1. PART-A
Q.1 Three straight lines l1, l2 and l3 have slopes 1/2, 1/3 and 1/4 respectively.All three lines have the same
y-intercept. If the sum of the x-intercept of three lines is 36 then find the y-intercept. [5]
[Ans. – 4]
[Sol. l1 : y =
2
1
x + c  x-intercept is – 2c
l2 : y =
3
1
x + c  x-intercept is – 3c
l3 : y =
4
1
x + c  x-intercept is – 4c
 – 2c – 3c – 4c = 36  – 9c = 36  c = – 4 Ans. ]
Q.2 Findthegeneralsolutionoftheequation














...
x
sin
)
1
..(
x
sin
x
sin
x
sin
1
....
x
sin
......
x
sin
x
sin
x
sin
1
n
n
3
2
n
3
2
=
x
tan
1
4
2

where x  k +
2

, k  I. [Ans. n + (–1)n
6

, n  I] [5]
[Sol. Nr of LHS =
x
sin
1
1

; Dr of LHS =
x
sin
1
1

hence
x
sin
1
x
sin
1


=
x
sec
4
2 = 4 cos2x = 4(1 – sin x)(1 + sin x)
hence 4(1 – sin x)2 = 1
(1 – sin x)2 =
4
1
 (1 – sin x) =
2
1
or –
2
1
 sin x =
2
1
or sin x =
2
3
(rejected)
 sinx= sin
6

 x = n + (–1)n
6

, n  I Ans. ]
Class: XI(PQRS & J) Time: 90 Minutes Max. Marks:90
M
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Instruction
1. The question paper contains 15 question. All questions are compulsory.
2. Onlyanswers are to be written in the same order in which theyappear in the question paper.
3. Each subjectivequestionshouldbegin aftertheend oftheprevious question afterdrawingaline.
4. Sub part in respect of a subjective question should be doneat the same place(if applicable).
5. Use of Calculator,Logtable and Mobileis not permitted.
6. Legibilityandclarityinansweringthequestionwillbeappreciated.
7. Put a cross ( × ) on the rough work done by you.
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2. Q.3 In a triangleABC if 2 cos
2
B
cos
2
C
=
2
1
+
2
A
sin
a
c
b





 
then find the measure of angleA.
A. [5]
[Sol. Given 2 cos
2
B
cos
2
C
=
2
1
+
2
A
sin
a
c
b





 
or cos 




 
2
C
B
+ cos 




 
2
C
B
=
2
1
+
2
A
sin
A
sin
C
sin
B
sin





 
sin
2
A
+ 




 
2
C
B
cos =
2
1
+
2
A
cos
2
A
sin
2
2
A
sin
2
C
B
cos
2
C
B
sin
2 




 





 
sin
2
A
+ 




 
2
C
B
cos =
2
1
+
2
A
cos
2
C
B
cos
2
A
cos 




 
sin
2
A
+ 




 
2
C
B
cos =
2
1
+ 




 
2
C
B
cos  sin
2
A
=
2
1
 A/2 = 30°  A = 60° Ans. ]
Q.432/1 Let p & q be the two roots of the equation, mx2 + x(2  m) + 3 = 0. Let m1, m2 be the two values of m
satisfying
p
q
q
p
 =
2
3
.Determinethenumericalvalueof
m
m
m
m
1
2
2
2
1
2
 . [Ans. 99 ] [5]
[Sol. mx2 + (2 – m) x + 3 = 0
p + q =
m
2
m 
; pq =
m
3
now m1 and m2 satisfies
3
2
p
q
q
p

 
3
2
pq
q
p 2
2


3
2
pq
pq
2
)
q
p
( 2



m
2
m
3
.
3
2
m
6
m
2
m
2








 

m
8
m
)
2
m
(
2
2


m2 – 4m + 4 = 8m  m2 – 12m + 4 = 0
 m1 + m2 = 12 and m1 m2 = 4
now 2
1
2
2
2
1
m
m
m
m
 = 2
2
1
3
2
3
1
)
m
m
(
m
m 
= 2
2
1
2
1
2
1
3
2
1
)
m
m
(
)
m
m
(
m
m
3
)
m
m
( 


=
16
12
.
12
123

=
16
11
.
122
= 99 Ans. ]
Decide yourself
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3. Q.5 If Cr denotes the combinatorial co-efficient in the expansion of (1 + x)n, n  N then using algebraic
approach prove that
C0 +
2
C1
+
3
C2
+ ........ +
1
n
Cn

=
1
n
1
2 1
n



. [5]
[Sol. Let S = C0 +
2
C1
+
3
C2
+ ........ +
1
n
Cn

generalterm Tr =
1
r
Cr
n

Tr = )!
1
r
(
)!
r
n
(
!
r
!
n

 = ]
)!
1
r
(
)!
r
n
)[(
1
n
(
)
1
n
(
!
n




 Tr =
1
n
1










)!
1
r
(
)!
r
n
(
)!
1
n
(
Tr =
1
n
C 1
r
1
n



 S =
1
n
1





n
0
r
1
r
1
n
C
S =
 
1
n
C
C
.....
C
C
C 0
1
n
1
n
1
n
2
1
n
1
1
n
0
1
n




 





=
1
n
1
2 1
n



Hence proved.]
Q.6 Let 'A' denotes the real part of the complex number z =
i
9
i
7
19


+
i
6
7
i
5
20


and 'B' denotes thesum of the imaginaryparts of the roots ofthe equation
z2 – 8(1 – i)z + 63 – 16i = 0
and 'C' denotes the sum of the series, 1 + i + i2 + i3 + ..... + i2008 where i = 1
 .
and 'D' denotes the value of the product (1 + )(1 + 2)(1 + 4)(1 + 8) where is the imaginary
cube root of unity.
Find the valueof
D
C
B
A


. [Ans. 6] [5]
[Sol. A = Re (z)
now z =
82
)
i
9
)(
i
7
19
( 

+
85
)
i
6
7
)(
i
5
20
( 

=
82
7
i
82
171 

+
85
30
i
35
i
120
140 


=
82
i
82
164 
+
85
i
85
170 
= 2 + i + 2 – i
z = 4 + 0i  4
A 
Let  = x + iy
 = a + ib
 +  = (x + a) + i(y + b)= 8 – 8i
 y + b = – 8
 sum of the imaginaryparts of the roots of the equation = – 8
 8
B 
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4. 'C' S = 1 + i + i2 + i3 + ........ + i2008 =
i
1
)
i
1
( 2009


=
i
1
i
1


= 1  1
C 
1
D  obvious
hence
D
C
B
A


=
2
8
4 
= 6 Ans. ]
PART-B
Q.7
(a) If  and  are the roots of the equation x2 + 5x – 49 = 0 then find the value of cot(cot–1 + cot–1).
(b)178/6 Prove that the sum to 'n' terms of the series,
tan1
1
3
+ tan1
1
7
+ tan1
1
13
+ tan1
1
21
+...... = cot–1 




 
n
2
n
[3+3]
[Sol.(a)  +  = – 5;  = – 49
let y = cot(cot–1 + cot–1) =
)
cot(cot
)
cot(cot
1
)
)·cot(cot
cot(cot
1
1
1
1










=




 1
=
5
50


= 10 Ans.
[Sol.(b) S = tan 1 1
1 1 2
 .
+ tan 1 1
1 2 3
 .
+ tan 1 1
1 3 4
 .
+.......
Now , Tn = tan 1 1
1 1
 






n n
( )
= tan 1 ( )
( )
n n
n n
 
 






1
1 1
= tan 1 (n + 1)  tan 1 n
T1 = tan12  tan11
T2 = tan–13 – tan–12
M
Tn = tan–1(n + 1) – tan–1n
——————————
Sn = tan1 (n + 1) 
4

Sn = tan–1
)
1
n
(
1
1
)
1
n
(




= tan–1 





 2
n
n
= cot–1 




 
n
2
n
Hence proved. ]
Q.8 In acute angledtriangleABC, a semicircle with radius ra is constructedwith its base onBC andtangent
totheothertwosides. rb andrc aredefinedsimilarly. Ifris theradius oftheincircleoftriangleABCthen
prove that
r
2
=
c
b
a r
1
r
1
r
1

 [6]
[Sol.
2
b
r
2
c
r a
a



=  (where  is the area of triangleABC) [T/S, ph-3]
ra(b + c) = 2
|||ly rb(c + a) = 2
rc(a + b) = 2
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5. adding
c
b
a
r
2
r
2
r
2 




= 2(a + b + c)
c
b
a
r
1
r
1
r
1

 =


 c
b
a
=

s
2
=
r
2
i.e.
r
2
=
c
b
a r
1
r
1
r
1

 Hence proved ]
Q.9 Findthe equationof the circlepassingthrough thepoint (–6, 0) ifthepowerof the point (1, 1)w.r.t. the
circle is 5 and it cuts the circle x2 + y2 – 4x – 6y – 3 = 0 orthogonally. [6]
[Ans. x2 + y2 + 6x – 3y = 0] [T/S, Q.2, Ex-2, circle]
[Sol. Let the circle be
x2 + y2 + 2gx + 2fy + c = 0
passes through the point (– 6, 0)
 36 – 12g + c = 0
12g – c = 36 .....(1)
and power of point (1, 1) w.r.t. circle is 5
 1 + 1 + 2g + 2f + c = 5
2g + 2f + c = 3 ....(2)
it cuts the circle x2 + y2 – 4x – 6y – 3 = 0 orthogonally
 2(–2g – 3f) = c – 3
– 4g – 6f = c – 3
– 4g – 6f = 12g – 39 (using(1))
+ 16g + 6f = 39
and 2g = 2f + 12g – 36 = 3
14g + 2f = 39
42g + 6f = 117
– 16g – 6f = – 39
———————
26g = 78  g = 3,  c = 0
6 + 2f = 3  2f = – 3  f = – 3/2
 required equation of circle is x2 + y2 + 6x – 3y = 0 Ans. ]
Q.10 Tendogs encounter8 biscuits. Dogs donot sharebiscuits. In howmanydifferent ways can the biscuits
be consumed
(a) if weassume that thedogsare distinguishable,but thebiscuits arenot.
(b) if we assume that both dogs andbiscuits are different and anydog can receive anynumber of biscuits.
(c) if dogs andbiscuits are different andeverydog can get atmost one biscuit. [2+2+2]
[Ans. (a) 17C8; (b) 108; (c) 10C8 · 8! ]
[Sol. (a) dogs different D1, D2 , ......., D10
biscuitsalike
 
 

8
B
.........
B
B
8 alike object to be distributed in 10 different dogs 
 

8
0
.........
0
0  
 

9
Ø
.........
Ø
Ø
Total ways = 17C8 Ans.
when anynumber of biscuits can be had byanydog.
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correct (1 Mark forreasoning and 1 Mark forAns.)

6. (b) dogs aredifferent D1, D2 , ......., D10
biscuitsaredifferent B1, B2 , ......., B8
1st biscuits can be given in10 ways
2nd biscuits can be given in10 ways
hence total ways = 108 Ans.
(c) Select 8dogs in 10C8 andgive them 1 biscuit each. Distribution of biscuits in 8! ways.
Total ways = 10C8 · 8! Ans. ]
Q.11 Given 

35
1
k
k
5
sin = tan 





n
m
, where angles are measured indegrees, and m and n arerelativelyprime
positive integers that satisfy
n
m
<90, find thevalue of (m + n). [6]
[Ans. 177]
[Sol. LHS: S = sin 5 + sin 10 + sin 15 + .......... + sin 170 + sin 175
S 





2
5
sin
2 = 2 sin
2
5
[sin 5 + sin 10 + ......... + sin 175]
T1 = cos
2
5
– cos
2
15
T2 = cos
2
15
– cos
2
25
  
T35 = cos
2
345
– cos
2
355
————————






2
5
sin
2 · S = cos
2
5
– cos
2
355
= 2 sin
2
180
· sin
2
175
= 2 sin
2
175
S =
2
5
sin
2
175
sin
=







2
5
90
cos
2
175
sin
=
2
175
cos
2
175
sin
= tan 





2
175
= tan 





n
m
 m = 175 and n = 2  m + n = 177 Ans. ]
PART-C
Q.12 Apointmovingaroundcircle(x +4)2 +(y+2)2 =25withcentreCbrokeawayfromiteitheratthe point
A or point B on the circleand moved along atangent to the circle passing through the point D(3, – 3).
Findthefollowing.
(i) Equation ofthe tangents atAand B.
(ii) Coordinates of the pointsAand B.
(iii) Equation of the circle circumscribing the DAB and also the intercepts made by this circle on the
coordinateaxes. [T/S, Q.12, Ex-1, circle] [2.5 + 2.5 + 2.5]
[Ans. (i) 4x + 3y = 3; 3x – 4y = 21 (ii) A(0, 1) and B (–1, – 6);
(iii) x2 + y2 + x + 5y – 6, x intercept 5; y intercept 7 ]
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7. [Sol.
(i) Equationoftangent from point (3, –3) to thegiven circle is
y + 3 = m(x – 3)
mx – 3m – y – 3 = 0
and also 2
m
1
3
2
m
3
m
4





= 5
(1 + 7m)2 = 25(1 + m2)  1 + 49m2 + 14m = 25 + 25m2  12m2 + 7m – 12 = 0
 (4m – 3)(3m + 4) = 0
 m = 3/4 or m = – 4/3
 equation of tangent at pointAand B are
y + 3 = –
3
4
(x – 3) and y + 3 =
4
3
(x – 3)
3y + 9 = – 4x + 12 4y + 12 = 3x – 9
4x + 3y = 3 3x – 4y = 21
(ii) Equation ofnormals to these 2tangents are
y + 2 =
4
3
(x + 4) and y + 2 = –
3
4
(x + 4)
4y + 8 = 3x + 12 3y + 6 = – 4x – 16
3(3x – 4y + 4 = 0) 4(4x + 3y= – 22)
9x – 12y = – 12 16x + 12y = – 88
16x + 12y = 12 9x – 12y = 63
—————— ——————
x = 0;  y = 1 25x = – 25
x = – 1;  y = – 6
 points Aand B are (0, 1) and (–1, – 6) Ans.
(iii) CirclecircumscribingDABwill havepointsAandBas its diametricalextremities
x2 + y2 – x(–1) – y(–5) – 6 = 0
x2 + y2 + x + 5y – 6 = 0 Ans.
x-intercept = c
g
2 2
 = 6
)
4
1
(
2  = 5 Ans.
y-intercept = c
f
2 2
 = 6
)
4
25
(
2  = 7 Ans. ]
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8. Q.13 Let f (n) =  
 






n
0
r
n
r
k
r
k
. Find the total number of divisors of f (9). [7.5]
[Sol. 

n
r
k
r
k
C = rCr + r + 1Cr + r + 2Cr + ....... + nCr
= 1 + r + 1C1 + r + 2C2 + r + 3C3 + ....... + nCn – r
now, n+1Cn – r = n+1Cr+1
 f (n) = 



n
0
r
1
r
1
n
C = n+1C1 + n+1C2 + n+1C3 + ..... + n+1Cn+1
= n+1C0 + n+1C1 + n+1C2 + ..... + n+1Cn+1 – 1
f (n) = (2n+1) – 1
f (9) = 210 – 1 = 1023 = 3 · 11 · 31
hence number of divisors are (1 + 1)(1 + 1)(1 + 1) = 8Ans. ]
Q.14 Givenbelow is apartial graph of an even periodic function f whose period is 8. If [*] denotes greatest
integerfunctionthenfindthevalueoftheexpression.
f (–3) + 2 | f (–1) | + 











8
7
f + f (0) + arc cos 
)
2
(
f + f (–7) + f (20) [7.5]
[Sol. f (–3) = f (3) = 2 [ f (x) is an even function,  f (– x) = f (x) ]
again f (–1) = f (1) = – 3
 2 | f (–1) | = 2 | f (1) | = 2 | – 3 | = 6
fromthegraph, – 3 < 





8
7
f < – 2
 











8
7
f = – 3
f (0) = 0 (obviouslyfromthegraph)
cos–1
 
)
2
(
f  = cos–1
 
)
2
(
f = cos–1(1) = 0
f (–7) = f (– 7 + 8) = f (1) = – 3 [f (x) has period 8]
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




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9. f (20) = f (4 + 16) = f (4) = 3 [ f (nT + x) = f (x) ]
sum = 2 + 6 – 3 + 0 + 0 – 3 + 3
 sum = 5 Ans. ]
Q.15 AsquareABCDlyinginI-quadrant hasarea36sq. units andis such thatits sideABis parallelto x-axis.
VerticesA, B and C are on the graph of y = logax, y= 2 logax and y = 3 logax respectivelythen find
the value of 'a'. [7.5]
[Sol. AB : y = c (c > 0)
Length of the side of square = 6
A has y-coordinate = c and it lies on y= logax
 x-coordinate = ac
 pointAis (ac, c)
|||ly B is (ac/2, c)
and BC  AB
 C has x-coordinate = ac/2 and it lies on y = 3 logax
y = 3 logaac/2 =
2
c
3
 point C is 





2
c
3
,
a 2
c
| AB | = 6
 ac – ac/2 = 6 a > 0, c > 0
let ac/2 = t
t2 – t – 6 = 0  (t – 3)(t + 2) = 0  t = 3 or t = – 2 (rejected)
 t = 3  ac/2 = 3  ac = 9
also | BC | = 6
2
c
3
– c = 6;  c = 12
 a12 = 9  a6 = 3;  a = 6
3 Ans. ]
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