1. REVIEW TEST- 10 / 7
PAPER CODE : A
P A P E R - 1
Class : XII & XIII
Time : 3 hour Max. Marks : 240
INSTRUCTIONS
1. The question paper contains 60 questions and 10 pages. Each question carry 4 marks and all ofthem are
compulsory. There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer.
Please ensure that the Question Paper you have received contains all the QUESTIONS and
Pages. If you found some mistake like missing questions or pages then contact immediately to the
Invigilator.
2. Indicate the correct answer(s) for eachquestionbyfilling appropriate bubble(s) inyour OMR sheet.
3. Use onlyHB pencilfor darkening the bubble(s).
4. Use ofCalculator, Log Table, Slide Rule and Mobile is not allowed.
5. For example ifonly'B' choice is correct then, thecorrect method for filling the bubble is
A B C D
For example if only'B & D' choices are correct then, the correct method for filling the bubbles is
A B C D
The answer ofthe question in anyother manner (such as putting , cross , or partialshading etc.)
willbe treated as wrong.
USEFUL DATA
Atomic Mass:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Ba = 137,
Co = 59, Hg = 200, Pb = 207, He = 4, F=19.
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1.

2. XII & XIII MATHEMATICS REVIEW TEST-10 / 7
Select the correct alternative(s). (One or More than one is/are correct) [15 × 5 = 75]
There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer.
Q.41 PointAliesonthe line y= 2xand thesumofits abscissaand ordinate is 12. Point B lies onthe x-axis and
the lineAB is perpendicular to the line y= 2x. Let 'O' be the origin. The area ofthe triangleAOB is
(A) 20 (B) 40 (C) 60 (D*) 80
[Sol. Let A (x1, y1)
x1 + y1 = 12
y1 = 12 – x1
A lies on y= 2x
y1 = 2x1
12 – x1 = 2x1
x1 = 4. Hence A  (4, 8)
equation ofAB is x + 2y =  ....(1)
(1) passes through (4, 8)
  = 20
x + 2y = 20  B is (20, 0)
Area of AOB =
2
8
·
20
= 80 sq. units.]
Q.42 Suppose f (x) = sin x and g (x) = 1 – x . Whichofthe following composite functions have the same
range?
(A)fog (B*) gof (C) fof (D*) gog
[Sol. (A) f [g(x)] = f (1 – x ) = sin(1 – x )
domain is x  0 ; range [–1, 1]
(B) g[f (x)] = 1 – )
x
(
f = 1 – x
sin
domain 2k  x  2k + ; range [0, 1]
(C) (fof)(x) = f[f(x)] = f(sin x) = sin(sin x)
Domain x  R; range [– sin 1, sin 1]
(D) (gog)(x) = g [g (x)] = 1 – )
x
(
g = 1 – x
1
domain is 0  x  1; range is [0, 1]
hence (B) and (D) have the same range ]
Q.43 Giventhe vectors
k̂
j
ˆ
î
2
u 



k̂
2
ĵ
î
v 



k̂
î
w 


Ifthe volume of the parallelopiped having – u
c

, v

and w
c

as concurrent edges, is 8 then 'c' can be
equalto
(A*) 2 (B*) – 2 (C) 4 (D) cannot be determined
[Sol. V = – c2 ]
w
v
u
[



= – c2
1
0
1
2
1
1
1
1
2




= – c2[2(1 – 0) – 1(1) + (–2 – 1)] = – c2[2 – 1 – 3] = 8
 2c2 = 8  c = 2 or – 2 Ans. ]

3. Q.44 Let A = the number ofways ofselecting a committee of5 froma group of9 persons.
B = the number ofpermutations ofthe word PRELEPCO takenallat a time.
C =the number ofwaysinwhich8 peoplecanbe arranged ina line iftwo particular people must
stand next to each other.
The value of
C
AB
equals
(A) 63 (B*) 126 (C) 360 (D) none
[Hint: A = 9C5; B = !
2
·
!
2
!
8
; C = 7! · 2!]
Q.45 If 349(x + iy) =
100
3
2
3









2
i
and x = ky then the value of k equals (x, y  R)
(A) 3 (B) – 3 (C)
3
1
(D*) –
3
1
[Sol. Let r cos  =
2
3
and r sin  =
2
3
 r2 =
4
9
+
4
3
= 3  r = 3
also tan  =
2
3
·
3
2
=
3
1
;   =
6

 349(x + iy) = 












 


100
6
sin
i
6
cos
3 = 350 




 


3
50
sin
i
3
50
cos
equating realand imaginaryparts
349 x = 350 cos
3
50
and 349y = 350 sin
3
50

y
x
= cot
3
50
= cot
3
2
 x = –
3
1
y ; hence K = –
3
1
Ans. ]
Q.46 Let x, ybe reals satisfying sin x+ cos y=1 and siny+ cos x= – 1. Then which ofthe following must
be correct?
(A*) sin(x + y) = 0 (B) cos(x – y) = 0 (C) cos(x + y) = 1 (D*) cos 2x = cos 2y
[Sol. Given sin x + cos y = 1 ....(1) and sin y + cos x = – 1 ....(2)
squaring equation(1), sin2x + cos2y + 2sin x · cos y = 1 ....(3)
squaring equation(2), cos2x + sin2y + 2 sin y · cos x = 1 ....(4)
adding equation (3) and (4), we get
2 + 2sin(x + y) = 2
 sin(x + y) = 0  (A) is correct
hence cos (x + y) = 1 or – 1  (C) need not be correct
hence x + y = k
 y = k – x  2y = 2k – 2x  cos 2y = cos(2k – 2x)
 cos 2y = cos 2x  (D) is correct ]

4. Q.47 Two pointsA(x1, y1) and B(x2, y2) are chosen on the graphof f (x) = ln x with0 < x1 < x2. The points
C and Dtrisect line segmentABwithAC < CB.ThroughC a horizontallineis drawn to cut the curveat
E(x3, y3). If x1 = 1 and x2 = 1000 then the value of x3 equals
(A*) 10 (B) 10 (C) (10)2/3 (D) (10)1/3
[Sol. Usingsectionformula
a =
3
1000
·
1
1
·
2 
= 334
b =
3
1000
n
·
1
0
·
2 l

 b =
3
1000
n
l
....(1)
now line y= b intersects the curve y= lnx
 b = ln x ....(2)
from(1) and (2)
3
1000
n
l
= ln x  ln (1000)1/3 = ln x
 x = (1000)1/3 = 10 Ans. ]
Q.48199/4 Suppose f is a differentiable realfunctionsuch that f (x) + f '(x)  1 for all x, and f (0) = 0, then the
largest possible value of f (1), is
(A) e–2 (B) e–1 (C*) 1 – e–1 (D) 1 – e–2
[Sol. Given f ' (x) + f (x)  1
multiplying byex
f ' (x) ex + f (x) ex  ex
 
)
x
(
f
·
e
dx
d x
 ex
integrating between0 and 1
 

1
0
x
dx
)
x
(
f
e
dx
d
 
1
0
x
dx
e
1
0
x
)
x
(
f
e 
1
0
x
e
e · f (1) – e0 · f (0)  e – 1
f (1) 
e
1
e 
]
Q.49 Let S be a set consisting offirst five prime numbers. SupposeAand B are two matrices oforder 2 each
with distinct entries S. The chance that the matrixAB has atleast one odd entry, is
(A) 0 (B) 20% (C) 80% (D*) 96%
[Sol. S = {2, 3, 5, 7, 11}
Total ways in whichAand B can be chosen = (5C4 · 4!)2 = (5!)2
P(E) = 1 – P(A and B does not contain 2)
1 – 2
2
)
!
5
(
)
!
4
(
= 1 –
25
1
=
25
24
= 96% Ans. ]

5. Q.50 A curve is represented parametrically by the equations x = t + eat and y = – t + eat when t  R and
a > 0. Ifthe curve touches the axis ofx at the pointA, thenthe coordinates of the pointAare
(A) (1, 0) (B) (1/e, 0) (C) (e, 0) (D*) (2e, 0)
[Sol. x = t + eat ; y = – t + eat
dt
dx
= 1 + aeat ;
dt
dy
= – 1 + aeat;
dx
dy
= at
at
ae
1
ae
1



at the pointA, y = 0 and
dx
dy
= 0 for some t = t1
 1
at
ae = 1 ....(1) ;
also 0 = – t1 + 1
at
e ;  1
at
e = t1 .....(2), putting thisvalue in(1)
we get, at1 = 1  t1 =
a
1
; now from(1) ae = 1  a =
e
1
hence xA = t1 + 1
at
e = e + e = 2e  A  (2e, 0) Ans. ]
Q.51 Suppose g (x) = 2x + 1 and h (x) = 4x2 + 4x + 5 and h (x) = (fog)(x). The area enclosed bythe graph
ofthe functiony= f (x) andthe pair oftangents drawn to it fromthe origin, is
(A) 8/3 (B*) 16/3 (C) 32/3 (D) none
[Sol. Given g (x) = 2x + 1; h (x) = (2x + 1)2 + 4
now h (x) = f[ g (x) ]
(2x + 1)2 + 4 = f (2x + 1)
let 2x + 1 = t  f (t) = t2 + 4
 f (x) = x2 + 4 ....(1)
solving y = mx and y = x2 + 4
x2 – mx + 4 = 0
put D = 0
m2 = 16  m = ± 4
tangents are y = 4x and y = – 4x
A=  

2
0
2
dx
]
x
4
)
4
x
[(
2 =  
2
0
2
dx
)
2
x
[(
2 =
2
0
3
)
2
x
(
3
2



 =
3
16
sq. units Ans. ]
Q.52vector Suppose a vector V

satisfies V
)
V
·
V
(



= V
169

and 4 )
î
·
V
(

= 3 )
ĵ
·
V
(

= )
k̂
·
V
(

where )
î
·
V
(

>0
then )
î
·
V
(

+ )
ĵ
·
V
(

+ )
k̂
·
V
(

equals
(A) 16 (B*) 19 (C) 21 (D) 23
[Sol. Given, V
|
V
| 2


= 169 V

 |
V
|

= 13 [To be put in teaching notes]
let 4 )
î
·
V
(

= 3 )
ĵ
·
V
(

= )
k̂
·
V
(

=  (say); Also k̂
)
k̂
·
V
(
ĵ
)
j
ˆ
·
V
(
î
)
î
·
V
(
V







hence, )
î
·
V
(

=
4

; )
ĵ
·
V
(

=
3

; )
k̂
·
V
(

= 
 2
|
V
|

=
16
2

+
9
2

+ 2 = 169
144
)
144
16
9
(
2



= 169

6.   = 12 or – 12 (since î
·
V

> 0, hence rejected)
hence î
·
V

= 3; ĵ
·
V

= 4 and k̂
·
V

= 12
 k̂
)
k̂
·
V
(
ĵ
)
j
ˆ
·
V
(
î
)
î
·
V
(
V







)
k̂
12
j
ˆ
4
î
3
(
V 



 sum = 19 Ans. ]
Q.53 Suppose the function gn(x) = x2n + 1 + anx + bn (n  N) satisfies the equation 


1
1
n dx
)
x
(
g
)
q
px
( = 0
for alllinear functions (px+ q) then
(A) an = bn = 0 (B*) bn = 0; an = –
3
n
2
3

(C) an = 0; bn = –
3
n
2
3

(D) an =
3
n
2
3

; bn = –
3
n
2
3

[Sol. We have 





1
1
n
n
1
n
2
dx
)
b
x
a
x
)(
q
px
( = 0
equating the oddcomponent to be zero and integrating we get
3
n
2
p
2

+
3
p
a
2 n
+ 2bnq = 0 for all p, q
hence bn = 0 and an = –
3
n
2
3

]
Q.54 Suppose f: R  R is a differentiable function satisfying f (x + y) = f(x) + f(y) + xy(x + y) for every
x, y  R. If f ' (0) = 0 then which of the following hold(s) good?
(A*) fis an odd function
(B*) fis a bijective mapping
(C) fhas a minima but no maxima
(D*) fhas aninflectionpoint
[Sol. Given f (x + y) = f (x) + f (y) + xy(x + y) ....(1)
f ' (x) =
h
)
x
(
f
)
h
x
(
f
Lim
0
h



=
h
)
h
x
(
xh
)
h
(
f
Lim
0
h



[ using (1) ]
= 0
h
Lim

x(x + h) + 0
h
Lim
 h
)
h
(
f
= x2 + f ' (0) ( as f (0) = 0 )
 f ' (x) = x2
f (x) =
3
x3
+ C
but f (0) = 0  C = 0
Hence f(x) =
3
x3
 A, B, D ]

7. Q.55ellipse Consider the parabola y2 = 4x and the ellipse 2x2 + y2 = 6, intersecting at P and Q. Which of the
following hold(s) good?
(A*) The two curves are orthogonal.
(B) The area enclosed bythe parabola and the double ordinate, commonto ellipse and parabola is 4/3.
(C*) The commondouble ordinate is the latus rectumofthe parabola.
(D*) Iftangent and normalat the point P onthe ellipse intersect the x-axis at T and G respectivelythen
area ofthe triangle PTG is 4.
[Sol. C1 : 2x2 + y2 = 6 [T/S, Q.9, Ex-2, Ellipse as subjective]
and C2 : y2 = 4x
solving (1) and (2)
P = (1, 2) and Q = (1, – 2)
Hence commonchord passes focus  (C) is correct.
also A = 2 
1
0
dx
)
x
2
( =
1
0
2
3
x
3
2
·
4 


=
3
8
 (B)is not correct
Also
1
C
P
dx
dy



·
2
C
P
dx
dy



= – 1 (A) is correct
now tangent at P on C1
2(x · x1) + yy1 = 6
2x + 2y = 6  x + y = 3
normalis x – y = – l it passes through (1, 2) ]
Q.56 TriangleABC has BC = 1 andAC = 2. The maximumpossible value ofthe angleAis
(A*)
6

(B)
4

(C)
3

(D)
2

[Sol. Using cosinerule
cos  =
x
4
1
4
x2


=
x
4
3
x2

= 






x
3
x
4
1
=

















 3
2
x
3
x
4
1
2
hence cos  is minimumif x= 3
 minimumvalue of cos  = 2 ·
4
3
=
2
3
maximumvalue of =
6

Ans. ]
Q.57complex Let the equation of a line in the Cartesian plane is Ax + By + C = 0 where A, B, C  R and
A2 + B2  0. If the corresponding equation on the complex plane is 0
z
z 




 where  is
complex and  is real, then the correct statement(s) is/are
(A*)  =
2
B
A i

and  = C (B*) If 

 the line is parallelto imaginaryaxis.
(C*) If 

 = 0 line is parallelto realaxis. (D*) slope m of the line is i






.
[Sol. Let z = x + iy; y
x
z i

 [Illustration, Complex No.]
now Ax + By + C = 0
2
)
z
z
(
A 
+
i
2
)
z
z
(
B 
+ C = 0

8. 2
)
z
z
(
A 
–
2
)
z
z
(
B i

+ C = 0
C
z
2
B
A
z
B
A






 






  i
2
i
= 0
comparing this with 0
z
z 





we have  =
2
B
A i

and  = C  (A) is correct
If 

 then 




 






 
2
i
i B
A
2
B
A
 B = 0
hence line is x = –
A
C
 | | to y-axis  (B) is correct
If 

 = 0 then
A– Bi + A+ Bi = 0  A = 0
hence line is y = –
B
C
|| to x-axis.  (C) is correct
Again m = –
B
A
....(1)
we have A – Bi = 2
A + Bi = 
2
A = 


and – 2Bi = 2( –  )
B = –
i



m = –
B
A
= i






 (D) is correct
hence (A), (B), (C), (D) are correct.]
Q.58misc on def & inde Let f (x) = 

x
1
t
dt
e
2
and h (x) =  
)
x
(
g
1
f , where g (x) is defined for allx, g'(x) exists for
allx, and g (x)  0 for x > 0. If h'(1) = e and g'(1) = 1, then the possible values which g(1) can take
(A*) 0 (B) – 1 (C*) – 2 (D) – 4
[Sol. Given f (x) = 

x
1
t
dt
e
2
; h (x) =  
)
x
(
g
1
f ; g (x)  0 for x > 0
now h (x) = 


)
x
(
g
1
1
t
dt
e
2
differentiating h'(x) =
 
)
x
(
'
g
·
e
2
)
x
(
g
1
h'(1) = e (given)
  )
1
(
'
g
·
e
2
)
1
(
g
1
= e
  2
)
1
(
g
1 = 1
1 + g (1) = ± 1  g (1) = 0 or g(1) = – 2  (A), (C) ]

9. Q.59prob P and Q are two points on the upper halfofthe ellipse 2
2
2
2
b
y
a
x
 = 1. The centre ofthe ellipse is at the
origin 'O'and PQ is parallelto the x-axis suchthat the triangle OPQ has the maximumpossible area.A
point is randomlyselected frominside ofthe upper halfoftheellipse. The probabilitythat it lies outside
thetriangle, is
(A*)


 1
(B)



2
1
2
(C)



2
1
(D)



4
1
[Sol. A =
2
sin
b
cos
a
2 

= ab sin  cos  =
2
2
sin
ab 
Hence Ais maximum if  = /4
and Amax =
2
ab
; Also sample space
2
ab

 Favourable and come =
2
ab

–
2
ab
=
2
ab
( – 1)
Probability =
2
ab
·
ab
2
)
1
(



=


 1
Ans. ]
Q.60 Asquare OABC is formed by line pairs xy= 0 and xy+ 1 = x + y where 'O' is the origin.Acircle with
centre C1 inside the square is drawn to touchthe line pair xy= 0 and another circle with centre C2 and
radius twice that of C1, is drawn to touch the circle C1 and the other line pair. The radius ofthe circle
with centre C1 is
(A)
 
1
2
3
2

(B)
 
1
2
3
2
2

(C*)
 
1
2
3
2

(D)
2
3
1
2 
[Sol. diagonalofthe square = 2
also d = 2
r + 3r + r
2
2
2 = r
2
3 + 3r
r =
 
1
2
3
2

Ans. ]