1. REVIEW TEST- 8
PAPER CODE : A
P A P E R - 1
Class : XII (ABCD)
Time : 3 hour Max. Marks : 225
INSTRUCTIONS
1. The question paper contains 75 questions and 20 pages. Each question carry 3 marks and all of them are
compulsory.There is NEGATIVE marking. 1 mark will be 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 for each question byfilling appropriate bubble in your OMR sheet.
3. Use onlyHB pencil fordarkeningthe bubble.
4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed.
5. The answers of the questions must be marked byshading the circles against the question by dark HB
pencilonly.
For exampleifonly'B'choice is correct then, thecorrect method for fillingthe bubble is
A B C D
thewrongmethodfor fillingthebubbleare
(i) A B C D
(ii) A B C D
(iii) A B C D
The answer ofthe questions in anyother manner will betreated as wrong.
USEFUL DATA
Atomic weights: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,
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1.

2. Select the correct alternative. (Only one is correct) [75 × 3 = 225]
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
Q.51cont&deri )
x
tan
arc
(
x
)
x
2
sin
1
(
n
·
)
1
2
(
Lim
x
sin
0
x



l
equals
(A*) ln 4 (B) ln 2 (C) (ln 2)2 (D) 1
[Sol.
)
x
(tan
x
)
x
2
sin
1
(
n
·
x
sin
x
sin
)
1
2
(
Lim 1
x
sin
0
x 


 l
= x
1
0
x
1
0
x
)
x
2
sin
1
(
n
Lim
·
x
tan
x
sin
Lim
)
2
n
( 



l
l
=
x
)
1
x
2
sin
1
(
Lim
)
1
)(
2
n
(
0
x



l ] = 2 ln 2 = ln 4 Ans. ]
Q.52ph-3 In an acute triangleABC, ABC = 45°,AB = 3 andAC = 6 . The angle BAC, is
(A) 60° (B) 65° (C*) 75° (D) 15° or 75°
[Sol.
3
)
135
sin( 


=
6
45
sin 
sin(135° – ) =
2
3
135° –  = 60° or 120°
 = 75° or 15°
 = 15° (not possible as the  is acute)
  = 75° Ans.]
Q.53mat If A =






1
1
1
1
and det. (An – I) = 1 – n, n  N then the value of , is
(A) 1 (B*) 2 (C) 3 (D) 4
[Sol. An – I = 











1
2
2
2
1
2
1
n
1
n
1
n
1
n
hence | An – I | = (2n – 1 – 1)2 – (2n – 1)2 = (2n – 1 – 1 – 2n – 1) (2n – 1 – 1 + 2n – 1) = 1 – 2n
Hence  = 2 Ans. ]
Q.54cont&deri Thefigureshows arighttrianglewithits hypotenuseOBalongthe
y-axis and its vertex A on the parabola y = x2. Let h represents the
length of the hypotenuse which depends on the x-coordinate of the
point A.The value of )
h
(
Lim
0
x
equals
(A) 0 (B) 1/2 (C*) 1 (D) 2
[Sol. Let A = (t, t2); mOA = t; mAB = –
t
1
equationofAB, y – t2 = –
t
1
(x – t2)
put x = 0
h = t2 + 1 (as x  0 then t  0)
now )
h
(
Lim
0
t
= )
t
1
(
Lim 2
0
t


= 1 Ans. ]

3. Q.55p&c Number ofways in which 7green bottles and 8blue bottles can be arrangedin a row if exactly1 pair
of green bottles is side byside, is (Assume all bottles to be alike exceptfor the colour).
(A) 84 (B) 360 (C*) 504 (D) none
[Sol. G G G G G G G / B B B B B B B B
one gap out of nine can be taken in 9C1 ways
now greenremaining5
gapsremaining8
5 gaps for remaining 5 green can be selected in 8C5
Hence Total ways 9C1 · 8C5 = 9 · 3
·
2
·
1
6
·
7
·
8
= 72 · 7 = 504 Ans.
Alternatively: 9C6 × 6 = 504 Ans. (think ! how)]
Q.56prop of def The integral, 



4
5
4
dt
)
t
cos
|
t
sin
|
t
sin
|
t
cos
|
( has the value equal to
(A*) 0 (B) 1/2 (C) 1/ 2 (D) 1
[Sol. I = 


2
/
4
/
dt
t
cos
t
sin
2 + 




2
/ zero
dt
)
t
cos
t
(sin
)
t
cos
t
sin
(

 

 
 + 



4
5
dt
t
cos
t
sin
2
= 


2
/
4
/
t
2
sin dt – dt
t
2
sin
4
5



thesetwointegrals cancels  Zero ]
Only for 12th
Q.57st.line Consider 3 non collinear points A, B, C with coordinates (0, 6), (5, 5) and (–1, 1) respectively.
EquationofalinetangenttothecirclecircumscribingthetriangleABCandpassingthroughtheoriginis
(A) 2x – 3y = 0 (B) 3x + 2y = 0
(C) 3x – 2y = 0 (D*) 2x + 3y = 0
[Sol. Note that theABC is right angled atA
 equationofcircle
(x + 1)(x – 5) + (y – 1)(y – 5) = 0
x2 + y2 – 4x – 6y – 5 + 5 = 0
x2 + y2 – 4x – 6y = 0
Hencethecirclepasses throughthe origin
 tangent at (0, 0) is 4x + 6y = 0
2x + 3y = 0 Ans. ]
Only for 13th
Q.57def If f (x) is continuous and 
9
0
dx
)
x
(
f = 4 then the value of the integral 
3
0
2
dx
)
x
(
·
x f is
(A) 9 (B) 8 (C) 4 (D*) 2
[Sol. Let I = 
3
0
2
dx
)
x
(
·
x f ; put x2 = t  2x · dx = dt
= 
9
0
dt
)
t
(
2
1
f = 4
·
2
1
= 2 Ans. ]

4. Direction for Q.58 and Q.59 (2 questions together)
SquareABCD has the verticesA(1,4), B(5, 4), C(5, 8) and D(1, 8). From a point Poutside the square,
a vertex of the square is said to be visible if it can beconnected to P bya straight line that does not pass
through thesquare.
Thus, from anypoint P outside, either two or threeof the vertices ofthe square are visible. The visible
areaofP is theareaof onetriangleorthesum oftheareas ofthetwotriangles formed byjoiningP to the
two orthree visible vertices ofthe square.
Q.58st.line Visible area of the point Pwith coordinates (2, –6) in square units is
(A) 10 (B*) 20 (C) 30 (D) 40
[Sol. The visible area of P(2, –6) is the area of ABP.
ABPhas baseAB of length 4, and its height is the distance
fromABto the point P, whichis 10, sinceAB is parallel to the
x-axis.
Thus, the area of ABP is bh
2
1
= )
10
)(
4
(
2
1
= 20 sq. units,
i.e. the visibleareaof P is 20 square units.]
Q.59st.line Visible area of the point Q(11, 0) in square units is
(A*) 20 (B) 30 (C) 40 (D) none
[Sol. The visible area of Q(11, 0) is the sum of the areas of
QBC and QBA
QBC has base BC of length 4, and its height is the
distancefromQtothelinethroughBandC,which is6.
Thus, the area of QBC is bh
2
1
= )
4
)(
4
(
2
1
= 8 sq. units
so the visible area of Q is the sum of these two areas, or 20 sq. units.]
Q.60max&min If the function f (x) = ax e–bx has a local maximum at the point (2, 10) then
(A) a = 5; b = 0 (B*) a = 5e, b = 1/2 (C) a = 5e2, b = 1 (D) none
[Sol. f (2) = 10, hence 2ae–2b = 10  ae–2b = 5 ....(1)
f ' (x) = a [e–bx – bx e–bx] = 0
f ' (2) = 0
a(e–2b – 2be–2b) = 0
ae–2b (1 – 2b) = 0  b = 1/2
from (1) if b = 1/2; a = 5e or a = 0 (rejected)
 a = 5e and b = 1/2 Ans. ]
Q.61mat Ifan idempotent matrix is also skew symmetric then it must be
(A)aninvolutarymatrix (B)anidentitymatrix
(C)anorthogonalmatrix (D*)anullmatrix.
[Sol. A2 = A
(A2)T = AT  (AT)2 = AT  (– A)2 = – A
A2 = – A  A = – A  A = 0 ]

5. Q.62tang&Nor If the curves y = x3 + ax and y = bx2 + c pass through the point (–1, 0) and have a common
tangent line at this point then the value of (a+ b + c2)is
(A) 0 (B) 1 (C) – 3 (D*) – 1
[Hint: f (–1) = 0; g (–1) = 0; f ' (–1) = g ' (–1)
 a = – 1; b = – 1; c = 1; hence (a + b + c2) = – 1 Ans. ]
Q.63limit of sum  



















n
1
k
1
n n
k
tan
1
1
n
1
tan
Lim has the value equal to
(A)
2
)
1
(cos
n
1 l

(B)
2
)
1
(sin
n
1 l

(C)
2
)
1
cos
1
(sin
n
1 
l
(D*)
2
)
1
cos
1
(sin
n
1 
 l
[Sol.
 






 





 n
1
k
n
1
n n
k
tan
1
1
n
1
Lim
·
n
1
tan
n
Lim =  
1
0
x
tan
1
dx
=
2
)
1
cos
1
(sin
n
1 
 l
Ans. ]
Q.64leibintz rule Let f (x) be a continuous function suchthat f (x) > 0 for all x  0 and  101
)
x
(
f = 1 + 
x
0
dt
)
t
(
f .
The value of  100
)
101
(
f is
(A) 100 (B*) 101 (C) 5050 (D) 100
1
)
101
(
[Sol. Given  101
)
x
(
f = 1 + 
x
0
dt
)
t
(
f
differentiating
101 ·  100
)
x
(
f · f '(x) = f (x)
 101 ·  99
)
x
(
f · f '(x) = 1 (as f (x) > 0)
integrating
 
100
)
x
(
)
101
( 100
f
= x + C
but f (0) = 1  C =
100
101

100
101
 100
)
x
(
f = x +
100
101
put x = 101
100
101
 100
)
101
(
f = 101 +
100
101
=
100
)
101
)(
101
(
  100
)
101
(
f = 101 Ans. ]

6. Direction for Q.65 and Q.66 (2 questions together)
An equilateral triangleABC has its centroid at the origin and the base BC lies alongthe line x + y= 1
Q.65st.line Area of the equilateral ABC is
(A*)
2
3
3
(B)
4
3
3
(C)
2
2
3
(D)
4
3
2
Q.66st.line Gradient of the other two lines are
(A) 3 , 2 (B)
3
1
,
3 (C) 1
2
,
1
2 
 (D*) 3
2
,
3
2 

[Sol. GD =
2
1
; AG = 2
parametric through(0, 0)

cos
x
=

sin
y
= – 2 (think !)
cos  = –
2
1
; sin  = –
2
1
x = – 1; y = – 1
hence coordinates of D are (– 1/2, – 1/2)
AD = p =
2
3
;
Area of equilateral triangle =
3
p2
=
3
1
·
2
9
=
2
3
3
Ans.
equationoflinethroughA, y + 1 = m(x + 1) ....(1)
(1) makes an angle of 60° with BC
m
1
1
m


= 3  m = 2 – 3 or 3
2  Ans. ]
Q.67def  





2
0
n
n
dx
x
sin
1
n
Lim equals
(A)  ln 2 (B*)
2

ln 2 (C) –
2

ln 2 (D) none
[Sol. L =  





2
0
n
n
dx
x
sin
1
n
Lim ; put n =
t
1
= 

 






 

2
0
t
0
t
dx
t
)
x
(sin
1
Lim ; L = 

 






 
2
0
)
x
(sin
n
t
0
t
dx
)
x
(sin
n
·
)
x
(sin
n
t
e
1
Lim l
l
l
but
)
x
(sin
n
t
e
1
Lim
)
x
(sin
n
t
0
t l
l


= – 1;  L = – 
 2
0
dx
)
x
(sin
n
l =
2

ln 2 Ans. ]

7. Q.68complex If x =
2
3
1 i

then the value of the expression, y = x4 – x2 + 6x – 4, equals
(A*) – 1 + i
3
2 (B) 2 – i
3
2 (C) 2 + i
3
2 (D) none
[Sol. x =
2
3
1 i

= – 2
 y = 8 – 4 – 62 – 4
= 2 –  – 62 – 4
= 52 –  – 4
= 3
4
1 2
zero
2







  
 

= + 4 






 
2
3
1 i
– 3 = 2(1 – 3
i ) – 3 = – 1 + i
3
2 Ans. ]
Only for 12th
Q.69prob When a missile is firedfrom a ship, theprobabilitythat it is interceptedis 1/3. Theprobabilitythat the
missile hitsthetarget, giventhatitis notinterceptedis3/4.Ifthreemissilesarefiredindependentlyfrom
theship,theprobabilitythat all three hitstheirtargets, is
(A) 1/12 (B*) 1/8 (C) 3/8 (D) 3/4
[Sol. R: Missileis intercepted
P(R) =
3
1
; P )
R
( =
3
2
H:Missilehitsthetarget
P(H) = P(H  R) + P(H  R ) = P(R) · P(H/R) + P( R ) · P(H/R)
=
3
1
· (0) +
3
2
·
4
3
=
2
1
Hence P(H H H) =
2
1
·
2
1
·
2
1
=
8
1
Ans. ]
Only for 13th
Q.69ellipse For each point (x, y) on the ellipse with centre at the origin and principal axes along the coordinate
axes,thesum ofthedistancesfromthepoint(x,y) tothepoints(± 2, 0)is8. The positivevalueofx such
that (x,3)lies on theellipse, is
(A)
3
3
(B*) 2 (C) 4 (D) 3
2
[Sol. c = 2; a = 4
equationoftheellipseis ( 1
c
a
y
a
x
2
2
2
2
2


 )
4
16
y
16
x 2
2

 = 1 
12
y
16
x 2
2
 = 1
when y = 3 then
12
9
16
x2
 = 1 
16
x2
=
4
1
 x = 2 Ans. ]

8. Only for 12th
Q.70 The contents of three Urns w.r.t. the Red, White and Green balls is as shown in the table given.
2
1
3
III
1
2
3
II
1
3
2
I
G
W
R
Urn
A coin when tossed is twice as likely to come heads as compared to tails. Such a coin is tossed two
times. If bothheads and tails arepresent then 3 balls are drawn simultaneouslyfrom theUrn-I, if head
appears on both the occassions then 3 balls are drawn in a similar manner from Urn-IIand if no head
appears in both the tosses then 3 balls from the Urn-IIIare drawn in the same manner.
The probabilitythat 3 drawn balls are 1 each ofdifferent colours, is
(A) 10% (B) 15% (C*) 30% (D) 90%
[Hint: P(H) =
3
2
; P(T) =
3
1
; P(H T or T H) =
9
4
; P(H H) =
9
4
; P(T T) =
9
1
3
6
1
1
1
3
1
2
C
C
·
C
·
C








9
1
9
4
9
4
=
20
6
=
10
3
= 30% Ans.]
Only for 13th
Q.70 Let f (x) = Min. 








4
x
5
,
4
x
3
2
1 2
2
for 0  x  1. The maximum value of f (x) is
(A) 0 (B) –
4
1
(C*)
16
5
(D) none
[Sol. We have
2
1
–
4
x
3 2
=
4
x
5 2
 2 – 3x2 = 5x2  8x2 = 2  x =
2
1
or –
2
1
 f is maximum when x = 1/2
fmax =
4
1
·
4
5
=
16
5
Ans. ]
Q.71aucLet 'a' be a positive constant number. Consider two curves C1: y = ex, C2 : y =ea – x. Let S be the area
of the part surrounding byC1, C2 and the y-axis, then
2
0
a a
S
Lim

equals
(A) 4 (B) 1/2 (C) 0 (D*) 1/4
[Sol. Solving ex = ea – x, we get
e2x = ea  x =
2
a
S =  

2
a
0
x
x
a
dx
)
e
e
·
e
( =   2
a
0
x
x
a
)
e
e
·
e
( 
 
= (ea + 1) – (ea/2 + ea/2) = ea – 2ea/2 + 1 = (ea/2 – 1)2

9.  2
a
S
=
2
2
a
a
1
e







 
=
2
2
a
2
a
1
e
4
1







 

2
0
a a
S
Lim

=
4
1
Ans. ]
Q.72max&min Let f (x) be a continuous function with continuous first derivative on (a, b), where b > a, and let
)
x
(
Lim
a
x
f


  ; )
x
(
Lim
b
x
f


 –  and f '(x)+ f 2 (x)  –1, for all x in (a, b) then minimum value of
(b – a) equals
(A*)  (B)
2

(C) 2 (D) 1
[Sol. f ' (x) + f 2(x)  – 1
 f 2(x) + 1  – f ' (x) in (a, b)
1  –
)
x
(
f
1
)
x
(
'
f
2

in (a, b)
 
b
a
dx  –  
b
a
2
dx
)
x
(
f
1
)
x
(
'
f
b – a  –  
 b
a
1
)
x
(
f
tan
= – 




 






 

2
2
 (b – a)   Ans. ]
Q.73 Considera pyramidP-ABCDwhose baseABCDis asquareandwhose vertex Pis equidistant fromA,
B, C and D. IfAB = 1 and APD = 2, then the volume of the pyramid is
(A)
6
sin 
(B)
6
cot
(C)

sin
6
1
(D*)


sin
6
2
cos
[Sol. x =

sin
2
1
....(1)
V =
3
1
(12) · h =
3
h
....(2)
now h2 = x2 –
2
2
1






 h2 =

2
sin
4
1
–
2
1
h2 = 








2
2
sin
sin
2
1
4
1
h =


sin
2
2
cos
; hence V =


sin
2
2
cos
3
1
=


sin
6
2
cos
Ans. ]

10. Q.74prop of def The value of the definite integral  



n
2
0
1
dx
)
x
(sin
sin
,
x
sin
.
max equals (where n I)
(A)
2
)
4
(
n 2


(B)
4
)
4
(
n 2


(C*)
4
)
8
(
n 2


(D)
4
)
2
(
n 2


[Sol. I =  



n
2
0
1
dx
)
x
(sin
sin
,
x
sin
.
max [to be put in Teaching Notes]
=











 






 2
2
2
0
dx
)
x
(sin
dx
)
x
(
dx
x
n = n
















 






2
4
2
1
2
8
2
2
2
2
= n 











2
8
3
2
8
2
2
2
=
4
)
8
(
n 2


Ans. ]
Q.75inde
 
 dx
x
sin
·
)
x
101
sin( 99
equals
(A*)
100
)
x
)(sin
x
100
sin( 100
+ C (B)
100
)
x
)(sin
x
100
cos( 100
+ C
(C)
100
)
x
)(cos
x
100
cos( 100
+ C (D)
101
)
x
)(sin
x
100
sin( 101
+ C
[Sol. I =  
  dx
x
(sin
·
)
x
x
100
sin( 99
=  
  dx
)
x
(sin
)
x
sin
·
x
100
cos
x
cos
)
x
100
(sin( 99
=  dx
)
x
(sin
·
x
cos
)
x
100
sin( 99
II
I

 

 


 
 +  dx
)
x
(sin
)·
x
100
cos( 100
=
100
)
x
)(sin
x
100
sin( 100
–  dx
)
x
)(sin
x
100
cos(
100
100 100
+  dx
)
x
)(sin
x
100
cos( 100
=
100
)
x
)(sin
x
100
sin( 100
+ C Ans. ]