Rt Solutions-Practice Test Papers 1 to 14 Sol
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7/30/2019 Rt Solutions-Practice Test Papers 1 to 14 Sol
1/61PAGE # 1
Practice Test Paper-1Q.1
[Sol. Given (2 + sin ) (3 + sin ) (4 + sin ) = 6.As, L.H.S 6. Equality can hold only if sin = 1
=
2
3,
2
7 sum =
2
3+
2
7= 5 = k (Given)
So, k = 5 Ans.]
Q.2
[Sol. M1
M2
M3
M4
M5
M6
| G1
G2
G3
| H1
H2
H3
H4
H5
H6
H7
Number of ways of selecting 4 books = 6C4
+ 6C3 10C
1+ 6C
2 10C
2+ 6C
1 10C
3= 1610. Ans.
Alternatively: Total number of ways of selection of four books
total number of ways of selecting four non maths books.
= 16C410C4 = 1610 Ans.]
Q.3
[Sol. Let f '(x) = (x 1) (x + 1) = (x21)
Integrating on both sides, we get f(x) =
x
3
x3
+ c
Now, f(1) = 1 1 =3
2+ c ......(i)
1
1
1
3
1
x
y
O
Similarly, f (1) = 3 3 =3
2+ c ......(ii)
From (1) and (2), we get 2c = 2 c = 1 and = 3
f(x) =
x3
x
3
3
+ 1 = x
3
3x + 1So, f(2) = 8 6 + 1 = 3 Ans.]
Paragraph for Question 4 to 6
[Sol. Let P ' (x) = A(x 1)(x 2)(x 3)
or P ' (x) = A(x36x2 + 11x 6)
Integrating both sides with respect to x, we get
P(x) =4
A(x48x3 + 22x224x) + B .....(1)
Given, P(0) = 0 B = 0 ......(2)
and P(
1) = 55 A = 4 .....(3)So, from (1), we getP(x) = x48x3 + 22x224x
= x )24x22x8x( 23
= x(x 4)rootsrealnon
26x4x
(i) Clearly, area of triangle formed by extremum points of P(x) = 1122
1
[HINTS & SOLUTIONS]
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PAGE # 2
(ii) As, P(x) + P(x) = functioneven
24 x44x2
So,
1
1
dx)x(P)x(P = 1
0
24dxx44x22 =
1
0
24 dxx22x4 =
1
0
35
3
x22
5
x4
=
3
22
5
14 =
15
11034 =
15
1134=
15
452Ans.
(iii) From above graph, P(x) has two relative minimum points and one relative maximum point
(A) is incorrectClearly, from above graph, range of P(x) = [9, ) (B) is correctAs, real roots of equation P(x) = 0 are 0 and 4. So sum of real roots = 4 (C) is incorrectAs, P'(x) = 0 has 3 distinct real roots, so P''(x) = 0 has two distinct real roots.
P(x) has two inflection points (D) is incorrect.]
Q.7
[Sol. Statement-1: Let det.(B) 0, then B1 exists.Now, AB = O ABB1 = O A = Oso det.(B) = 0.
Similarly, suppose det.(A) 0, then A
1exists.Now, AB = O A1AB = O B = OBut B is not a null matrix, so det.(A) = O
Statement-1 is true.Statement-2: Obviously Statement-2 is true. For example :
Let M =
20
10and N =
00
55then, MN = O, but neither M = O nor N = O. Ans.]
Q.8
[Sol. Statement-1: The equation of plane through A (1, 1, 1), B (1, 1, 1) and C (1, 3, 5) is
3x z 2 = 0, which clearly, passes through (2, k, 4) k R.Obviously, Statement-2 is true and not explaining Statement-1. ]
Q.9
[Sol. We have
g(x) =
1,0xif,0
1,0x,1x
sin1xx
sinx22
Clearly, g(x) is differentiable x R.(As sum and product of differentiable functions is also differentiable function.)
Now, g '(x) =
1,0x,0
1,0x,1xsin)1x(21xcosxsinx2xcos
Clearly, g'(x) is discontinuous at both x = 0 and x = 1.
Also, g(0) = 0 = g(1) (Given)
So, Rolle's theorem is applicable for g(x) in [0, 1].
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PAGE # 3
Q.10
[Sol. A vector coplanar with kji2a
, kjib
and perpendicular to k6j2i5c
will be
along cba
= acbbca
= kji29kji18 = kj39k9j27
Vector will be along kj3 .
This vector will lie in the plane which will be parallel to it.
Normal of plane will be perpendicular to vector. Ans.]
PART-B
Q.1
[Sol.
(A) Obviously each digit must be 0 or 1 1
Hence number o five digit numbers are = 1 2 2 2 2 = 16 Ans.
(B) Let u = x2 du = 2x dx
2b
12b
du1u
1
2
1Limit = 1tanbtan
2
1Lim 121
b
=
422
1=
8
= L
L120=
8
120
= 15 Ans.
(C) ax + 3y z = a .....(1)
2ax y + z = 2 .....(2)
bx 2y + z = 1 a .....(3)
For no solution, D = 0
D =
12b
11a2
13a
= 0 [Note: D1 = a]
D = 2b a = 0
Also Dx
=12a1
112
13a
= a(1) 3(2 1 + a) 1(3 a)
Dx
= a 3 3a + 3 + a = a
Dx 0 (Given a [1, 8] )
Hence for no solution
b =2
a, a I
sum =2
1(1 + 2 + ......... + 8) =
2
36= 18 Ans.
Note: when a = 0, b = 0 system will have infinite solution.]
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PAGE # 4
PART-C
Q.1
[Sol. We have, f(x) = x33ax2 + 3(a21) x + 1
f '(x) = 3 1aax2x 22 = 3 )1a(x)1a(x = a + 1
= a 1
M m
a 1 a + 1
a 1 > 2 and a + 1 < 4 a > 1 and a < 3So, a ( 1, 3) a1 = 1 and a2 = 3.Hence, (a12 + a22) = (1)2 + (3)2 = 10. Ans.]
Q.2
[Sol. Given, 2xy + 6 4x 3y = 0
2x (y 2) 3 (y 2) = 0 (2x 3) (y 2) = 0
Clearly, image of R (3, 4) in y = 2 is P(3, 0) and in x =2
3is Q (0, 4). y = 2
(3, 4)R(0, 4)
(3, 0)2
3x O
Y
X
4,
2
3
(3, 2)
Inradius of the PQR is, r = s =
2
543
34
2
1
= 1. Ans.]
Q.3
[Sol. Given, f(x) = ex x
0
x2ye)1xx(dy)y('fe ..........(1)
differentiating both the sides
f '(x) = ex ex f '(x) + x
0
yxdy)y('fee 1x2ee1xx xx2
f '(x) = f '(x) + x
0
2xyx xxedy)y('fe
0 = f(x) + (x2x + 1) exex (x2 + x) [Substituting x
0
yxdy)y('fe = f (x) + (x2x + 1) ex from (1)]
f(x) = ex (x2 + x x2 + x 1)
f(x) = ex (2x 1) f(1) = e.Also, f '(x) = ex 2 + (2x 1) ex f '(1) = 2e + e = 3eand f '' (x) = 2ex + (2x 1)ex + 2ex
f '' (1) = 2e + e + 2e = 5e.
Hence f(1) + f '(1) + f '' (1) = 9e k = 9. Ans.]
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Practice Test Paper-2Q.1
[Sol. Given, h(x) =)x(g
)x(f
h'(x) =)x(g
)x('g)x(f)x('f)x(g2
Clearly, h'(c) = 0 (As, f ' (c) = 0 and g ' (c) = 0)
So, h'(c) =)c(g
)c('g)c(f)c('f)c(g2
[f '(c) > 0; g'(c) < 0 ; f (c) > 0 ; g(c) > 0]
h'(c) > 0|||ly h'(c+) < 0
So, h(x) has a local maximum at x = c. Ans.
Aliter-1: Nr i.e. f (x) is maximum at x = c
and Dr i.e. g (x) is minimum at same x i.e. x = c
Hence)x(g
)x(fis maximum at x = c
Aliter-2: For example, let f (x) = 2x1
1
and g (x) = x2 and c = 0.
So, h (x) =)x1(x
122 = )x(g
)x(f h ' (x) = 222
2
)x1(x
)x21(x2
Clearly, for x < 0, h ' (x) > 0
and for x > 0, h ' (x) < 0 h(x) has a local maximum at x = c]
Q.2
[Sol. Given that 0ckbka21
Now, OQRArea
PQRArea
=
4
cb2
1
acab2
1
P(a)
Q(b)
b a c a
O(0)
cb
ck1bkckbk12121
= 4 (1 + k
1) (1 + k
2) k1k2 = 4
Hence, k1
+ k2
= 3. Ans. ]
Q.3
[Sol. | A | = x + y + z
det.(adj.(adj. A)) = 2
)1n(A.det
= (det. A)4 = 28 34 = 124
| A | = 12
x + y + z = 12Using beggar method x + y + z = 9.
Number of matrix = 11C2 = 55. Ans.]
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PAGE # 6
Paragraph for Question 4 to 6
[Sol.
(i) The given lines (L1
and L2) are parallel and distance
between them (BC or AD) is =5
515 = 2 units.
Let BAC = AB = BC cosec = 2 cosec
B
B1
C
A
D
A1
L2
L1
L
P(4, 3)
and AA1
= AD sec = 2 sec
Clearly, area (||gmAA1BB1)= (AB)(AA
1) = 4 sec cosec
=2sin
8, which is least for =
4
Let slope of line L be m.
So, 1 =
m4
31
4
3m
(4m + 3) = (4 3m) m =
7
1or 7
But m > 0 (Given)
The equation of line L, is (y 3) =7
1(x 4) x 7y + 17 = 0 Ans.(i)
(ii) If line L : x 7y + 17 = 0 is orthogonal to circle x2 + y2 6x + 4y 9 = 0, then line Lmust be normal to the given circle. So, centre of circle (3, 2) must satisfy the line L.Hence, 3 7( 2) + 17 = 0 17 + 17 = 0 = 1 Ans.(ii)
(iii) Equation of circle S having the ends of diameter at (0, 1) and (2, 0) is
x(x + 2) + y(y + 1) = 0 i.e. x2 + y2 + 2x + y = 0
So, equation of circle S', is (x2 + y2 + 2x + y) + (x 7y + 17) = 0
As it passes through (1, 2), so 171412)1(2)2()1(22
= 0
9 + 4 = 0 =4
9
Hence, the equation of circle S', is
(x2 + y2 + 2x + y) 4
9(x 7y + 17) = 0 4x2 + 4y2x + 67y 153 = 0 Ans.(iii)]
Q.7
[Sol. Equating the coefficient of x20, we get
220 a20 = 1 20 20 12a (C)
put, x = 2
1
, we get
0 1020
q2
p
4
1b
2
a
0q2
p
4
1b
2
a1020
b2
a and 0q
2
p
4
1
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PAGE # 7
a + 2b = 0 (B)
b =2
1220 20
.
put, x = 0 we get
1 b20 = q10
1
2020 20
2
12
= q10; 10
20
20
q2
121
;
10
20q
2
1 q =
4
1
Using, 0q2
p
4
1
04
1
2
p
4
1 p = 1
Hence B, C, D. Ans.]
Q.8
[Sol. x
y
3/2
O 44
x= 4
y=3
. Now, verify alternatives. ]
PART-B
Q.1
[Sol.
(A) IN =
0
x9 dxex2
=
0 II
x
I
9
dxex
2
I.B.P.
IN
=
0
x8
2
ex2
+
0
x7 dxex2
8 2
IN
= 0 + 4ID
IN
= 4ID
D
N
I
I= 4 Ans.
(B)MB
x4 13x2 + 36 0 (x29) (x2 4) 0 x [3, 2] [2, 3]Now, let
f(x) = x33x f '(x) = 3(x21) > 0 x [3, 2] [2, 3] f
max.(x = 3) = (3)33(3) = 27 9 = 18. Ans.
(C)MB
Any circle through (2, 2) and (9, 9) is
(x 2) (x 9) + (y 2) (y 9) + (y x) = 0 .........(1)For the point of intersection with x-axis, we put y = 0 in (1), we get
(x 2) (x 9) + 18 x = 0
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PAGE # 8
Put disc. = 0 (11 + )2 4 + 36 = 0 = 23, 1
x =2
1 = 6
So, the absolute value of the difference of x-coordinate of the point of contact = | 6 (6) | = 12 Ans.
(D) y = cos1(3x 4x3) =2
sin1(3x 4x3) =
2
xsin3 1
because sin1(3x 4x3) = 3 sin1x if x
1,
2
1
Hencedx
dy=
2x1
3
2
3xdx
dy
=
4
31
3
= 6 Ans.
Alternatively (D): Clearly,dx
dy= 23
2
)x4x3(1
)x123(1
)x43(x1
)1x4(3
dx
dy
2
2
2
3x
= 6 Ans.]
PART-C
Q.1
[Sol. x
0
x
0 KingUsing
dtxttandtttan)t(f)x(f = 0
x
0
x
0
dtttandtttan)t(f)x(f = 0
Differentiate w.r.t. x
f '(x) + 1)x(f tan x = 0
1)x(f
)x('f
+ tan x = 0
integrate w.r.t. x
ln 1)x(f + ln (sec x) = Cf(0) = 0 C = 0.
Hence
1xcos
1)x(f
f(x) = cos x
1 f(x) = 0 cos x = 1
Hence only solution in
2
,2
is x = 0.
Hence number of solution is 1.
Note that domain of f(x) is
2
,2
Ans.]
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PAGE # 9
Q.2
[Sol. We have, f '' (x) = 12x 4
f ' (x) = 6x24x + cAs, f ' (1) = 0 c = 2 f ' (x) = 6x24x 2 f (x) = 2x32x22x + As, f (1) = 0 = 2Hence, f (x) = 2(x3x2x + 1)
or f (x) = 2(x 1)2 (x + 1).
Now, M (x = 2, y = 6) and f ' (2) = 14.
So, the equation of normal at M is (y 6) =14
1(x 2)
For x-intercept, put y = 0
we get x = 86 Ans.]
Q.3
[Sol. Let y = f 2(x) g2(x)
Now,dx
dy= 2 )x('g)x(g)x('f)x(f = 2 )x(f)x(g)x(g)x(f = 0
f2(x) g2(x) = k (constant)So, f2(3) g2(3) = k (5)2 (4)2 = k k = 9, x RHence, f2() g2() = 9. Ans.]
Practice Test Paper-3Q.1
[Sol. We have
(1 + x)10 = 10C0
+ 10C1
x + 10C2
x2 + 10C3
x3 +........ + 10C9
x9 + 10C10
x10 .......(1)
Also (x 1)10 = 10C0
x1010C1
x9 + 10C2
x8 +........ 10C9
x + 10C10
.......(2)
Multiplying (1) and (2), we get
102 1x = 1010109910110010 xCxC.......xCC
10
109
1091
10100
10 CxC.......xCxC .......(3)
Comparing the coefficients of x10 in (3) , we get
10C5
(1)5 = 210
102
9102
2102
1102
010 CC.......CCC Ans.]
Q.2
[Sol. f '(x) = f(x) = 1)x(f
)x('f
ln (f(x)) = x + c
x = 0, f(0) = 1 c = 0 f(x) = ex
g(x) = ex (x + 1)2ex = ex (x2 + 2x)
1
0
xdx)x(g)x(fe =
1
0
2xdxx2xe =
1
0
2x xe = e Ans.]
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PAGE # 10
Q.3
[Sol. Let, I =
dx
Add&King
xsin1xsin1
x
2
2
2I =
dx
xsinxsin11
xsin122x
22
2
22
So, I =
dx
xsin112
xsin11x
2
22
=
dxx2
1 2=
3
3. Ans.]
Paragraph for question nos. 4 to 6
[Sol.(i) angle between banda is obtuse
hence ba
0 (log3x)2 + 2log
3x 1 0 x (0, )
42 + 4 0 [1, 0]
Volume =6
1
111
11
12
=6
1|23 |
Since, [1, 0]
Volume
2
1
,3
1
(ii) ca
= cb
0c)ba(
cba
kj2i)xlog(3
= kj)xlog(i)x(log33
+ kji
xlogxlog33
; 2 = log3x + and 1 = 1
= 2 log
3x = 0 and log
3x = 2
= 0 = 2, log
3x = 2, = 0
kj2a
ki2b
kjic
7|ji|
|k2ji3|
|ca|
|cb|
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PAGE # 11
(iii) c)ba(
= b2a
b2ab)ca(a)cb(
1cb
, 2ca
log3x + log
3x 1 = 1 ... (1)
log3
x + 2 + 1 = 2
log3x =1 ... (2)
log3x = 1, = 1
kj2ia
kjib
kjic
accbba
= 2cba
= 4. Ans.]
Q.7
[Sol. f(x) =
x xe
1
e
1
22t1
dt
t1
dt xe1
1 ]ttan xe
11 ]ttan =
4)e(tan
4)e(tan x1x1
f(x) = )e(tanx1
+ )e(cotx1
= 2
Now verify alternatives. Ans.]
Q.8
[Sol. cos C =2
1=
64
c6416 2 c2 = 48 c = 34
Also,Asin
a=
Csin
c
Asin
4=
23
34 A = 30
As C = 60, A = 30 B = 90
Now, r =s
=
2
cba
ac2
1
=
cba
ac
=
8344
344
=
3412
316
=
33
34
Now verify alternatives.
(C) r = 33
34
(D) the length of internal angle bisector of angle C is3
8
Hence A, B are correct.]
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PART-B
Q.1
[Sol.
(A) Given, C : y2 = px3 + q .......(1)
2ydx
dy= 3px2
dx
dy=
y2
px3 2
.......(2)
Put x = 2 and y = 3 in (1), we get 9 = 8p + q .......(3)
Also, from (2), we get)3,2(M
dx
dy
=6
)4(p3= 2p = 4 .......(4)
From (2) and (3), we get p = 2, q = 7.Hence, (p q) = 2 ( 7) = 2 + 7 = 9. Ans.
(B) We have
cosec210
1cosec21
011
V = 4 cosec2 1 2cosec
= 1eccos2
eccos42
=
4
5
4
1eccos4
2
2
V eastl = 14
4
4
5
16
94 ]
(C) We have11
xxsin
1x
xxcos= 0
(sin x x) ( cos x x2) = 0 Either sin x = x or cos x = x2
Hence, number of real roots are 3. Ans.
(D) Given, tan A = 4
1
, tan B = 3
2
, tan C = 5
1
and tan D = d
Now, A + B + C + D = 2
tan (A + B) = tan (C + D) ab1
ba
=1cd
dc
3
2
4
1
1d5
1=
6
11
d5
1
Solving 39 d = 65
d =3
5 3 | tan D | = 3
3
5= 5. Ans.]
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PAGE # 13
PART-C
Q.1
[Sol. Let O be the centre of polygon
Area of rectangle = 4 OA1A
2= 6 .........(1)
and Area OA1A
2=
n
1area of polygon .........(2)
A1
A2
Ak + 1
AkO
(1) and (2) 4
6
= 60n
1
n = 40. Ans.]
Q.2
[Sol. Using tan1 + tan1 + tan1 = tan1
1In L.H.S. we get
tan1
2)c)(bx()bx(ax)ax(c1
bxaxcbxaxc
where c =
8
x
x
1.
Now, 1 = ax
8
x
x
1+ abx2 + bx
8
x
x
1 1 = 222 x
8
bbabxx
8
aa x R
1 = (a + b) + x2
8
baab
On comparing, we geta + b = 1 ..........(1)
and ab =8
ba =
8
1.......(2) (Using (1))
Now, a2 + b2 + 2ab = 1 a2 + b2 +4
1= 1
Hence, 4(a2 + b2) = 3. Ans.]
Q.3
[Sol. Using cosine law in BKC,
2
Bcos =
22
232
14
184
=26
2
93
=212
15=
24
5
B
A K C
c= 2m
m 1
a = 22
23x
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PAGE # 14
Now,
x =2
Bcos
ca
ac2
=
24
5
2m2
)m222(
2
23=
1m
m4
24
5 3 =
1m
m5
3m + 3 = 5m
m =
2
3
AB = 2m c = 3 . Ans.]
Practice Test Paper-4Q.1
[Sol. S =
80tan1
1.......
10tan1
1
0tan1
1333
Note that
90tan1
1
tan1
133 = 1
=
33 cot1
1
tan1
1=
3
3
3 tan1
tan
tan1
1=
3
3
tan1
tan1.
Hence, S = 1 + (1 + 1 + 1 + 1) = 5 A. Ans.]
Q.2
[Sol. Given, f '(x) = 5)x(f2 or5y
dy
= 2dx. On integrating, we get
ln )5y( = 2x + c, if x = 0 ; y = 0 c = ln 5
ln
5
5y= 2x y + 5 = 5e2x y = 5e2x5 f(x) = 5(e2x1)
Now, f(x) + 5 sec2x = 5(e2x + tan2x) = 0
So, no solution exist (0, 2). Ans.]Alternative : f '(x) 2 f(x) = 10
dx
dy2y = 10 ; Integrating factor = e2x
y e2x = 5e2x 5
f(0) = 0 ; c = 5
y = 5e2x 5 f(x) = 5 (e2x1)Hence, f(x) + 5 sec2x = 0 e2x + tan2x = 0 No solution. Ans.]
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Q.3
[Sol. I = 2
2
12
1
dx1xx
xtan.......(1) put x =
t
1
I =
2
21
2
1
dt
1tt
t
1tan
.........(2)
Now, (1) + (2) gives
2I =
2
2
122
2
3
2
1x
dx
2
I =
2
2
1
1
3
1x2tan
3
2
4
=
0332
=36
2 A. Ans. ]
Paragraph for question nos. 4 and 5
[Sol. 0nar 1
nanr1
1
A
(1,1,1)
L
r = (1, 1, 1)+ (1, 1, 1)
n = i + j1^ ^
jikzjyix = 2x + y = 2 ............ plane
1
Now, kiAB ; A(1, 1, 1); B(0, 1, 0)
Now,
111
101
kji
nVAB2
n = AB V2
V = i j k^ ^ ^
B( j )^
2
1k11j10in2
= kj2in2
Hence, equation of plane 2 is kj2ijr = 0 or x + 2y z = 2
(i) Vector along the line of intersection of planes 1
and 2
is21
nn
which is also perpendicular
to the normal vector of plane 1.
21
nn
= kji
So, required vector =
3
kji6 = 2 kji or 2 1,1,1
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(ii) If is the acute angle between 1 and 2 then
cos =21
21
nn
nn
=
62
k2j2iji =
62
3=
2
3 = cos1
2
3=
6
. Ans.]
Q.6
[Sol. A is true by the Extreme Value Theorem, B is true by the Intermediate Value Theorem and C is not
always true, D is true because g is continuous.]
Q.7
[Sol.mb
As, cos1x 0 x [ 1, 1]and tan1x > 0, x in (0, )So, a solution must be a positive number only.
Now, cos1x = tan1x
cos1x = cos1
2x1
1 x2 = 2
x1
1
x4 + x21 = 0 x2 =
2
51
x =2
51 (0, 1)
Now verify alternatives.]
Q.8
[Sol. Suppose , be two distinct roots in (0, 1) of the equation x33x + p = 0Let f(x) = x33x + p.
As f() = 0 = f()
f(x) satisfies hypothesis of Rolle's theorem in [, ] (0, 1). So f '(c) = 0 3c2 = 3 c = 1But c must lies between and .Hence p .So, S
1is false.
Also, S2
is obviously true. Ans.]
Q.9
[Sol. We have
f(x) = tr.(A ) =
3
1iii
a = x +x9 + 2
As,
x
9x
2x
9x
x +x
9 6 (as x > 0 is given)
so, f(x) 8 f
Min.(x = 3) = 8
Also, S2
is true and explaining S1also. Ans.]
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PAGE # 17
PART-B
Q.1
[Sol.
(A) Let the point of intersection of curves y2 = 2ax (a > 0) and xy = 24 be P(x1, y1).
Now, y2 = 2ax )y,x(P 11
dx
dy
=
1y
a= m
1... (1)
Also, xy = 24
1
1
)y,x(P x
y
dx
dy
11
= m
2... (2)
As, m1 m
2= 1 x
1= a ...(3)
As, x1y
1= 24 y1 = a
24
So, P
a
24y,ax
11
Also, Point P must satisfy
y12 = 2ax
1
2a
32= 2a2 a4 = 16 a = 2. Ans.
(B) f (x) is differentiable in R
Hence )x(fLimx
must exist and is finite
y = f (x) must have a horizontal asymptoteO
y=1x
y=f(x)(0,4)
as x then only )x(fLimx
will exist
)x('fLimx
= 0
)x('f)x(fLimx
= 3 )x(fLimx
= 3 Ans.
(C) y = mx + (n + 2 x) or y + x (n + 2) mx = 0 )2n(y + (1 m) (x 0) = 0 m R Fixed point is the point of intersection of lines y = n + 2 and x = 0.Family of lines which passes through (0, n + 2)
Hence n + 2 = 3 n = 1. Ans.]
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PAGE # 18
PART-C
Q.1
[Sol. Let the 3 consecutive terms are
a d, a, a + d ; d > 0
hence a2 2ad + d2 = 36 + K ....(1)
a2 = 300 + K ....(2)
a2 + 2ad + d2 = 596 + K ....(3)
now (2) (1) gives
d(2a d) = 264 ....(4)(3) (2) gives
d(2a + d) = 296 ....(5)
(5) (4) gives
2d2 = 32 d2 = 16 d = 4 or d = 4 (rejected as increasing A.P.)Hence from (4)
4(2a 4) = 264 2a 4 = 66 2a = 70 a = 35 K = 352 300 = 1225 300 = 925 Ans.]
Q.2
[Sol. Given, f (y) = f (x) + y
y2 + ay + b = x2 + ax + b + y y2 + y(a 1) x2ax = 0As, y R, so D 0 x R (a 1)2 + 4(x2 + ax) 0 x R 4x2 + 4ax + a22a + 1 0 x RNow, D 0
16a2 16(a2 2a + 1) 0 2a 1 0 a 2
1
So, amax.
=2
1
Hence, maximum value of 100a = 50 Ans.]
Q.3
[Sol. For number of solutions of given equation to be 9, so n must be 5.
Now (3 + cosec2x) + ysin2
2 = 5
possible only if cosec2x = 1 and sin2y = 0
possible values of x =2
,
2
3,
2
5,
2
7
possible values of y = 0, , 2, 3, 4
Hence number of ordered pairs (x, y) = 4 5 = 20 Ans.]
Q.4
[Sol. Put f (x) = t f (t) = 4 t = 2, 0, 4Hence f (x) = 2 2 values of x.
or f (x) = 0 2 values of x.or f (x) = 4 3 values of x.
Hence, the number of real solutions of the equation f(f(x)) = 4 are 7. Ans.]
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PAGE # 19
Practice Test Paper-5
Q.1
[Sol. )x(fLim
2x
=h
(sinh)sinLim
0h = 1 and )x(fLim
2x
= 1h
)1(sinLim
0h
.
So, )x(fLim
2x
does not exist. Ans.]
Q.2
[Sol. 1, x, y G.P. Let x = k, y = k 2
and x, y, 3 A.P. 2y = x + 3 2k2 = k + 3
2k2 k 3 = 0 k = 1,2
3.
x + y = k2 + k =4
1
2
1k
2
.
x + y|max
=4
15
4
14 at k =
2
3. Ans.]
Q.3
[Sol. Possible cases :
(i) 0 0 0 0 0 3 !5
!6
(when digit 0 comes at first place then number of arrangement =!4
!5)
Number of six digit numbers with 0 0 0 0 0 3 =!5
!6
!4
!5= 1
(ii) Similarly for 0 0 0 2 2 1
!2!2
!5
!2!3
!6= 30
(iii) 1 1 1 1 1 2 !5
!6= 6
Total = 1 + 30 + 6 = 37. Ans.]
Paragraph for question nos. 4 to 6
[Sol. We have, M(0) =
3
200
02
10
004
3
= diag
3
2,
2
1,
4
3
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222
2
3
2,
2
1,
4
3diag)0(M
333
3
3
2,
2
1,
4
3diag)0(M and so on
Now, n32 )0(M.........)0(M)0(M)0(M
=
n2n2n2
3
2.....
3
2
3
2,
2
1......
2
1
2
1,
4
3......
4
3
4
3diag
So, P =
.....3
2
3
2,........
2
1
2
1,.....
4
3
4
3diagLim
222
n
=
32
1
3
2
,
2
1
1
2
1
,
4
3
1
4
3
diag = diag (3, 1, 2)
Also, Q = diag
2
1,1,
3
1
PQ =
100010001
= I3
(PQ)m = I3 m N
Rm
= M(x) R100
= M(x)
Sk= M(x) k2 )PQ(.....)PQ()PQ( S
360= M(x) 360 I
3= 360 M(x)
(i) adj (adj P) = PP2n
= 6 P = 6P
Padjadj.Tr = Tr (6P) = 36.
(ii) Tr . (R100
) = f(x) = )x(M.Tr
f(x) = x23
2
2
1x
4
1xx3 2
= 3x2 + 2x +12
23.
2
0
dx6
41)x(cosf)x(sinf
= dx6
41
12
23xcos2xcos3
12
23xsin2xsin3
2
0
22
= dxxcosxsin22
0
= 4
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(iii) Tr. (S360
) = g(x) = 360 )x(M.Tr
=
12
23x2x3360
2=
36
23x
3
2x3360
2
=
9
1
36
23
3
1x3360
2
=
36
19
3
1x3360
2
g(x)|min = 360 3 3619 = 570. Ans.]
Q.7
[Sol. We have
22x2
x1
e.1x)x(f
f ' (x) =
x
32
23
e.1x
1x5x3x1x
1
01
graph of g(x) = x 3x + 5x 13 2
Let g (x) = x33x2 + 5x + 1,
g ' (x) = 3x26x + 5, discriment < 0
so g (x) has only one real root.
Also g (1) g (0) < 0 so one root of g(x) = 0
belong to the interval (1, 0) say
Now sign scheme of f '(x) = x
32
)x(g
23
e.)1x(
1x5x3x1x
At x = 11
ve +ve
At x =
+ve ve
Clearly f (x) has two points of extremum, maxima at x (1, 0) and minima at x = 1.]Q.8
[Sol.
(A) ba
= ca
either cb
or cba
.
(B) ba
= ca
either cb
or cb||a
(C) True
(D) True. Ans.]
PART-B
Q.1
[Sol.
(A) Given cos = 12 sinsin = 12
cos
= cos12
12 cos = sin cos2 = sin + cos cos + sin = 2 cos but given that cos + sin = k cos
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PAGE # 22
Hence, K = 2
L = 11 2 (As, K + L = 11 (given)) 9 < L < 10 2 < log
3L < 3
Sum of integers = 2 + 3 = 5. Ans.
(B) OC = 2 OC2 = 4 h2 + k2 = 4 x2 + y2 = 4.
O 22
C(h,k)2
Y
X Ans.
(C) sin x21sin 1 = sin
3
where = sin1x
1 2x = sin21
cos2
3= 2
xx12
32
2 4x = xx13 2 x3x332 2
(2 3x)2 = 3(1 x2) 4 12x + 9x2 = 3 3x2
Hence, (12x 12x2) = 1.
(D) g (10) = 10
0
f (t) dt = 5
2
0
f (t) dt = 5
1
0
1
0
td)t2(ftd)t(f
[As, f is periodic with period 2.]
a2
0
a
0
a
0dx)xa2(fdx)x(fdx)x(f
=
1
0
1
0
td)t(ftd)t(f (Since f(x) is periodic with period 2 hence f (2 t) = f ( t)) .
=
1
0
1
0
td)t(ftd)t(f (f ( t) = f(t), as f is odd.)
= 0. Ans.]
PART-CQ.1
[Sol. x2 + y2 + y 1 + k (x + y 1) = 0 .......(1)
Hence C passes through the intersection of x2 + y2 + y1= 0 and y = 1 x
Solving
x2 + (x 1)2 + (1 x) 1 = 0
2x23x + 1 = 0
2x22x x + 1 = 0
2x (x 1) (x 1) = 0
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PAGE # 23
x = 1 or x =2
1
y = 0 or y =2
1
Hence, points are (1, 0) and
2
1,
2
1
Now minimum radius will be when the two points are the extremities of the diameter of the circle C.
diametermin
=4
1
4
1 =
2
1
Hence, r |min
=22
1=
8
1 R = 8. Ans.]
Q.2
[Sol. = b82
1 =
2
1 8 a =
2
1 10 c
a = b and c =5
a4.......(1)
cos A =bc2
acb222
=bc2
c2
=
5
a4a225
a16 2
cos A =5
2 sin A =
5
21
From (1)
sin C =5
4sin A =
25
214
=2
1ab sin C =
2
1 h
c c
=2
1a2
25
214=
2
1 10 c
=25
214
16
c252
= 10 c
c2 =21
40 c c =
21
40=
q
p
p + q = 40 + 21 = 61 Ans.
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PAGE # 24
Alternatively: = aha21
a
2= h
a= 8
a
= 4 .....(1)
|||lyb
= 4 .....(2) and
c
= 5 ......(3)
(1) and (2) a = b
andc
a=
4
5=
c
b
a = b =4
c5
s =2
cba =
4
c7
s a =4
c5
4
c7 =
4
c2=
2
c= s b
s c = c4
c7 =4
c3
2 = s(s a)(s b)(s c) =4
c3
2
c
2
c
4
c7.....(4)
Also =2
1 c 10 = 5c
2 = 25c2 .....(5)
(4) and (5) 25c2 =4
c3
2
c
2
c
4
c7 25 =
64
c21 2 c2 =
21
6425
c = AB =21
85=
21
40=
q
p
p + q = 40 + 21 = 61 Ans.]Q.3
[Sol. Let I = 100
1
dxx
)x(f=
100
10
10
1
dxx
)x(fdx
x
)x(f=
100
10
10
1
dxx
x
100f
dxx
)x(f
Put,x
100 = t dxx
1002
= dt tx
dx = dt tdt
xdx
I =
1
10
10
1
dtt
)t(fdx
x
)x(f= 2
10
1
dxx
)x(f= 2 5 = 10. Ans.]
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PAGE # 25
Practice Test Paper-6
Q.1
[Sol. As, ar.(ABC) = ar. (ACD) + ar. (BCD)
2
1(3) (4) =
2
1(4) (CD) sin 60 + 30sinCD3
2
1
A
C B
D
60
30
b=4
a=3
c=5
CD =334
24
=
39
)334(24 = )334(
13
8 . Ans.]
Q.2
Sol.
100
0k !3k!2k!1k
3k=
100
0k 2k3k2k1!1k
3k
=
100
0k2 9k6k!1k
3k=
2
3k!1k
3k=
100
0k !1k3k
1
=
100
0k !3k13k2k =
100
0k !3k1
!2k1 =
!1031
!21
S =2
1 B. Ans.]
Q.3
[Sol. Clearly, = 0
111 11k
1k1
= 0 1k Ans.]
Paragraph for Question 4 to 6
[Sol.
(i) Given 1, b, c are in A.P. 2b = 1 + c ......(1)Also, 2, 5b, 10c are in G.P.
25b2 = 20c or 5b2 = 4c .......(2)Now, putting c = (2b 1) from (1) in (2), we get
5b2 = 4(2b 1) 5b2 + 8b 4 = 0 b = 2 (As b I)
So, c =
5 f (x) = x22x 5 = (x 1)26Clearly in interval x [0, 4],m = minimum value of f (x) = f
min.(x = 1) = 6
and M = maximum value of f (x) = fmax.
(x = 4) = (4 1)26 = 9 6 = 3.
Hence, (M + m) = 3 + (6) = 3 Ans.(i)
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(ii) Given g (x) =
upwardopeningparabolais)x(gofGraph
222 3x4ax1a
Also, g (0) = 3
g (x) < 0 is true, for atleast one real x provided a2 + 4 0 2 a 2So, the number of integral values of a are 5 (i.e., a = 2, 1, 0, 1, 2). Ans.(ii)
(iii) We have y = f (x) + h(x) = (x22x 5) + x2 (p 3)x + p = x2 + (1 p)x + (p 5)
As, range of y is [0, ), so put D = 0
(1 p)2 = 4(p 5) 1 2p + p2 = 4p 20 021p6p
RppositiveAlways
2
So, p Ans.(iii)]
Q.7
[Sol. Statement-1 is true and Statement-2 is false. Statement-2 could be true only if f(x) is continuous.
Rolle's Theorem can also be used for validity of Statement-1. Ans.]
Q.8
[Sol. Statement 1: B1
+ B2
+ B3
+ B4
= 10
B1
+ B2
+ B3
+ B4
= 6
Using beggar total number of ways 9C3
Statemen-2: Number of ways of choosing any 3 places from 9 different places is 9C3
Hence S-1 is true, S-2 is true and statement-2 is the correct explanation for statement-1. Ans.]
Q.9
[Sol. S-1 : Let r be the common ratio of geometric progression. So,
S = 8r1
r
x
(Given)
x = 8(r r2) ........(1)
As, r ( 1, 1) {0} so r r2
4
1,2 {0} .......(2)
From equations (1) and (2), we getx = 8 (r r2) (16, 2] {0}
Hence, S - 1 is false and obviously S-2 is true. Ans.]
Q.10
[Sol.(A)O x=a
y
x
point of non-differentiability.
(B) y = x3
f '(0) = 0
O
y
xBut at x = 0, f(x) is increasing.
(C) f "(c) can be zero also. e.g. f(x) =x4.
(D) false e.g. f(x) = x3 at x = 0 is increasing.]
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PAGE # 27
Q.11
[Sol. As, a1
+ a2
+ a3
= 126
and a2
=2
aa31
=
3
126= 42.
The numbers in A.P. are 42 d, 42, 42 + d.
Let g1
= A, g2
=AR; g3
= AR2
AR = 34 and A + AR2 = 85
R34 + 34R = 85 34R2 85R + 34 = 0
34R268R17R + 34 = 0 (R 2)(2R1) = 0For R = 2, A = 17
g1
= 43 + d = 17 d = 26 (Rejected)
for R =2
1, A = 68
g1
= 43 + d = 68 d = 25
Common difference of A.P. = 25 and common ratio of G.P. is2
1Ans.]
PART-B
Q.1
[Sol.
(A) We know that | adj A | = | A |2 for a 3 3 matrix
Given adj A = KAT |adj A| = |KAT| = K3 | A | (|AT| = | A | ) K3 | A | = | A |2
K3 = | A | ; Now det A = 1 (1 4) 2(2 4) + 2 (4 + 2) = 27 k3 = 27K = 3.
(B) Given, f (x) = (2x + 1)50 (3x 4)60
f ' (x) = 220(2x + 1)48 (3x 4)58 (2x + 1)(3x 4)(3x 1)
12
13
43
+ +
Sign scheme of f '(x)
Least positive integer is k = 2(C) By applying continuity and differentiability at x = 1, we get a = 3, b =1.
Hence, (2a + b) = 2 (3) 1 = 5. Ans.]
PART-C
Q.1
[Sol.mb
Sn
= 1 2 + 2 3 + 3 4 + ......... upto n terms
=
n
1n
)1n(n =
n
1n
n
1n
2 nn =2
)1n(n
6
)1n2()1n(n
Sn
=3
)2n()1n(n ......(1)
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PAGE # 28
Also, n1
= .......6543
1
5432
1
4321
1 upto (n 1) terms
=
1n
1n )3n()2n()1n(n
1=
1n
1n )3n()2n()1n(n
n)3n(
3
1
n1
=
1n
1n )3n()2n()1n(
1
)2n()1n(n
1
3
1
n1
=)2n()1n(n3
1
18
1
n1=
nS9
1
18
1 [using (1)]
n1
=n
n
S18
2S 18S
n
n1 Sn + 2 = 0
Hence, 118S1nn
= | 2 | = 2. Ans.
For objective: Put n = 2 ]
Q.2
[Sol. We have f '(x) =h
)x(f)hx(fLim
0h
f '(x) =h
h
02x2f
2
h2x2f
Lim0h
=
h
2
)0(f)x2(f
2
)h2(f)x2(f
Lim0h
=h2
)0(f)h2(fLim
0h
= f '(0) = k (say)
f '(x) = kf(x) = kx + c ; f(0) = 0 c = 0 f(x) = kx
Now, I =
2
0
2dxxsinkx =
2
0
222dxxsinxsinkx2xk
=
2
0
32
3
xk
2
0
2
0
2dxxsindxxsinxk2 =
2k2
3
8k
32
=
k43
k8 32.
Hence I(k) =
k)4(k
3
8 23 .
This is a quadratic in k and its minimum value occurs when k = 316
34
= 2
4
3
f(x) = x4
32
; f( 42) = 3. Ans.]
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PAGE # 29
Practice Test Paper-7
Q.1
[Sol. m = 5! 2 5! = 10 4! 5!
n = 4! 5!
Hence m = 10n k = 10 Ans.]
Q.2
[Sol. Let k
zj
yi
xr
; x + y + z 12, y 3 and x, z 1(Give y = 3 and 1 to each x and z, add one extra beggar)
x' + y' + z' + u' = 7 Number of possible r
= 10C
3(Using beggar Method) Ans.]
Q.3
[Sol. Let f() = b f1(b) = f ( + h) = b k f1(b k) = + hf (h) = b + k f1(b + k) = h
x
y
O
y = f(x)
( , f( ))
Now L.H.D. of f1(x) at x = f () = b
(f1) ' (b) =b)kb(
)b(f)kb(fLim
11
0k
=
)(f)h(f
hLim
0h
=
h
)(f)h(f
1Lim
0h =
)('f
1 = r
1
||ly Now R.H.D. of f1(x) at x = f () = b
(f1) ' (b+) =b)kb(
)b(f)kb(fLim
11
0k
=
)(f)h(f
)h(Lim
0h
=
h
)(f)h(f
1Lim
0h
=)('f
1
=l
1
Aliter: Verification by taking an example
y = f (x) =
0xforx2
0xforx3
f ' (0) = 3 = l
f ' (0+) = 2 = r
g (y) = x = f1(y) =
0yfor2
y
0yfor3
y
x
y
O
y = f(x)
y = 3x
y = 2x
g ' (0+) = f ' (0+) =3
1=
l
1
g ' (0) = f ' (0+) =21 =
r1
Note: Ifl and r are positive, then L.H.D and R.H.D of f1 arer
1&
1
land if l and r are negative
then L.H.D. and R.H.D arel
1&
r
1. (Think) Ans.]
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Q.4
[Sol.
x= 1(0,0)
2( ,0)
X
Y
y = sinx
( /2, 1)y = 1 + cos x
y=x +x 10x3 2
Graph of f(x)
1
Clearly, from above graph, f(x) has local maximum at x =2
and absolute maximum at x = 1. Ans.]
Q.5
[Sol. f(x) =
xsinsin
2xcoscos
211
=
2x3,)2x(4
x2
5
2
3x)2x(
2)x2(
2
2
3x,
4x
2x
2
3xx
2x2
2
x2
,x2
x2
x2
2x0,x
2
222
22
2
2
f(x) =
xsinsin
2xcoscos
2
11
O
y
/2
3 /2 2x
2/4
2/4
f(x) =
2,0xx
2
2
,
f(x) =
x2
x2
f(x) =
2
x2
; x
,2
f(x) = ( x)24
2, < x
2
3,
f(x) =4
2 (2x)2
2
3< x 2
At x = , f(x) is not differentiable
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PAGE # 31
At x = , it is local as well as global minimum.
Range :
4,
4
22
,
I =
2
0
dx)x(f =
2
0
2
dxx2
Using king
I =
2
0
2 dxx =2
0
3
3
x
=
24
3Ans.]
Q.6
[Sol. Given,7
b
13
a =
15
k= k (say)
a = 13k, b = 7k, c = 15 k
Now, cos A = bc2
acb222
= )15()7(2
16922549
= 1052
105
= 2
1
A =A = 3
cos B =ac2
bca 222 =
)15()13(2
49225169 =
26
23
cos C =ab2
cba 222 =
)7()13(2
22549169 =
26
1.
As, cos C < 0,
C is obtuse. So ABC is obtuse.
Also,k35
k26k35s2
a2s2s
as
as
srr
1
=k35k9 =
359
Also, 2s
=
)k35()k35(
4Asinbc2
1
=2k3535
2
3k15k74
2
1
=3535
3105= 35:33
Also, tan
2
CB=
cb
cbcot
2
A= 3
k15k7
k15k7
= 322
8= 3
11
4.
Now, verify alternatives. Ans.]
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PAGE # 32
PART-C
Q.1
[Sol. Equation of S is (x 1)2 + (y 1)2 + (x + y 2) = 0x2 + y2 + x(2) + y(2) + 2(1 ) = 0 ........(1)
S=0
P(1,1) y=2xx2 + y2 + 2x + 2y 2 = 0 ........(2)
Given (1) and (2) are orthogonal, so
122
2)1(2
)2(2
= 2 (1 ) 2 2 + 2 = 2
4 = 4 = 1Hence equation of S is x2 + y2x y = 0.
Now, length of tangent from (2, 2) is 2244 = 2. Ans.]
Q.2
[Sol. Given a1
> 0.
Let d = common difference of A.P.
Now, 3a8
= 513
3 (a1 + 7d) = 5(a1 + 12d) 2a1 + 39d = 0 .......(1)As, a1
> 0 so d < 0.
Now, Sn
= d1na22
n1
= d1nd392
n = n40n
2
d 2 = 40020n2
d 2
Clearly, Sn
will maximum when n = 20. (As d < 0). Ans.]
Q.3
[Sol. Given, y = x + 2 .......(1)
Equation of CD
y 2 = 1 (x 0)
y = 2 x .......(2)
Area of triangle ABC =
168
13
14
3
8
120
2
1
=168
3148
120
6
1
P (0,2)1
(0,2)
y=2x
45 y = 2x + 10
P2
3
14,
3
8
x
y
O
C
DP (8, 6)3
y= 2x+10
m= 1
135
=168
2200120
6
1
= 20486
1 =
3
64
6
168 = .
213
64
. Ans.]
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PAGE # 33
Q.4
[Sol. Equating the components,
3x + 2
y + 4
z = x ; 2
x + 2
z = y & 4
x + 2
y + 3
z = z
hence (3 )x + 2y + 4z = 02x y + 2z = 04x + 2y + (3 )z = 0
for non trivial solution
324
22423
= 0
Use : C1 C
1 C3 and C3 C3 2C2
)1(21
)1(20
02)1(
= 0 ( + 1)2
121
20
021
( + 1)2[1(4) 2(2)] = 0 ( + 1)2 ( 8) = 0Hence = 1 or 8 sum = 7. Ans. ]
Q.5
[Sol. Given, ac2ycoscbxsinbar and cbar = 0
cba
(sin x + cos y + 2) = 0
But cba
0, so sin x + cos y = 2, which is possible when sin x = 1 and cos y = 1
For (x2 + y2) to be minimum, x2 =4
2and y2 = 2.
Hence, the minimum value of 220
(x2 + y2) =
2
2
2 4
20=
4520
2
2= 25. Ans.]
Q.6
[Sol. F(x) =
1x,2
)1(g)1(f
1x,2
)1(g)1(f
,11,x),x(g
1x1),x(f
continuous at x = 1 F(1
) = F(1
+
) = F(1)
f(1) = g(1+) =2
)1(g)1(f
4 + a = 1 + b =2
)b1()a4( =
2
ba5 .
b a = 3. .......(1)continuous at x = 1 F(1) = F(1+) = F(1)
b 1 = 4 a =2
)1b()a4(
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PAGE # 34
b 1 = 4 a =2
ab3
a + b = 5 .......(2)(1) and (2)
a = 1, b = 4 a2 + b2 = 17. Ans.]
Practice Test Paper-8
Q.1
[Sol. Given, z3 + iz2 = (1 + i) z1 z3z1 = i (z1z2)
12
13
zz
zz
= i
12
13
zz
zz
= 6
i
e
A(z )1
C(z )3B(z )2
/6
| z3z
1| = | z
2z
1| ABC is isosceles and A =
6
. ]
Q.2
[Sol. Urn-IBnWm
1
1 ball1
Urn-IIBnWm
2
2 ball1
Mixed lot item1
P(item belongs to Lot-I) =2
1= P(item belong to Lot-II)
P(item from the mixed lot is W) =
22
2
11
1
nm
m
nm
m
2
1
Ans.]
Q.3
Sol. F(x) = dx)x(P = x2013x20128x2 + 8x + C, where C is constant of integration.F(x) = x(x 1) (x20118) + C
F (0) = F(1) = F (81/2011) = C
F'(x) = 0 has atleast two real roots. (Using Rolle's Theorem)
Note that P(x) = 0 has exactly two real roots in
2011
1
8,0x ]
Q.4
Sol. Equation of normal at P is
(y 1) = 2(x 1) 2x + y = 3
At x =4
1, y =
2
7
0,
4
1
y,
4
1
4
1x
P(1, 1)
C
y
radius =22
2
71
4
11
=4
25
16
25 =
4
55Ans.
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PAGE # 35
Paragraph for Question no. 5 to 7
[Sol. Let f(x) = ax3 + bx2 + cx + d
f(0) = 5 d = 5So, f(x) = ax3 + bx2 + cx + 5
f '(x) = 3ax2 + 2bx + cNow, f '(2) = 0 12a 4b + c = 0 .....(1)
and y = f(x) passes through P (
2, 0), so 0 =
8a + 4b
2c + 5 ......(2)Also, f '(0) = 3 c = 3 .....(3)
On solving, we get a =2
1, b =
4
3
f(x) =23 x
4
3x
2
1
+ 3x + 5
(i)x
y
5
4
3
2
1
1O1
2
52
5
y = 3
Clearly, from above graph, we get, number of solutions of equation |)x(|f = 3 are 4. Ans.
(ii) Equation of normal at Q(0, 5) is
(y
5) = 3
1(x
0) x + 3y = 15. Ans.
(iii) Required area =
1
1
dx)x(f = dx5x3x4
3x
2
11
1
23
= dx5x4
32
1
0
2
=2
19. Ans.]
Q.8
Sol. Put x = tan
f(x) = sin1 2sin
=
20;
2sinsin
02
;2
sinsin
1
1
f(x) =
x0;xtan2
1
0x;xtan2
1
1
1
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PAGE # 36
f '(x) =
x0;)x1(2
1
0x;)x1(2
1
2
2
f '(1) =4
1; f ' (1) =
4
1 (A) is correct
Range of f (x) =
4
,0 . Hence (B) is incorrect.
Also f '(x) is an odd function. (C) is correct
x
)x(fLim
0x
2
1
2
1)0(fand
2
1)0(f,As (D) is incorrect]
Q.9
[Sol. The system will always have a solution (x, y) unless the two lines are parallel.
i.e. m = 2m
1 m = 1 ]
PART-C
Q.1
[Sol. A : outcome of the disease is positive
B1
: He has disease d1
B2
: He has disease d2
3
1)B(P
i
B3
: He has disease d3
P(A / B1
) =
10
8
P(A / B2) =
10
6
P(A / B3) =
10
4
P(B1/A) =
10
4
10
6
10
8
3
110
8
3
1
=18
8=
9
4 (p + q) = 4 + 9 = 13. Ans.
P(B2/A) =
18
6
P(B3/A) =
18
4]
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PAGE # 37
Q.2
[Sol. As, 32 rrn
=212
431
kji
= k5j10i10
3
kj2i2n
d = .nonOPofojectionPr
= 3
kj2i2kj2i3
=
3
146 =
3
9= 3. Ans.
Aliter: Equation of plane OQR is 2x 2y + z = 0
So, distance of P from plane OQR =144
1)2(2)3(2
=
3
9= 3. Ans.]
Q.3
[Sol. Equation of normal at P1
(4 cos 1, 3 sin
1) is
7sin
y3
cos
x4
11
... (1)
Also, equation of CQ1
is
y =
1
1
cos
sinx ... (2)
Solving (1) and (2), we get
1cos
x4
1sin
3
1
1
cos
sinx = 7
y
xC
(0, 0)
P (4 cos , 3sin )1 1 1 x +y =162 2
K1
1
Q1 1 1(4 cos , 4sin )
1
cos
x
= 7 x = 7 cos 1, y = 7 sin 1
So, K1 = (7 cos 1, 7 sin 1) CK1 = 7Similarly, CK2
= CK3
= ..... = CKn
= 7
n
1ii
175CK 7n = 175 n =7
175= 25. Ans.]
Practice Test Paper-9
Q.1
[Sol. Obviously for c (0, 1), f (x) lies obove the g (x)
also x2 = cx3 x = 0 or x =c
1
hence c1
0
32 dxcxx = 3c12
1
x
y
x2x3
O (0,0) x=1c
or 3c12
1=
3
2 c3 =
8
1 c =
2
1
Hence,
2
c
1
c
1= 2 + 4 = 6 Ans.]
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PAGE # 38
Q.2
Sol. Equation of chord of contact with respect to point (4, 2) is
2a
x4 2
b
y2= 1 and with respect to point (2, 1) is 1
b
y
a
x222
.
Now, according to given condition,
2
2
2
2
b1
a
2
b2
a
4
= 1 4
4
a
b=
4
1
2
2
a
b=
2
1
Now, e =2
3
2
11
a
b1
2
2
Ans.]
Q.3
[Sol. P(E) = P( R R W W B or R R R W W or W W W R R)
=
!2!2
!5 5
3
1+
!2!3
!5 5
3
1+ 5
3!2!3
!5
Simplifying P(E) = 53
50 k = 50 Ans.]
Q.4
Sol. 2
= 12
3
= 22 =
14
Hence
2
321
]
Paragraph for Question no. 5 to 7
[Sol. C1
: z + z = 2 | z 1 |
2x = 2 | x 1 + iy |
x2 = (x 1)2 + y2
y2 = 2x 1 y2 = 2
2
1x
C2
: arg )i1(z =
Curve C2 is a ray emanating from (1, 1) and making an angle from the positive real axis. C
1and C
2has exactly one common point
C2
must be a tangent to C1.
C2
: y + 1 = m(x + 1)
Solving, C1
and C2
y2 = 2
1m
1y 1
0,
2
1
(1,1)
my2 = 2(y + 1 m) m
my22y + 3m 2 = 0
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PAGE # 39
Put D = 0 4 4m (3m 2) = 0
3m22m 1 = 0 (3m + 1) (m 1) = 0 m =3
1, 1.
m =3
1rejected m = 1 (As, a (0, ) )
C2
: y + 1 = x + 1 y = xPutting y = x in the curve C
1
x2 = 2x 1 (x 1)2 = 0 x = 1 P (1, 1)Complex number corresponding to P is z
0= 1 + i
(i) | z0 | = 2
(ii)
From above graph , (x1
+ y1) = 3 + 4 = 7 Ans.
(iii) Area of the shaded region
= 1
2
1
dx1x22 =
1
2
1
2
3
22
3 )1x2(2
(1,1)P(z )0
Q(z ')0
4
4
=3
2[1 0] =
3
2sq. units.]
PART-B
Q.1
[Sol.
(A) The equation of circle passing through A(1, 2), B(2, 3) and having least posssible perimeter, is(x 1) (x 2) + (y 2) (y 3) = 0 (Circle described on AB as diameter)
x2 + y2 3x 5y + 8 = 0 ..........(5)If above equation of circle intersects orthogonally the circle x2 + y2 + 2x + 2ky 26 = 0
then, using condition of orthogonality,
we get
k
2
51
2
32 = 8 26 3 5k = 18 5k = 15
k = 3 Ans.
(B) Using the limit y 0 2y
ycos1 =21
We have,
a
2
20x x
)xcos1cos(1
)xcos1cos(1
xcos1cos(1cos1Lim
=2
1
a
42
20x x
)]xcos1[(
)xcos1(
)xcos1cos(1Lim
= a
84
20x x
x
x
xcos1Lim
8
1
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PAGE # 40
l =a8
0xxLim
128
1
For finite limit 8 a 0 a 8 Ans.
(C) If matrix A is non-singular, so det. (A) 0
1053
842
2p31 0 2(p + 2) + 12 0 p 4
So, p R {4}. Ans.
(D)1t
1y
1t
t
dt
dy
I.F. = (t + 1) et
(t + 1) et y = e t + CPut t = 0 and y = 1
C = 0 At t = 1
2e1 y = e1 y =2
1]
PART-C
Q.1
[Sol. Since it is isosceles. So AB = AC
Now, tan2
=
a
r1
r1
= a tan
2= 2 tan 15 = 2 32
Now r2
= 4r1
2Csin
2Bsin
2Asin 111
I
B C
A
2a = 4
B1 C1
A1
r2
r1
90
90
r2
= 4r1
sin 15 sin 15 sin 60
r2
= 2r1
(2sin2 15) sin 60 = 2r1
(1 cos 30) sin 60 = 2r1 2
3
2
31
= 2 2 32
2
32
2
3= 2323 = 1237
Now, (2r2 + 7r1) = 2 1237 + 7 322 =
24 + 28 = 4. Ans.]
Q.2
Sol. Given,
0
22 6dx)x(gxsin)x(g8
6dx.xsin4)x(gdxxsin16
0
22
0
4
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PAGE # 41
6dxxsin4)x(g16
332
0
22
6dxxsin4)x(g6
0
22
g (x) =4 sin2x Max. of g (x) = 4 sin2 x is 4. Ans.]
Practice Test Paper-10
Q.1
[Sol. Given,
n
0rr
nC1r
2r=
n
0rr
nC1r
11r
= 2n +
n
0r
rn
1r
C= 2n + S (where S =
n
0r
rn
1r
C)
S =
1n
C........
3
C
2
C
1
Cn
n
2
n
1
n0
n
= 1
0
ndx)x1( =
1n
121n
Now, 2n + S = 2n +1n
1
(2n + 1 1) =
1n
1
13n2n
Now,
1n
123nn
=6
128
n = 5 Ans.]
Q.2
[Sol. As,
sin (cos1x) = sin
xsin2
1= cos (sin1x)
Hence,
1
1
11 dcottan , where = cos (sin1x) =2
(2) = . Ans.]
Q.3
[Sol. Putting (, ) in y2 = 4x, we get = 0, 4
but 0 (Given)So, = 4
P = (4, 4) t1 = 2
P( , )P(4,4)
y2
=4x
S(1,0)
Q(9,6)
(0,0)V
m2
m1
y
x
L(1,2)
Hence t2
= 3
112 t
2ttUsing
Q = (9, 6)
Now m1
=3
4; m
2=
4
3
= 90Hence equation of line passing through (1, 2) and inclined at an angle = 90 is x = 1. ]
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PAGE # 42
Paragraph for question nos. 4 to 6
Sol.
(i) No American together
A1A
2A
3A
4B, C, D, E
For A1
we have four (B, C, D, E) favourable cases out of total cases 7.
Hence, probability =7
4(say) A
1and B are paired and the remaining
are, A2A
3A
4C D E
for A2, favourable cases 3 out of total cases 5
Hence, probability =5
3say AA
2and C are paired and the remaining are
Now, A3
A4
D, E
for A3, favourable cases 2 out of total 3.
Hence probability =3
2.
P (No two Americans together) =35
8
3
2
5
3
7
4 (B)
(ii) P (delegates of the same country form both pairs)A
1A
2A
3A
4E, B, C, D
= 115
1
7
3
[For A1
we have 3 favourable and of 7 and two A2
only 1 favourable out of 5 are for the remaining, no
constraints.]
EBCD
=35
3(B)
(iii) P (delegates of the same country not forming any pair+ forming both pair + forming exactly one pair) = 1
P (forming exactly one pair) = 1
35
3
35
8=
35
24 (C). Ans.
Alternatively: A1, A2, A3, A4, B, C, D, E
n(S) =!4)!2(
!84 = 105
(i) The probability that no two delegates of the same country are paired =105
CCC1
2
1
3
1
4
=
105
234=
35
8
(ii) The probability that delegates of the same country form two pairs =105
33=
35
3
(As A1, A
2, A
3, A
4can be paired in 3 ways and B, C, D, E can be paired in 3 ways.)
(iii) The probability that exactly two delegates of the same country are paired together
=105
34C2
4 =
35
24Ans.]
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PAGE # 43
PART-B
Q.1
[Sol.
(A) (z + i)2 = (z i)2 or (z i)2 ; if (z + i)2 = (z i)2 4zi = 0 z = 0if (z + i)2 = (z i)2 = i2 (z i)2
z + i = i (z i) or i (z i)
(1 i) z = i2 i = 1 iz = 1.
if z + i = iz + i2
(1 + i) z = 1 i = (1 + i)
z = 1
Hence, z = 0 or 1 or 1 3 solutions Ans.
Aliter: Given, 1iz
iz4
iz
iz= (1)1/4
we get, z = 1, 0,
1 ]
(B) We have,21
t
2
t
2= 1 t1 t2 = 4 ; also t2 = t1
1t
2....(1)
t1t2
= 21
t 2 4 + 2 = 21
t 21
t = 2
Also, 22
t = 21
t + 21
t
4+ 4 squaring (1)
90S(a,0)
P(t )1
Q(t )2
x
y
x+a=0
O
22
t = 2 + 2 + 4 = 8
Now, SQ = a(1 +22t ) = a(1 + 8) = 9a and SP = a(1 +
21t ) = a(1 + 2) = 3a
SP
SQ= 3 SQ = 3SP = 3 Ans.]
(C) On differentiating both sides with respect to x, we get
0 sin2x f(sin x) cos x = cos x
f(sin x) =xsin
12
2)2(2
1f
2
Ans.
(D) Given, 4y2 + 2 cos2x = 4y sin2x 4y24y + 1 + cos2x = 0 (2y 1)2 + cos2x = 0
y =2
1and cos x = 0 x =
2
,
2
3
So, two ordered pairs are possible i.e.,
2
1,
2and
2
1,
2
3Ans.]
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PAGE # 44
PART-C
Q.1
[Sol. Given, Re(z) 2 = | z 7 + 2i |
(x 2)2 = (x 7)2 + (y + 2)2
(y + 2)2 = 10
2
9x
So, the given locus is that of a parabola with directrix
x = 2 and focus (7, 2).
x=2
x
2
P(z1)P(7, 3)
Q(z2)
Q(7,7)
(2, )O S(7,2)92
, 2
y
V
Clearly, minimum PQ = l(L.R.) = 10 Ans.]
Q.2
Sol. An =
2/1
0
n
dxx AAn = )1n(2
11n 2
n
An = )1n(2
1
n
A2n
n
=n)1n(2
1
O
2
1
x
y
Hence,
n
1n
nn
n
A2=
2
1
n
1n 1n
1
n
1=
2
1
1n
11
Given,2
1
1n
11 =
3
1 1
3
2=
1n
1
3
1=
1n
1
n = 2. Ans.]
Q.3
[Sol. We have, (3p2pq + 2q2) wvu
= 0
But, wvu
0 3p2 pq + 2q2 = 0Now divide by q2 (assuming q 0)
3t2 t + 2 = 0, where t =q
pand t R
but D < 0, hence no real values of t
only solution is p = 0 and q = 0Hence exacly one ordered pair of (p , q) i.e. (0, 0).
Aliter: We have, (3p2pq + 2q2) wvu
= 0
But, wvu
0 3p2 pq + 2q2 = 0 2p2 + p2 pq +2
2
q
+
4
q7 2= 0
2p2 +2
2
qp
+4
q72
= 0 p = 0, q = 0, p =2
q
This is possible only when p = 0, q = 0
i.e. exacly one ordered pair of (p , q)]
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Practice Test Paper-11
Q.1
[Sol. Given, f(x) = x3ax2 + 2x, x [0, 2]Now, f ' (x) = 3x22ax + 2
f(2) = (2)34a + 4 = 12 4a and f(0) = 0
Using LMVT, we have
02
)0(f)2(f
2
1'f
2
0)a412(2a
4
3
a4
11 = 6 2a a = 6
4
11 a =
4
13. Ans.]
Q.2
[Sol. Equation of line through O(0, 0, 0) and perpendicular to the plane 2x y z = 4, is
1
0z
1
0y
2
0x
= t (let)
Any point on it is (2t, t, t)
As above point lies on the plane 3x
5y + 2z = 6, so
6t + 5t 2t = 6 9t = 6 t =3
2.
Co-ordinates of point of intersection are
3
2,
3
2,
3
4 (x
0, y
0, z
0) [Given]
Hence, (2x03y0 + z0) = 4 Ans.]
Q.3
[Sol. Given, z + w = 0 ..........(1)
and z
2
+ w
2
= 1 .........(2) Putting w = z from (1) in (2), we get
2z2 = 1 z = 2
1.
For z =2
1, w =
2
1and for z =
2
1, w =
2
1.
So, from both possibility, we get
2wz . Ans.]
Q.4[Sol.
We must have |2 4| < 5 5 < 2 4 < 51 < ( 2)2 < 9 0 ( 2)2 < 9 3 < 2 < 3
D(z)
2 F1F2
1 < < 5 0 < < 5 = 1, 2, 3, 4 4 values. ][Note : = 0, 4 is not possible because, | z 2 | + | z 4| = 5 will represent a circle.]
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Paragraph for question nos. 5 to 7
[Sol.mb
Given, f(x) = x2 ex f '(x) = x ex (2 x)
+
0 2
Sign scheme of f '(x) Ox
y
4
1y
2e
4y,2x
x = 2
Graph of f(x) = x e2 x
As, g(x) =
xe2
02
dtt1
)t('f g'(x) =
)e41(
e2)e2('fx2
xx
=
)e41(
)e1(ee8x2
xe2x2 x
.
Now, verify alternatives. ]
Q.8
[Sol.
(A) P(A/B) = P(B/A) P(A) = P(B)
Now, P(A B) = P(A) + P(B) P(A B) P(A B) = P(A) + P(B) P(A B)
P(A B) = 2P(A) 1 > 0; P(A) >2
1 True
(B) P(B) =4
3; P(A/ B) =
2
1
)B(P
)BA(P =
2
1 P(A B) =
4
3
2
1=
8
3
Now, P(A) + P(B) P(A B)= P(A B) 1
P(A) +43
83 1
P(A) 8
5
P(A)]max.
=8
5 True
(C) P(AC BC)C = C4443)aa()aa( = (a4)
C = a1
+ a2
+ a3
= P(A + B)
P(AC BC) = (a1
+ a3
+ a4)C = a
2= P(AB)
Hence P(AC BC) + P(AC BC) = P(A) + P(B)] = 6 233121
= 65 .
(D) Given, A is subset of B.
Now, P(B/A) =)A(P
)AB(P =
)A(P
)A(P= 1.
S
AB
]
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Q.9
[Sol. As, 2213 rrrr
= 122221 rrrrrr
= )k3ji(6)kji2(6 = k12i6
Row 3 vector 556
)k2i(6 = k2i or k2i
A =
201112
311
or
201112
311
Tr.(A) = 0, 4.
Also 3232 rrrr = |rr||r||r| 3232 = 5656 = 30.
Since 321 randr,r
are coplanar they are linearly dependent.
133221rrrrrr
= 2321rrr
= 0]
PART-C
Q.1
[Sol. Clearly, the given curves intersect at x = 0,
1k
k2
Required area = dxk
xkxx
1k
k
0
22
2
= 2
k
1k6
1
Now, for maximum area,2
k
1k
will be minimum k +k
1= 2 k = 1. Ans.]
Q.2[Sol. Let = 3h 2, = 3k
3
2= h and
3
= k
As, (h, k) lies on given circle, so
h2 + k22h 4k4 = 0
43
4)2(
3
2
99
)2(22
2 + 2 2 12 = 44 ( 1)2 + ( 6)2 = 44 + 37 Locus of (, ) is (x 1)2 + (y 6)2 = 81,which represents circle whose radius = 9.]
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PAGE # 48
Q.3
[Sol. Let f n(x) =
x2
x
t dten
)x('fn =
nn x)x2(ee2
For maxima and minima, )x(f'n
= 0 nn x)x2( ee2
nnn x)x2( ee2
Taking log on both sides, we get
ln 2 2nxn = xn ln 2 = xn (2n 1) xn =12
2nn
l x =
n
1
n12
2n
l= a
n
Also, f n''
n
1
n 12
2nx
l< 0 f
n(x) is maximum at x =
n
1
n12
2n
l.
Now, ln an
=n
12
2nn
n
ll
=n
)12(n)2n(nn
lll
Hence L = )a(nLim nn
l
=n
)12(n)2n(nLim
n
n
lll=
n
2
112n
n
)2n(nLim
n
n
n
lll
L =
n2
11n2nn
0Lim
n
n
ll
=
ln 2
Hence, eL = 2 Ans.]
Q.4
[Sol.mb
We know that | adj. A1 | = | A1 |2 = 2|A|
1
11A.adj.det =|A.adj|
11 = | A ||
2 = 22 = 4 Ans.]
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PAGE # 49
Practice Test Paper-12Q.1
[Sol. The equation of the tangent is )i..(..........12
3
b
y
2
1
a
x
Auxiliary circle is x2 + y2 = a2.................(ii)
C is the centre.
Combined equation of CL, CM is obtained by homgenising (ii) with (i), i.e.,
x2 + y2a2 0b2y3
a2x
2
Since LCM = 90
1 4
1+1 0
b4
a32
2
4
7
b4
a32
2
90
M
P( =60)
(0, 0)C
y
x
L
7b2 = 3a2 7 a2 (1e2) = 3a2
Hence e =7
2Ans. ]
Q.2
[Sol. Originally the number of white balls in the bag vary from 0 to 100.
P(B0) = P(B
1) = P(B
100) =
101
1
ballW100hasBag:B
ballW2hasBag:BballW1hasBag:B
ballWnohasBag:Binitially
100
2
1
0
A
B0 B1
B2 B100
After the white has been dropped in the bag
A = A B0
+ A B1
+ ........+ A B100
P(A) = P(B0) P(A/B0) + P(B1) + P(A/B2) + ...... + P(B100) P(A/B100)
=101
1
101
101......
101
2
101
1=
101
51
2
102101
101
1
101
1
. Ans.]
Q.3
[Sol. From above figure,
P(C (A B)') = 1
5
1
15
1
10
1 BA
S
C=
30
62330 =
30
19Ans.
Q.4
[Sol. (a + b + c2) (a + b2 + c) = | a + b + c2 |2 = (a2 + b2 + c2ab bc ca) (a b)2 + (b c)2 + (c a)2 = 2 (a = b, | b c | = 1) or (b = c, | a c | = 1) or (c = a, | a b | = 1)| b c | = 1
(b, c) = {(2, 1), (1, 2), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6)}
Required probability =666
103
=36
5Ans.]
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PAGE # 50
Paragraph for question nos. 5 to 7
[Sol. Let dr's of line L be , so equation of line L is
kcjbiatr
......(1)
If L intersects L1
at P, so shortest distance between them is zero.
0 =
kj2ikcjbiakj2ikcjbiakji2
3
4,
3
10,
3
10
Q
(5, 5, 2)P
0O
L : r =2
L : r =1
8i3
3j + k
+ (2i j+ k)
(2i + j k)
+ (i 2j + k)
121
cba
kji
kji2
= 0 a + 3b + 5c = 0 .......(2)
Similarly, L intersects L2
at Q, so shortest distance between them is zero.
0 =
kji2kcjbia
kji2kcjbiakj33
i8
112cba
kji
k
j
33
i8
= 0
3a + b 5c = 0 .......(3)
On solving (2) and (3), we get2
c
5
b
5
a
.
So, the equation of line L is k2j5i5tr
.
Any point on line L is A' (5t, 5t, 2t). If A' is the point of intersection of L and L1, so A' will also satisfy
L1, we get 1
1t2
2
1t5
1
2t5
2 (5t 2) = 5t 1 10t + 4 = 5t 1 t = 1So, co-ordinates of P are (5, 5, 2).
Similarly, if A' (5t, 5t, 2t) is the point of intersection of L and L2, so A' will also satisfy L2, we get
1
1t2
1
3t5
23
8t5
5t + 3 = 1 2t t =
3
2
So, co-ordinates of Q are
3
4,
3
10,
3
10
(i) Given M (1, 2, 3) and N (2 + , 1 2, 1)
MN = P.v of N P.v ot M = k)4(j)21(i)1(
Also, n
= normal vector of plane k3j4ir
= k3j4i .
Now, nMN
= 0 1 ( + 1) + 4 (1 + 2) + 3 ( 4) = 0
12 = 7 =12
7 Option (B) is correct.
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PAGE # 51
(ii) The normal vector of the plane through P (5, 5, 2) and Q
3
4,
3
10,
3
10and perpendicular to the
plane kjir
+ 1 = 0 is parallel to the vector = kjiPQn
=
111
3
2
3
5
3
5kji
=
111
255
kji
3
1
= k0j3i3
3
1 = ji
The required equation of plane, is 1 (x 5) + 1 (y + 5) + 0 (z 2) = 0
x + y = 0 or jir
= 0 Option (D) is correct
(iii) Volume of tetrahedron OPAB (where O is origin) = OBOAOP6
= |
502
310
255
|6
1
=6
1[5(5) + 5(6) + 2(2)] =
6
51=
2
17. Option (C) is correct.]
Q.8
[Sol. S = {HTH, THH, TTH, HHH, HTT, THT, TTT, HHT}
A = {HTH, HHH, HTT, HHT}
B = {HTH, TTH, HTT, TTT}
C = {HTH, THH, TTH, HHH}
D = {TTH, HTT, THT}
E = {HHH, TTT}
A B C = {HTH}
As, P(C) =2
1, P(D) =
8
3, P(E) =
4
1
2 P(D) =4
3= P(C) + P(E) (A) is correct
Also, A B C = {HTH, THH, TTH, HHH, HTT, TTT, HHT} A B C S, as THT is not included in A B C. A, B, C are not exhaustive (B) is incorrect
Also, P(A B C) =8
1P(A) P(B) P(C)
A, B, C are independent events. (C) is correct
Note that, P(A) = 2
1= P(B) = P(C)
A,B,C are equally likely. (D) is correct Ans.]
Q.9
[Sol. Given, + = p, = q ;1
+ = p1 ,
= q1
; and +1
= p2 ,
= q2
Now, q1
q2
=
= 1. and p1
+ p2
=
1
=
)1()(=
q
p(q + 1)
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PAGE # 52
Also, qq2
=
= 2, as is the root of x2 + px + q = 0.
So, 2 + p + q = 0 qqpqq 22 = 0
Note that (p1 p2)
2 = ()2 2
2
)(
)1(
= (p2 4q)
q
)1q( 2
(qp1 qp
2)2 = (p24q) (q + 1)2. Ans.]
PART-C
Q.1
[Sol. Let c = 14 x, a = 14, b = 14 + x
Now, cos B =13
5 (14 + x)2 = (14 x)2 + 1422(14 x) 14
13
5 x = 1.
So, a = 14, b = 15, c = 13.
Hence, r =s
=
21
84= 4. Ans.]
Q.2Sol. Sn
= C0C
1+ C
1C
2+ ... + C
n1C
n
Sn
= nC0
.nCn1
+ nC1
.nCn2
+ ... + nCn1
.nC0
= 2nCn1
Sn
= 2nCn1
Now,n
1n
S
S =
4
15
1nn2
n2n2
C
C
=4
15
4
15
!n2
)!1n()!1n(
)!2n(!n
)!2n2(
n)2n(
)1n2)(1n(
=8
158(2n2 + 3n + 1) = 15n2 + 30n
16n2 + 24n + 8 = 15n2 + 30n n26n + 8 = 0 (n 4) (n 2) = 0 n = 2 or 4
Sum of all values of n = 6 ]
Q.3
Sol. Given equation of planes are
P1
: x + y + 1 = 1 ... (1)
P2
: x + 2ay + z = 2 ... (2)
P3
: ax + a2y + z = 3 ... (3)
If 3 planes intersect in a line, then
01aa1a21
111
2
Applying C1C
1C
2and C
2C
2C
3
011aaa
11a2a21
100
22
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PAGE # 53
On expanding along C1, we get
(1 2a) (a21) (a a2)(2a 1) = 0 1 + 3a 2a2 = 0 a =2
1, 1.
But, for a = 1, planes P1
and P3
are parallel and for a =2
1, planes P
1and P
2are parallel.
Hence, we conclude that 3 planes will never intersect in a line for any real value of a.
Practice Test Paper-13
Q.1
Sol. Given, tan1 x + tan1 y + tan1(xy) =12
11...(1)
At x = 1;4
+ 2 tan1 y =
12
11 2 tan1 y =
3
2 y = 3
Differentiate both sides of equation (1) with respect to x, we get
2x1
1
+ 2y1
1
y ' + 2)xy(1
1
(xy' + y) = 0
2
1+
4
1y' +
4
1(y' + 3 ) = 0
2
1+
4
3+
2
1y' = 0 2 + 3 + 2y' = 0
So, y' = 1 2
3. Ans.]
Q.2
[Sol. As, f(0+) = f(0) = f(0) = 0, so f(x) is continuous at x = 0.
Further, f(0 + h) > f(0) and f(0 h) > f(0) where h is sufficiently small positive quantity.
Hence, f(x) has local minimum at x = 0. ]
Q.3
[Sol. We have, dyy
dxydyx2
y
xd = dy
y
x= y + C
As, y(1) = 1 C = 2 y
x= y + 2
Now, y() = 3 3
= 5 = 15. Ans.]
Q.4
[Sol. Given, f (x) =xtan1
9xtan4xtan2
2
=
xtan1
)xtan2(22
+
xtan1
xtan14
2
2
+ 5 = 2 sin 2x + 4 cos 2x + 5
Rf= 205,205
Hence, (M + m) = 10 Ans.
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PAGE # 54
Aliter : f(x) =xtan1
xtan2
2
+
xtan1
xtan42
2
+
xtan1
92
= sin2x + 2 sin 2x + 9 cos2x
= 1 + 4(1 + cos 2x) + 2 sin 2x = 5 + 2 sin 2x + 4 cos 2x. ]
Paragraph for question nos. 5 to 7
Sol. For the given ellipse, 116
y
25
x22
, 5
3
25
16
1e . So, eccentricity of hyperbola = 3
5
.
Let the hyperbola be, 1B
y
A
x2
2
2
2
... (1)
Then, B2 = A2
19
25=
9
16A2. Also, foci of ellipse are (3, 0).
As, hyperbola passes through (3, 0). So, 2A
9= 1 A2 = 9, B2 = 16
Equation of hyperbola is 116y
9x
22
(i) Vertices of hyperbola are (3, 0) (A) is correct.Focal length of hyperbola = 10 (B) is incorrect.
Equation of directrices of hyperbola are x = 5
9. (C) is incorrect.
(ii) Any point of hyperbola is P(3sec, 4tan).Equation of auxiliary circle of ellipse is x2 + y2 = 25.
Equation of chord of contact to the circle x2 + y2 = 25, with respect to P(3 sec, 4 tan), is3x sec + 4y tan = 25 ... (1)
If (h, k) is the mid point of chord of contact, then its equation is
hx + ky 25 = h2 + k225 hx + ky = h2 + k2 ... (2)As, equations (1) and (2) represent the same straight line, so on comparing, we get
22kh
25
k
tan4
h
sec3
sec =
22 kh
25.
3
h, tan =
4
k
kh
2522
Eliminating , we get,2
22kh
25
16
k
9
h22
= 1. (As, sec2tan2 = 1)
Locus of (h, k) is
22222
25
yx
16
y
9
x
(iii) Required area of quadrilateral = 2(A2 + B2) = 2(9 + 16) = 50 Ans.]
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PAGE # 55
Q.8
Sol. S = {1, 2, 3, ... n}
E1
= No. is divisible by 2.
E2
= No. is divisible by 3.
If n = 6k say n = 6
S = {1, 2, 3, 4, 5, 6}
P(E1) =
6
3=
2
1; P(E
2) =
6
2=
3
1
P(E1 E
2) =
6
1= P(E
1).P(E
2) (B) is correct.
If n = 6k + 2 say n = 8
S = {1, 2, 3, 4, 5, 6, 7, 8}
P(E1) =
8
4=
2
1; P(E
2) =
8
2=
4
1
Here, P(E1E
2) =
8
1= P(E
1) P(E
2) (C) is correct.
Note that : P(E1 E2) = 101
P(E1) P(E2)
Not independent dependent. ]
Q.9
[Sol. 2xydx
dy= x2 + y2 + 1
Put y2 = t
dx
dtx = x2 + t + 1
dx
dt
x
t=
x
1x2
I.F. =x
1
Hencex
t=
dx
x
1x2
2
= x x
1+ C
y2 = x21, C = 0 as y(1) = 0
Now, y (x0) = 3 3 = 20x 1
20
x = 4
x0
= 2 Ans.
Alternatively: We have, 2xydy = (x2 + 1)dx + y2dx
2
2
x
dxydyxy2 = dx
x
11
2
x
1xd
x
yd
2
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PAGE # 56
We get, cx
1x
x
y2
As, y (1) = 0 c = 0 y2 = x21
Now, y (x0) = 3 3 = 20x 1
20
x = 4
x0 = 2 Ans.]
PART-C
Q.1
[Sol.
P(HHH) =
33
6
1
3
11
18
5
2
1
18
13
=
8
1
18
5
8
1
18
13 =
8
1 p + q = 9 Ans.]
Q.2
[Sol. Clearly, d =22
4
3
2
1
2
1
4
3
=16
1
16
1 =
8
1=
22
1
So, | sin | = d22 | sin | = 1
y
x
0,
2
1
2
1,0
4
3,
2
1
2
1,
4
3
y=
x
O
=2
,
2
,
2
3.
Hence, the number of values of are 3.]
Q.3
[Sol. Given,dx
dyx2y = x4 y2 or y
x
2
dx
dy = x3y2. Now dividing by y2, we get
x
1
y
2
dx
dy
y
12
= x3 ........(i)
This is a Bernouli's differential equation, substituting t
y
2
, we get
dx
dt
dx
dy
y
22
. So, equation (i) becomes
x
t
dx
dt
2
1 = x3 t
x
2
dx
dt = 2x3
IF =2xnxn2
dxx
2
xeee2
ll
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PAGE # 57
So, general solution is given by
x2t =6
x26
+ C y
2x2=
3
x6
+ C y
2=
3
x4
+ 2x
C
If x = 1, y = 6 C = 0
3
x
y
2 4
y = 4x
6i.e. f(x) = 4x
6
Now,dxdy
= 24x5 = 5x
24. Hence,
5
1
3xdxdy
=
324
= 8. Ans.]
Q.4
[Sol. Given, y = x2
Now,axdx
dy
=
axx2 = 2a
Equation of tangent is (y a2) = 2a (x a) BOx
y
A (a, a )2
0,2
a
Put y = 0, we get
x = a 2
a=
2
a.
Also, equation of OA is y = ax
Area = a
0
2dx)xax( = k
2
a
2
a2
a
0
32
3
x
2
ax
=
4
ka3
3
a
2
a33
=4
ka3
6
a3
=4
ka3
k =3
2
q
p(Given) (p + q)
least= 5. Ans.]
Practice Test Paper-14
Q.1
[Sol. Tangent to the parabola
y2 = 4x is y = mx +m
1....(i)
m2x my + 1 = 0,
As, it touches the circle x2 + y2 = 1, soS(1,0)O
y
x
1
mm
1
24
m4 + m21 = 0
m2 = tan2 =2
411 =
4
152
2
15= 2 sin 18 Ans.]
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Q.2
Sol. Given, 3
2 III
7dx)x(''f)x3(
32)x('f)x3( +
3
2
7dx)x('f
0 (f '(2)) + f(3) f(2) + 7 f(3) = f '(2) + f(2) + 7 = 4 + (1) + 7 = 10. Ans.
Q.3
[Sol. Let A =
333231
232221
131211
aaa
aaa
aaa
, where aij {0, 1}
As, trace A = 1
So, any two elements in main diagonal are 0 and one element is 1. Also, non-diagonal elements an be 0
or 1.
So, number of matrices = 3 23 = 3 8 = 24. Ans.]
Q.4
Sol. As, z lies on the curve arg(z + i) =4
, which is a ray originating from (i) and lying right side of
imaginary axis making an angle4
with the real axis in anticlockwise sense.
O45
Re(z)
Im(z)
(4+3i)
(43i)
The value of | z(4 + 3i) | + | z (4 3i) | will be minimum when z, 4 + 3i, 4 3i are collinear. Minimum value = distance between (4 + 3i) and (4 3i)
= 22 )33()44( = 3664 = 100 = 10. Ans.]
Q.5
[Sol. Statement-1: We have baxa
0bxa
atbx
, for some scalar t.
atbx
= k)t31(j)t21(i)t2(
Now, xa
= 0 t = 21
2
k
2
i3x
x
=4
1
4
9 =
2
5 Statement-1 is false.
Obviously, Statement-2 is true. ]
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PAGE # 59
Q.6
[Sol. Option (B) is true.
S-1: As, f ' (1+) = 0 = f '(11) f is differentiable at x = 1.So, f is differentaible x R.
S-2: As, f(1 + h) < f(1) < f (1 h).
Where h is sufficienty small positive quantity, so f(x) is decreasing at x = 1.
f(x) has neither local maximum nor local minimum at x = 1.But, S-2 is not explaining S-1. Ans.]
Q.7[Sol. Given, = 2 tan = tan 2
tan =
2tan1
tan2
2x
y
0
0
=
202
0
00
y1x
1xy2
20
20
0
0 y)1x(
)1x(2
2x
1
3x02 y
02 = 3
(1,0) (2,0)x
y
M(x , y )0 0
(0,0)A B
Locus of M is hyperbola 3x2y2 = 3Now, verify alternatives. Ans.]
Q.8
[Sol. Given, f(x) =
0x,0
1x0,)x2(cos1xx1
Note: f (x) is not defined at x = 1 for 0 ......(1)f(x) is continuous at x = 0 but for continuous at x = 1.
f (1) = f(1)
2 hsinhLim 20h
= 0 20h
hLim
= 0
+ 2 > 0 > 2 ......(2)Hence, (1) (2) 0
Note: f (x) is differentiable in (0, 1). Ans.]
Q.9
[Sol. Now, slope of LM =0t
2
= t (given)
2 = t2 = 2 + t2
Let mid point of LM is (h, k)
(h, k)
y
x
slope = m = t(given)
(0, 0)
M(0, ) L (t, 2)
Now, 2h = t and 2k = + 2 = t2 + 4 On eliminating t, we get
k = 2(h2 + 1)
Locus of (h, k) is y = 2(x2 + 1) x2 =2
1(y 2)
Now, verify alternative. Ans.]
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PAGE # 60
PART-B
Q.1
[Sol.
(A) Clearly, ABC is equilateral.
Now, ar. (ABC) =2
zz4
3
=2
z4
33= 348 (Given)
B( z)
A(z)
Re(z)
C( z) 2
2 /3
2 /3
Im(z)
O(0,0)
64z2
8z . Ans.]
(B) Equation of chord of hyperbola 11
y
2
x22
, whose mid-point is (h, k) is
2
hxky =
2
h2
1
k2
(using T = S1)
As, it is tangent to the circle x2 + y2 = 4, so
22
22
k4
h
k2
h
= 2
2
22
22
k4
h4k
2
h
Locus of (h, k) is (x22y2)2 = 4(x2 + 4y2) = 4.
(C) We know that ac = (semi-minor axis)2 = 4
Now,
4
4
dx}x2{ = 2/1
0
dx}x2{16 =
2
18
2
18
dx}x2{ =
2
1
0
dx}x2{16 =
2
1
0
dxx216 =
2
1
0
dxx32
= 322
1
0
2
2
x
= 32
8
1= 4 Ans. ]
PART-C
Q.1[Sol Given, A = (I + B) (I B)1
Now, AT
= T1
)BI(BI
= T1
)BI(
(I + B)T
= 1T)BI( (I + BT) = (I + B)1 (I B)Also, AAT = (I + B) (I B)1 (I + B)1 (I B)
= (I + B) ((I + B) (I B))1(I B) = (I + B) ((I B) (I + B))1 (I B)= (I + B) (I + B)1 (I B)1 (I B) = I I = I2 = I.
| AAT | = | I | = 1 | A |2 = 1As, | A | > 0 | A | = 1.Hence, det.(2A) det. (adj A) = 8 det.( A) (det.( A))2 = 8(1) (1)2 = 7 Ans.]
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Q.2
[Sol. p = sinv
=|c|
cv
=
|c|
cv|c|v222
=3
4)3)(6( =
3
14
B(3,5,2) M Q
A(2,3,1)v=(1, 2, 1)
c=(1, 1, 1)
r
r r
r
= v ^ cp
30p2 = (30) 314
= 140 Ans.]
Q.3
[Sol. 2
4
5
2
0
4
1
0
4
5
4
1
dxxcosdxxsindxxcosdx)x(F
=
2
4
54
5
4
14
1
0
xsinxcosxsin
y
xO 1/4
5/4 2
=
2
1
2
2
2
1=
22
10
10
2
024
10dx)x(F
24
10dx)x(F
24
22
= 5. Ans.]