
72
log
3
3
2
3
3





 and henceresult ] ]
Q.711/log It is known that x = 9 is aroot of the equation log (x2 + 15a2) – log(a2) = log
8
2
ax
a 
. Find theother
root(s) ofthis equation.
[Sol.
2
a
ax
8
log
2
a
a
15
x
log
2
2





 .........(1)
 x2 + 15a2 = 8ax
or x2 – 8ax + 15a2 = 0 .........(2)
(x – 5a) (x – 3a) = 0  a =
x
5
or a =
x
3
 x = 9 satisfies (1) hence a =
9
5
or a = 3
but a =
9
5
is not possible (think!)  a = 3
substitutinga = 3 in (2)
x2 – 24x + 225 = 0  x = 9 or x = 15
 other root is x = 15 ]
Q.849/06 If  =
1
2n


, prove that 2n cos cos2 cos22 ........ cos2n–1 = 1. What the value of the product
whould be if  =
1
2n


. [Ans. –1]
[Sol. We have
2n ·





 
sin
2
cos
.
..........
.
2
cos
.
2
cos
.
cos
.
sin 1
n
2
=




 

sin
2
cos
......
..........
2
cos
.
2
cos
.
2
sin
.
2 1
n
2
1
n
=



 

sin
2
cos
.......
..........
2
cos
.
2
sin
.
2 1
n
2
2
2
n
=



 

sin
2
cos
.......
..........
2
cos
.
)
.
2
sin(
.
2 1
n
3
3
3
n
Thisexpressionfinallyconvergerto
=


sin
2
sin n
=









sin
1
2
2
sin n
n
=











sin
1
2
2
sin n
n
=













sin
1
2
2
sin n
n
=
















1
2
sin
1
2
sin
n
n
= 1 ]
Q.9 Evaluate the product 



1
n
2
1
r
)
r
tan( where 4n = . [Ans : 1]
[Sol. P = tan  × tan 2 × tan 3 × ........... × tan(2n – 1)
= tan  · tan(2n – 1)
n
4

tan (2n – 1)
n
4

= 1






 


n
4
2
tan = cot
n
4

tan n = tan
4

= 1 Ans. ]
Q.1059/06 Find the exact value of cosec10° + cosec50° – cosec70° [Ans : 6]
[Sol:




 70
sin
1
50
sin
1
10
sin
1




 20
cos
1
40
cos
1
80
cos
1











80
cos
.
40
cos
.
20
cos
80
cos
40
cos
20
cos
80
cos
20
cos
40
cos
= 8 [cos20°(cos40° + cos80°) – cos40° cos80° ]
= 8 [2cos20° cos60° cos20° – cos40° cos80°]
= 4 [2cos220° – 2cos40° cos80°]
= 4 [1 + cos40° – (cos120° + cos40°)]
= 4 .
2
3
= 6 Ans ]
Q.1153/06 If sin x + cos x + tanx + cot x + secx + cosecx = 7 then sin2x = a  b 7 where a, b  N. Find
the ordered pair (a, b). [Ans. a = 22 , b = 8]
[Sol. S  sin x ; C  cos x
S + C +
C
S
+
S
C
+
C
1
+
S
1
= 7 
SC
C
S
C
S
S
C
C
S 2
2
2
2





= 7
SC (S + C) + 1 + S + C = 7 S C  (S + C) (1 + SC) = 7SC – 1
(S + C)2 (1 + SC)2 = (7 SC – 1)2
(1 + 2 SC) (1 + SC)2 = (7 SC – 1)2
put sin x · cos x = y
(1 + 2y) (1 + y)2 = (7y – 1)2  (1 + 2y) (1 + y2 + 2y) = 49y2 – 14y + 1
1 + y2 + 2y + 2y + 2y3 + 4y2 = 49y2 – 14y + 1  2y3 – 44y2 + 18y = 0
2y [y2 – 22y + 9] = 0  2y = 36
484
22 
 (as y  0)
y = 11 ± 9
121 = 11 ± 7
4
sin x · cos x = 11 ± 7
4
2sin x · cos x = 22 – 7
8 + ve is rejected as sin 2x can have max. value as 1 ]