practice test papers xiii vxy 1 to 6 sol
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PRACTICE TESTSOLUTIONS
TARGET IIT JEE 2012
MATHEMATICS
CONTENTS
PRACTICE TEST1 ........................................................... Page 2
PRACTICE TEST2 ........................................................... Page 6
PRACTICE TEST3 ........................................................... Page 10
PRACTICE TEST4 ........................................................... Page 15
ANSWER KEY .................................................................... Page 19
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PAGE # 2
[STRAIGHT OBJECTIVE TYPE]1. a + b + c = 3 .......(1); abc = 1 .......(2)
31
31
31
cba = 1
1
31
31
31
3
3
313
313
31
cba3cba
= 0
Hence, either 31
31
31
cba = 0 or 31
31
31
cba which is not possible, think!. ]
2. x2(x 1) + a(x 1) = 0 (x2 +a)(x 1) = 0hence x = 1, for two roots x2 + a = 0 should give two coincident roots i.e. x2 = 0 or one root 1 andother different from 1 i.e. x2 = 1x = 1Hence a = 0 or a = 1 ]
3. Common difference is same9 p = 3p q 9 = 2q p = 5, q = 2Hence a = 5 and d = 4t2010 = 5 + (2009)4 = 5 + 8036 = 8041 Ans. ]
4. = 21
8 3 = 12
R = 4
abc= 124
855
A
B C
b = 53
44
c = 5
aR = 6
25 ]
5. Given,
2BA
sin
2BA
cos
2BA
cos
2BA
sin =
151
7878
baba
2BA
tan
2BA
tan
(Using Napier's analogy) . Ans.]
6. Given,
=
22
)()(
= 2
2
)()(
(Apply componendo dividendo and take square on both sides, we get)
PRACTICE TEST # 1Syllabus : Logarithms, Quadratic Equation and expression, Compound angle Trigonometric equation and
inequations, Solution of triangle, Sequence and Progression.1. Only one is correct : 3 marks each.2. One or more than one is/are correct : 4 marks each.3. Matrix Match : 3 marks for each row.4. Integer type : 5 marks each.Time : 60 min. approx. Marks : 82
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PAGE # 3
22
2
)()()(
= 22
2
)()()(
4)( 2
=
4)( 2
q4p2
=
c4b2
p2c = b2q
Hence, (p2c b2q) = 0. Ans.]
7. Given, (sin x + cos x) = (sin x cos x 1) (sin x + cos x)2 = (sin x cos x 1)2 (Squaring both sides) 2 sin x cos x = sin2x cos2x 2 sin x cos x 4 sin x cos x = sin2x cos2x sin x cos x(4 sin x cos x) = 0 sin x cos x = 0 (sin x cos x 4)
x , 23
, 3, 27
, .......... x = (2n 1) or x = (2n + 1) 2
, n N ]
[COMPREHENSION TYPE]Paragraph for question nos. 8 to 10
[Sol.
(i) f(x) = x2 6mx + m2 + 4m + 2 ; Vertex = 2m6
= 3m
f(3m) = 9m2 18m2 + m2 + 4m + 2 = 8m2 + 4m + 2 = 8
41
2m
m2 = 2
5 8
2
41
m
f(3m) is maximum if m = 41
.
(ii) f(x) > 0 x R x2 6mx + m2 + 4m + 2 > 0 x RD < 0 36m2 4(m2 + 4m + 2) < 0 32m2 16m 8 0 8(4m2 2m 1) 0 4m2 2m 1 0
m = 8
1642 =
8522
=
81642
=
451
451
,
451
m
451
4510
largest integral value of 4236.21
= 4236.3
Hence, m = 0(iii) Given minimum value of f(x) is 2 for x 0Case-I : when vertex is 0
i.e., 3m 0 m 0minimum f(x) occurs at x = 3m
O
f(x)
x
y
f(3m) = 8m2 + 4m + 2 = 28m2 4m 4 = 0
m = 1, m = 21
. But m = 21
(rejects) m = 1
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PAGE # 4
Case-II : when vertex is < 0 i.e. m < 0In this case, minimum occurs at x = 0
f(0) = m2 + 4m + 2 = 2m2 + 4m + 4 = 0(m + 2)2 = 0 m = 2
m 1,2 . ]
[MULTIPLE OBJECTIVE TYPE]11. a
n = 53 + (n 1)(2) = 55 2n
As, an < 0 55 2n < 0 n > 2
55
number of terms = 27
S27 = 227 [2 53 + 26 ( 2)] = 2
27 [106 52] = 542
27 = (27)2 = 729. Ans.]
12. As, 101 99 cos 7x + 20 sin 7x 101
101 21a + 11 101 112 21a 90 2190
a21112
5.3 a 4.2
a = 5, 4, 3, 2, 1, 0, 1, 2, 3, 4 10 integral values of a . Ans.]
13. Sum of roots = a
b < 0
Since a > 0 b > 0
Product of roots = a
c < 0 c < 0
Now verify alternatives. ]
14.
2x
sin2x
cos
2x
sin2x
cos
Now define modulus and note that
2x
4tan =
2x
4cot . Ans.]
15. yx y xlogy 10 = log10y ........(1)
and 4y xy ylogy 10 = 4 log10 x ........(2)
On putting, log10x = y1
log10y from (1) in (2), we get
ylogy 10 = y4
log10 y
y4y
log10y = 0 Either y = 1 or y = 4.
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PAGE # 5
If y = 1, then x = 1 and if y = 4, then x = 2 Possible ordered pairs are (1, 1) and (2, 4). Ans.]
[MATCH THE COLUMN]16. Obviously, when b 0, we have no real roots as all the terms becomes positive. Also for b = 2,
We have x2 2| x | + 1 = 0 21x = 0 1x x = 1.Hence, the given equation has two distinct real roots.
Also, when b < 2, then 02
4bbx
2
Hence, the given equation has four distinct real roots, as | b | > 4b2 .Clearly, the above given equation can never have three distinct real roots for any real value of b. ]
[INTEGER TYPE / SUBJECTIVE]17. If log2(x 1) < 0
Case-I: x 1 < 1 1 < x < 2, , so 1xlog)1x(log
2
2
= 1
The given equation becomes x2 3x + 1 = 1 (x 2) (x 1) = 0 x = 1, 2As 1 < x < 2so x = 1, 2 (both rejected)Case-II : When x > 2, log2(x 1) > 0 (x 1) > 1 x > 2
log2(x 1) > 0, so 1xlog)1x(log
2
2
= 1.
Equation becomesx2 3x + 1 = 1 x (x 3) = 0
but x 0 x = 3. Number of solution = 1. Ans.]
18. x2 4x + 4 < 0 (x 1) (x 3) < 0 x (1, 3).For B A 211 + p 0 p 1
1p
22 + p 0 p 41
Now, let f(x) = x2 2 (p + 7) x + 5f(1) 0 p 4
4p
f(3) 0 p 314
So, p [ 4, 1] a = 4; b = 1a + b = 5 | a + b | = 5. Ans.]
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PAGE # 6
19. Let y = px4x34x3px
2
2
(p + 4y) x2 + 3(1 y)x (4 + py) = 0As x is real, so D 0 9(1 y)2 + 4(p + 4y) (4 + py) 0 or (9 + 16p) y2 + (4p2 + 46) y + (9 + 16p) 0 y RSo, 9 + 16p > 0 and (4p2 + 46)2 4(9 + 16p)2 0or 4 (p2 + 8p + 16) (p2 8p + 7) 0or (p + 4)2 (p2 8p + 7) 0or p2 8p + 7 0 1 p 7 ........(1)Also, the equations
px2 + 3x 4 = 0 and 4x2 + 3x + p = 0 have a common root, then (on subtracting), we get(p + 4)x2 = (p + 4) x2 = 1 x = 1
For x = 1, p + 3 4 = 0 p = 1 and for x = 1, p 3 4 = 0 p = 7.So, p = 1, 7 (not possible) .........(2) From (1) and (2), we get
1 < p < 7So, the number of possible integral values of p are 5 (i.e., p = 2, 3, 4, 5, 6). Ans.]
20. n
1k1k1kkk
k
23236
=
n
1k1k1k
1k
kk
k
233
233
= 22
2
1
1
233
233
= 33
3
2
2
233
233
= :
= :
Sn
= 1n1n
1n
233
13
nn
SLim
= 3 1 = 2. Ans.]
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PAGE # 7
PRACTICE TEST # 2Syllabus : Straight line and Circle.1. Only one is correct : 3 marks each.2. One or more than one is/are correct : 4 marks each.3. Matrix Match : 3 marks for each row.4. Integer type : 5 marks each.Time : 60 min. approx. Marks : 82
[STRAIGHT OBJECTIVE TYPE]
1. radius of circle C = 2516 = 41
O(2,2)
C(2,3)A B4
C
. Ans.]
2. Here, points S, A, C, B are cyclic. Also SAC = 90.
S(2,0)
C(6,0)
A
BSo, required equation of circle is (x 2) (x 6) + y2 = 0 or x2 + y2 8x + 12 = 0. Ans.]
3. Given, 1by
a
x
222222 baab
baay
babx
2p = 22 baab
4p2 = 2222
baba
8p2 = 2222
baba2
a2, 8p2, b2 are in H.P. Ans.]
4. Family of lines are concurrent at (4, 5). As, maximum distance of any line passing through (4, 5) frompoint ( 2, 3) will be 22 )8()6( = 10So, number of required lines = zero. Ans. ]
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PAGE # 8
5.O
(0, 1)
0,
25
x
y
25
,0
y = 2 (3, 2)
0,
29
(1, 0)
x y = 5/2x y = 1
From above graph, 3 < a < 29
. Ans.]
6. Coordinates of G =
98
,1
1 : 2
GO H1 3
23
11 3
43,
, Now AG : GD = 2 : 1
31h2
= 1,3
10k2 = 9
8
1 89
,G
A(1, 10)
B CD(h, k)
2
1 (h, k) =
311
,1 ]
7. x2 + y2 4x 8y + 4 = 0, centre (2, 2); r = 3 and (1, 3) is inside the circle [D]As the chord is of minimum length which is possible if the line through (1, 3) is at a maximum distancefrom (2, 2).Hence the line will be perpendicular to CM and passing through M because any other chord through Mwill be at a closer distance than AB. Hence equation of AB
y 3 = m(x 1)
when m(mCM) = 1; m
2123
= 1; m = 1
y 3 = x 1x y + 2 = 0 Ans.
Alternatively:Centre of the given circle is (2, 2) and radius is 3.Now, chord which is of minimum length will be at maximum distance from centre. (note very carefully)So, required equation of chord is
(y 3) =
3212
(x 1) i.e., (y 3) = (x 1) x y + 2 = 0. Ans.]
[COMPREHENSION TYPE]Paragraph for question nos. 8 to 10
[Sol.(i) L = PQ = PR = 141 = 2
Also, R = 1Area of SRQ = area of PQR
OR
Q
S
P(1, 2)
Mx
y
(1, 0)
6
,3
5,
53
= 22
3
LRRL
= 4181
= 58
required
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PAGE # 9
Area of quadrilateral PQOR = 2 area of POQ = 2 21
1 2 = 2
required ratio = 258
= 54
(ii) P = (1, 2), R = (1, 0)equation of chord of contact QR is x + 2y 1 = 0 ......(i)equation of chord PQ = y = mx + 2m1It passes through (1, 2)
m = 43
equation of PQ is 3x 4y + 5 = 0 ......(ii)Solving (i) and (ii)
Q =
54
,
53
Let S = (x1, y1) x1 + 1 = 1 53
x1 = 53
Also y1 + 2 = 0 + 5
4 y1 = 5
6 S
56
,
53
Equation of family of circle x2 + y2 1 + (x + 2y 1) = 0.
It passes through
56
,
53
15151
2536
259
= 0
= 51
equation of circle which circumscribe QSR is 5x2 + 5y2 + x + 2y 6 = 0(iii) Perpendicular distance from (0, 0) to line x + 2y 1 = 0
OM = 51
= 51
QR = 2 511
= 54
For max. area perpendicular distance from A to line QR should be maximum and it will be equal to
= 1 + 51
= 515
Area of AQR = 515
54
21
= 155
2
]
[MULTIPLE OBJECTIVE TYPE]11. Solving three equations we get the co-ordinates of the vertices.
P(1, 1) Q (3, 4) R (5, 2)(A) Co-ordinates of orthocentre (H)
Equation of line PD (y 1) = 31
(x 1) x 3y + 2 = 0
P(1,1)
R (5, 2)
Q(3, 4)
H E
3x + y 13 = 0
3x +
4y
7 =
03x
2y
1 =
0
D
E
Equation of line QE (y 4) = 34
(x 3) 4x = 3y
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PAGE # 10
Co-ordinates of orthocentre (H)
98
,
32
(B) Co-ordinates of centroid =
3yyy
,
3xxx 321321
= (3, 1)
2 1O G C
G = 3OC2
2OG3C
co-ordinates of circumcentre
C=
2983
,
2329
C =
1819
,
625
P(1,1)
R(3,4)
Q
H
E
k2 = 21819
625
= 1847
value of 18k = (47)(C) Equation of line joining orthocentre and & centroid
G(3, 1) & H
98
,
32
(y 1) = 3
32
198
(x 3) (y 1) =
3791
(x 3) (y 1) = 21
1 (x 3)
21y 21 = x 3 x 21y + 18 = 0. ]
12. We get t4 + (n + 1) t2 + mt + k = 0 a
bc
d Let (t, t2) when t can a or b or c or d lie on the given circle. Now use theory of equations.]
13. A = (3, 2), B = (3, 6), C = (6, 2)a = 5, b = 3, c = 4A = 90 Ex. circle opposite to vertex A will have largest radius.
xa =
cbacxbxax 321
B(3, 6)
A(3, 2)C(6, 2)
y
xO
34 5
= 435643335
= 9 center = (9, 8)
ya =
cbacybyay 321
= 8 radius ra = s. tan 2
A
(x 9)2 + (y 8)2 = 62 = 6 1 = 6
Since ABC is a right angle triangle is hence
Circumcenter =
226
,
263
=
4,
29
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PAGE # 11
The In-circle will be smallest circle which touches the sides
x1 = cbacxbxax 321
= 435643335
= 4
y1 = cbacybyay 321
= 12426325
= 3
Center (4, 3)Orthocenter is at A (3, 2) Ans (A) (B) (C) ]
14. Centre lies on angular bisector of the tangent lines which are 5
)3y2x( =
5)3yx2(
x + y = 2 ; y = xalso centre lies on 3x + 4y 5 = 0
centre is
5y4x32yx
xy5y4x3
P
A
B x + 2y 3 = 0
2x + y 3 = 0
C
(3, 1)
75
,
75
]
15. Let the equation of chord be
cos
1x =
siny
= r
(x = 1 + r cos , y = r sin ) be any point on it. Putting x and y in the equation of circle, we get
r2
r 3r
(12 cos + 10 sin ) r 108 = 0
Sum of roots = r + (3r) = 2r = 12 cos + 10 sin ... (1)Also, product of roots = 3r2 = 108 ... (2) From (1) and (2), we get
36 = (6 cos + 5 sin )2
tan = 0 or tan = 1160
Equation of chord is y = 0 or y = 1160 (x + 1)
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PAGE # 12
[MATCH THE COLUMN]16.(A) C1 = (1, 1), r1 = 3.
C2 = (3, 4), r2 = 5
Also, equation of radical axis 4x 3y + 27
= 0 slope 34
Clearly (1, 1) lies on C2 . Also radical axis is always perpendicular to line joining centers of two circle. One and of diameter is (1, 1) therefore other end is (7, 7) x1 = 1, y1 = 1, x2 = 7, y2 = 7
41 (x12 + y12 + x1y1 + y22 + x22 + x2y2) = 13
(B) (2 4 + ) (2 2 + ) > 0 ( 2) ( 4) > 0 < 2 > 4 p = 2, q = 4 q + p = 6
(C) Equation | x + y | + | x y | = 2represents a square with vertices (1, 1) ; (1, 1) ; (1, 1) and (1, 1).Now, f(x, y) = x2 (x + y)2
this is maximum if x = 1 and y = 1
(1, 1)(1, 1)
(1, 1) (1, 1)
O
y
x
Maximum value = 8. Ans.]
(D) Equation of BC is y = 2, which is parallel to x-axis.A
B(1, 2)
I (4,6)
C(6, 2)(4, 2)
y=2B/2
x
y
O
34
2B
tan B > 2
and 22C
tan C > 2
.
But, in a triangle, two angles cannot be greater than 90 and hence there is no such triangle. Ans.]
[INTEGER TYPE / SUBJECTIVE]17. Equation of radical axis of two given circles is 2ax + a2 + 2by b2 = 9, which passes through (0, b).
So, 0 + a2 + 2b2 b2 = 9 a2 + b2 = 9. Ans.]
18. Since (2c + 1, c 1) is interior point of the circle, so(2c + 1)2 + (c 1)2 2(2c + 1) 4(c 1) 4 < 0
0 < c < 56
.......(1)Also, given point (2c + 1, c 1) lies on smaller segment made by the chord x + y 2 = 0 on circle,so (2c + 1, c 1) and centre of circle (1, 2) will be on opposite side of the line. So
(2c + 1) + (c 1) 2 < 0 or c < 32
......(2) (1) (2)
c
32
,0 Number of integral values of c are zero. Ans.]
-
PAGE # 13
19. Centre of circle S1 is (2, 4) and centre of circle S2 is (4, 2)Now, radius of circle S1 = radius of circle S2 = 4 (each) Equation of circle S2 is (x 4)2 + (y 2)2 = 16 x2 + y2 8x 4y + 4 = 0 .......(1)Also, equation of circle touching y = x at (1, 1) can be taken as(x 1)2 + (y 1)2 + ((x y) = 0or, x2 + y2 + ( 2) x ( + 2)y + 2 = 0 .......(2)As, (1) and (2) are orthogonal so, using condition of orthogonality, we get
22
2242
22
= 4 + 2
4 + 8 + 2 + 4 = 6 = 3 The equation of required circle is x2 + y2 + x 5y + 2 = 0. On comparing, we get A = 1, B = 5 and C = 2Hence, (A + B + C) = 8. Ans.]
20. Let circle x2 + y2 + 2gx + 2fy + c = 0 cuts given circle orthogonally then2g 4f = c 4 ......(1)
and 4g + 4f = c + 4 ......(2)
Solving (1) and (2), we get g = c and f = 44c3
Equation of line PQ is,
c2
25
x +
c
235 y + c + 1 = 0
It meets x and y axis at points
0,
25
c2
1c and
5c23
1c,0 respectively..
Let mid point is (h, k) then h = 5c41c
, 10c31ck
Required locus is 9x 13y + 25xy = 0So, l = 9 and m = 25. Hence (l + m) = 34. Ans.]
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PAGE # 14
PRACTICE TEST # 3Syllabus : Permutation & Combination, Binomial Theorem, Function and Inverse trigonometric function.1. Only one is correct : 3 marks each.2. One or more than one is/are correct : 4 marks each.3. Matrix Match : 3 marks for each row.4. Integer type : 5 marks each.Time : 60 min. approx. Marks : 82
[STRAIGHT OBJECTIVE TYPE]1. Let f(x) = sin1 x tan1 x
so, f '(x) = 0x1
1
x1
122
x ( 1, 1)
f(x) is an increasing function in x [ 1, 1]
So, Range of f(x) = )1(f),1(f =
42
,
42 =
4,
4
So, k
4,
4 . Ans.]
2. We know that0 cos1 x and 0 cos1 (x) Using A.M. G.M., We get
0 22
xcosxcosxcos.xcos
1111
f (x)
4
,02
. Ans.]
3. Clearly,
98f
=
398
sincos 1 =
911
sincos 1 =
911
2coscos 1
=
1813
coscos 1 = 18
13. Ans.]
4. For 2xx to be defined, 2xx 0 0 1x
Also,2
2
21
x41
xx
so, 0 2xx 21
0 sin1 6xx 2
. Ans. ]
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PAGE # 15
5. We have sin16x = x36sin21
sin (sin16x) =
x36sin2
sin 1
6x = x36sincos 1Squaring both the sides, we get
36x2 = 1 108x2 144x2 = 1 x = 121
But x = 121
(Rejected)
Hence, x = 121
. Ans.]
6. We have
1
x
11loglog)x(f41
313
1x
11log41
31
> 0 log3 1
x
1141
3
x
1141 41x
1 > 2
41
x < 21
x < 161
0 < x < 161
(As, x 0)
Hence, domain of f(x) =
161
,0 . Ans.]
7. Parabolas = 4 ; Circles = 5 ; Lines = 3 Different possibilities are as follows :P/P + C/C + L/L + P/L + C/L + P/C
= 4C2 4 + 5C2 2 + 3C2 1 + (4C1 3C1) 2 + (5C1 3C1) 2 + (4C1 5C1) 4
= 24 + 20 + 3 + 24 + 30 + 80 = 181. Ans.]
[COMPREHENSION TYPE]Paragraph for Question no. 8 to 10
[Sol. We have g(x) =
2
1
x1x2
sin2 = 2 +
21
x1x2
sin
As, sin1 2x1x2
2,
2
21
x1x2
sin = 2, 1, 0, 1.
-
PAGE # 16
Range of g(x) = {0, 1, 2, 3} for )x(gf > 0 x R(i) Put D = 0
4a2 = 4(a 2) 3 a2 = a + 2 (a2 + a 2) = 0 (a + 2) (a 1) = 0. a = 2, 1Sum = 2 + 1 = 1. Ans.
(ii) Number of integers in the range of g(x) are 4. Ans.(iii) f(0) > 0 and f(3) > 0
Now, f(0) > 0 a 2 > 0 a > 2 ......(1)and f(3) > 0 9 6a + a 2 > 0 ......(2)
a < 511
0 1 2 3 x-axis
f(x) = x 2ax a 22
(1) (2) a Hence, no real a exists. Ans.]
[MULTIPLE OBJECTIVE TYPE]
11. Clearly, 1xgxgf 2 = 1
1x1
2 and )x(fg = 1)x(f
1
= 2x
1
Now, verify alternatives. Ans.]
12.
(A) As, f(x) = 2
x [ 1, 1]
So, f 210
Because f(x) = 2
cos1 )x(sincos 1 + cos1 )x(cossin 1
= xcossincosxcos2coscos21111
= 2
.
(B) For example : Let f(x) = x + 2 and g(x) = x + 3Now, )x(gf = f(x + 3) = (x + 3) + 2 = (x + 5)Also, )x(fg = g(x + 2) = (x + 2) + 3 = (x + 5) )x(gf = )x(fg x RBut f(x) and g(x) are not inverse of each-other.
(C) We have f(x) = sin1 (sin ax)
Period = 2a2
(Given) | a | = 4 a = 4, 4Hence, sum of possible values of a = 4 + 4 = 0. ]
-
PAGE # 17
(D) As, cot1 (cot 6) = 6 and tan1 (tan 6) = 6 2 cot1(cot 6) + tan1 (tan 6) = 12 3 . ]
13. onto functions = !3!21
!2!1!1!4
= 36 a
bcd
123
A Bf [13th, 12-02-2012, P-1]
(A) x + y + z = 7 (x, y, z W) 9C2 = 36 (A) is correct.
(B) 9 279C2 = 36 (B) is correct.
(C) From above information, it is not possible (think!) Number of ways = 0 C is incorrect.(D) 4500 = 22 32 53
Total divisors = 3 3 4 = 36 (D) is correct.]
14. As, (10C0)2 + (10C1)2 + ....... + (10C10)2 = 20C10.(A) T
r + 1 = 20C
r ( 1)r xr
coefficient of x10 = 20C10(B) 20C1 + 20C2 + ....... + 20C20 = (220 1).(C) Using gap method, number of ways = 20C10.(D) 20C10. ]
15.(A) f(x) = 2 tan1x + 2 tan1x + 2 tan1x = 6 tan1x.(B) f(x) = 2 tan1x 2 tan1x + 2 tan1x = 2 tan1x.(C) f(x) = 2 tan1x + 2 tan1x + 2 tan1x = 2tan1x(D) f(x) = 2 tan1x 2 tan1x + + 2 tan1x = 2 tan1 x.
[MATCH THE COLUMN]16.
(A) dxdy
= 3x2 + 2(a + 2) x + 3aPut D 0 a2 + 4a + 4 9a < 0 a2 5a + 4 < 0 (a 4) (a 1) < 0 a [1, 4] a = 1, 2, 3 & 4.
(B) tan1 (2 tan x) + tan1 (3 tan x) = 4
xtan61
xtan52
= 1 6 tan2x + 5 tan x 1 = 0 6 tan2 x + 6 tan x tan x 1 = 0
6 tan x (tan x + 1) 1 (tan x + 1) = 0 tan x 1 or tan x = 61
(C) Coefficient of x7 in 11
2
bx1
ax
= 11C5 5
6
ba
and coefficient of x7 in 11
2bx1
ax
= 11C6 6
5
ba
.
ab = 1. Ans.
-
PAGE # 18
(D) We have
x
3x
x
3x1
x
3x
x
3x
tan 1 =
x
6tan 1
2
x
9x
x
6x
6x
6 x2 2x
9 = 0 (x2)2 = 9 x2 = 3. Ans.]
[INTEGER TYPE / SUBJECTIVE]
17.
10
1k
1
)1k(k1k1k
tan
10
1k
1 )1k(tan tan1(k)
T1 = tan12 tan11T2 = tan13 tan12 T10 = tan111 tan110
S = tan111 tan11 = tan1 1210
= cot1 1012
cot
1012
cot 1 = 10
12 = 5
6 = b
a a + b = 11 Ans. ]
18. Let a = cos1x and b = sin1ySo, a2 + b = 1 .........(1)and a + b2 = 1 .........(2) From (1) and (2), we get
a2 + b = a + b2 (a2 b2) (a b) = 0 (a b) (a + b 1) = 0 Either a = b or a + b = 1Case-I : When b = a,
Now, equation (1) becomes
a2 + a 1 = 0 a = 2
51
But a [0, ] and b
2,
2 a = b = 215
(x, y) =
2
15sin,
215
cos
Case-II :When a + b = 1,Now, equation (1) becomesa2 + (1 a) = 1 a(a 1) = 0
-
PAGE # 19
So, a = 0 or a = 1If a = 0, b = 1 (x, y) = (1, sin 1) and a = 1, b = 0 (x, y) = (cos 1, 0)Hence, number of ordered pairs (x, y) are 3
i.e.,
2
15sin,
215
cos , (1, sin 1) and (cos 1, 0). Ans.]
19. 2, 3, 4, 5, 6, 7, 8, 9Select any 3 digit in 8C3 wayssay (2, 5, 8).They can be arranged in two ways as per the condition givene.g. 5, 2, 8 or 8, 2, 5Now the remaining 5 digit are 3, 4, 6, 7, 9There are 6 gaps between them. Select one gap in 6C1 ways for these blocks. Arrange these 5 nos. ineither ascending or descending order so that none of these digit is less than both digits on its left and right.Hence the total number of ways = 8C3 2 6C1 2 = 56 2 6 2 = 56 24 = 1344 Ans.]
20. Let N = a1 a2 a3 a4 a5
3 ways 4 ways 3 ways 5 ways(i.e. 1/3/5/7/9)
2 ways(either 2 or 6)
Total numbers = 3 4 3 5 2 = 360. Ans.]
-
PAGE # 20
PRACTICE TEST # 4Syllabus : Logarithms, Quadratic Equation and expression, Compound angle Trigonometric equation and
inequations, Solution of triangle, Sequence and Progression, Straight line and Circle, Permutation &Combination, Binomial Theorem, Function and Inverse trigonometric function.
1. Only one is correct : 3 marks each.2. One or more than one is/are correct : 4 marks each.3. Matrix Match : 3 marks for each row.4. Integer type : 5 marks each.Time : 60 min. approx. Marks : 82
[STRAIGHT OBJECTIVE TYPE]
1.1xsinxsin
12
is always defined x R.
And sin1 1x1
2 is defined when 1 1x2
<
x2 1 1 or x2 1 1 x2 2 or x2 0
Domain = ,202, . Ans.]
2. Point (2, 3) lies on the line 2x 3y + 5 = 0Using parametric form of C1C2
133
3y
1322x
= 132 . P(2,3)C1
C22x3y+5=0 co-ordinates of centre are (2, 9) and (6, 3)
Greatest value of | h | + | k | = 11. Ans.]
3. Let the perimeter of the pentagon and decagon be 10 x. The each side of the pentagon is 2x and its area
is 5x2 cot
5 . Also, each side of the decagon is x and its area is
10
cotx25 2
.
decagonregularofAreapentagonregularofArea
=
10cotx
25
5cotx5
2
2
=
18cot36cot2
=
52
.
[Note : Area of regular polygon having n sides ]
=
ncot
n
na2
where a = length of each side. Ans.]
-
PAGE # 21
4. Given, 4 sin4x = 1 cos4x 4 sin4 x = (1 cos2x ) (1 + cos2x) 4 sin4x = sin2 x (2 sin2 x) sin2 x (4 sin2 x 2 + sin2 x) = 0sin2x [5 sin2x 2] = 0 sin x = 0 or sin2x = 5
2
sin x = 0 give x = and sin x = 52
given 4 solutions in (0, 2) Total solutions = 5. Ans.]
5. Let the number a, b, 12 are in G.P. b2 = 12a ........(1)Also, a, b, 9 are in A.P. 2b = a + 9 .......(2) (1) and (2) eliminate a b2 24b + 108 = 0 b = 6, 18As, G.P. is decreasing b = 18 a = 27Hence, ab = 486. Ans.]
6.N
C(7, 5)M P(2, 7)
90
PC = 13and r = 15
longest chord will be the diameter through P and its length = 30 and smallest and will be perpendicular toabove diameter and passes through P
i.e., length of smallest chord = MN = 22 CPCN = 562 . Ans.]
7. y = 1xtanxtan1xtanxtan
24
24
(y 1) tan4x + (y 1) tan2 x + (y + 1) = 0Case I : y 1 tan x R D 0
(y 1)2 4 (y 1) (y + 1) 0 35
y < 1
Case II : y = 1 It didn't satisfy above equation
35
y < 1. Ans.]
[COMPREHENSION TYPE]Paragraph for question nos. 8 to 10
Sol. As, tan = 43
711711
sin = 5
3
Area = OA OB sin = 9
B
A
C
y=x
y=17x
x
y
O(0,0)
-
PAGE # 22
OA OB 53
= 9
OA OB = 5 3 OA = 5 and OB = 3.
Coordinate of B is = (0 + 3 cos 45, 0 + 3 sin 45) =
23
,
23
coordinate of A is =
50150,
50750
=
21
,
27
Mid point of diagonal m =
22
,
25
, then line OC is y = x52
.
0 = 2, b = 5 a + b = 7.Also, AB = 10 (using distance formula). Ans.]
[MULTIPLE OBJECTIVE TYPE]
11. We have (y 3) = m (x 2) are other sides. Then, 3
tanm11m
.
m = (2 + 3 ) or ( 3 2).Find two values of m and the equation of sides.
A(2,3)
CB60 60
xy+3=0
slope = m
Also, Area of triangle = 32
= cosec 60. Ans.]
12.
(A) tan1(x2) = cot1
2x
1 is obviously true for all x R {0}.
(B) cos1
2
2
x1x1
[0, ) for all x R cos1
2
2
x1x1
.
(C) Domain of f (x) is {1} ; range of f (x) is {0} f (x) = 0 x domain of f (x)
(D) tan2x + cot2x 2 for all x R
2n
n I.
(D) is incorrect (f is not defined for any x. Domain is ) ]
13. Here, exponent of 2 is 21
and exponent of 3 is 51
and L.C.M. of 2 and 5 is 10.
So, only those terms will be rational in which power of both 2 and 51
3 are multiples of 10.
Here, index of
5
1
32 is 10, therefore only first and last terms in the expansion of 10
51
32
will be rational.
-
PAGE # 23
Now, sum of rational terms = t1 + t11 = 10C0 0
51
1032
+ 10C0
10
51
032
= 25 + 32 = 32 + 9 = 41. Ans.]
14.
(A) f(x) = )x2sin1(x2sinx2cos2
)x2cos1(x2cosx2sin2
2
22
2
22
= 2cos22x + 2 sin2 2x= 2 A is correct.
(B) g(x) =
2x
sin2x
cos
2x
sin2x
cos
2x
tan1
2x
tan1
22
2
=
2x
tan1
2x
tan1
2x
tan1
2x
tan1 = 1 B is correct.
(C) h(x) = )xsinx(cosx2sinxcosxsin)xsinx(cos
442
2222
=
xcosxsin4xcosxsin
22
22 = 4
1
h(x) = 41
C is also correct.
(D) l(x) = 2x
cos2
)xcos1(xtan2
= tan x. Ans.]
15. As,a
Asin = b
Bsin =
c
Csin = k (say)
a
Acos = b
Bcos =
c
Ccos = k' (say)
tanA = tanB = tanC = 'k
k
ABC is equilateral a = b = c. Ans.]
-
PAGE # 24
[MATCH THE COLUMN]16.(A) We have )3x(log1 1x log3(x + 1) < log3(2x 3)
log3(x + 1) + log3(x 3) < log3(2x 3) )3x()1x(log3 < log3(2x 3)
1 3
0 4
(x + 1) (x 3) < (2x 3) 0 < x2 2x 3 < 2x 3 x (3, 4) No natural value of x exist. Ans.
(B) To cancel f(2x + y) and f(3x y) from both LHS and RHS equating the argumentwe have 2x + y = 3x y
x = 2y
y = 2x
.
Hence put y = 2x
to get f (x) + 2x5 2
= 2x2 + 1 f (x) = 1 2
x2
Hence f ( 4) = 7 4f = 7 Ans.
(C) S =
1i 1jji2
ij =
2
1ii2
i
=
2
32 ........23
22
21
Let P = ........23
22
21
32 ..........(1)
2P
= ........22
21
32 ..........(2)
2P
= ........21
21
21
32 (By (1) and (2))P = 2
S = 4. Ans.(D)CASE 1: a1 = 1 ; a2 = 3
2 m < 4 (i) 27 m < 81 (ii) hence no solution
CASE 2: a1 = 3 ; a2 = 1 8 m < 16 (iii) m = 8 is the only possible integral value 3 m < 9 (iv) Number of integral value = 1. Ans.]
-
PAGE # 25
[INTEGER TYPE / SUBJECTIVE]17. Let common difference of a1, a2, a3 be d1 (d1 > 0)
a1 + a2 + a3 = 15 a2 = 5, a1 = 5 d1, a3 = 5 + d1Let common difference of b1, b2, b3 be d2 (d2 > 0) b1 + b2 + b3 = 15 b2 = 5, b1 = 5 d2, b3 = 5 + d2 (a2 b2) + (b1 a1) = 1 (a2 a1) (b2 b1) = 1 d1 d2 = 1 d1 = d2 + 1
87
bbbaaa
321
321 87
)d5)(d5(5)d5)(d5(5
22
11
87
d25d25
22
21
25 8 8 (d2 + 1)2 = 25 7 7 22d
25 = 8 ( 22d + 2d2 + 1) 7 22d 22d + 16 d2 17 = 0 d2 = 17 or 1 d2 = 1 (d2 > 0) d1 = 2 a1, a2, a3 will be 3, 5, 7 a1 a2 a3 = 105and b1, b2, b3 will be 4, 5, 6 b1 b2 b3 = 120 b1 b2 b3 a1 a2 a3 = 15. Ans.]
18. 18
logxcosxsinlog2
1122
sin1x cos1x = 18
2 sin1x
xsin2
1 =
18
2
(sin1x)2 xsin.21
+ 18
2
6xsin
3xsin 11
= 0
x1 = 3sin
=
23
; x2 = sin 6
= 21
x12 + x2
2 = 14
421
23 2
2
. Ans.]
19.y
x3 /4/2 /4
/4
O
y=|cot 2x|
y=tan (tan x) /41
Number of solution = 1 N = 1 Ans.
13 cosec x + 13 sec x = 4 2
-
PAGE # 26
21
21
23
21
cosec x +
21
21
23
21
sec x = 2
sin (60 45) cosec x + cos (45 30) sec x = 2cos x sin 15 + sin x cos 15 = 2sin x cos xsin 2x = sin (x + 15) 2x = x + 15 or 2x = 180 x 15 x = 15 3x = 165 x = 55 M = 2Hence, (N + M) = 1 + 2 = 3. Ans.]
20. Total number of ways = 10C3 = 120.Number of ways to select 3 consecutive gates = 10.Number of ways to select 2 consecutive and 1 separated gate = 10(10 4) = 10 6 = 60.Hence, the number of ways to select 3 gates so that all are separated
= 120 (10 + 60) = 120 70 = 50. Ans.]
-
PAGE # 27
PRACTICE TEST # 5Syllabus : Calculus1. Only one is correct : 3 marks each.2. One or more than one is/are correct : 4 marks each.3. Matrix Match : 3 marks for each row.4. Integer type : 5 marks each.Time : 90 min. approx. Marks : 76
[SINGLE CORRECT CHOICE TYPE]Q.[Sol. The required area will be equal to the area enclosed by y = f(x), y-axis between the abscissa at
y = 1 and y = 3.
Hence, A =
0
1
1
0dxxf3dx1)x(f
.
11x
y3
1
=
0
1
1
0
33
25dxxx2dx2xx
. Ans.]
Q.2
[Sol. ln c + ln x + ln y = yx
.
dxdy
y1
x
1
= 2ydxdy
xy = dx
dyyx
y1
2 dxdy
=
1yx
yx
1yx
yx
dxdy
so (z) = )1z(z1z
. Ans.]
Q.3[Sol. g(x) = x + 1 and h(x) = 2 (x2 + 2x + 1) + 2 = 2(x + 1)2 + 2
h(x) = 2(g(x))2 + 2 = 2 (g(x))2 + 1) = f(g(x)) f(x) = 2(x2 + 1). Now solving with y = mx.
Q P (1,4)
O(0,2)
2x2 mx + 2 = 0 D 0 so m = 4
Required area = 2 1
0
2 dxx42x2 = 4
1
0
2dx)1x(
= 103)1x(34
= 3
4 sq. units. Ans.]
-
PAGE # 28
Q.4[Sol. Let y = vx
dxdy
= v + x dxdv
v + x dxdv
= 2
222
vx2xvx
2v1v2
dv =
x
dx
ln (1 v2) = ln x + ln c
ln
2
2
x
y1 x c = 0
(x2 y2) c = x it is passing through (2, 1) 2(x2 y2) = 3x. Ans.]
Q.5
[Sol. f(x) = x x 2,0 A = 2 4A4 2 . Ans.]
Q.6[Sol. (x3y3 xy)dy = 2dx
x3y3dy yxdy = 2dx2dx + yx dy = x3y3dy
3
23 yxy
dydx
x
2
Let 2x
1 = t dy
dtdydx
x
23
dydt
+ yt = y3
2/ydyy 2ee.F.I
cdyeye.t 2/y32/y22
ce2ye2e.t 2/y
22/y2/y 222
2/y
22/y
2
2
e
c)2y(et
t = (y2 2) + c 2/y2e
1 = (xy)2 + cx2 2/y2e 2x2at x = 1, y = 0
1 = c 2 c = 3
1 = x2 (y2 + 3 2/y2e 2) ]
-
PAGE # 29
Q.7
[Sol. Let r (x) = )4x(
x2x)2x)(1x( 2
122
x2x)4x()2x)(1x(2Lim)x(r Lim
2xix
]
Q.8[Sol. [2x 1] is discontinuous at three points
x = 25
, 23
, 2
f(x) may be continuous if f(x) = ax3 + x2 + 1 = 0 at x = 25
, 23
, 2. g(x) can be zero at only one point
for a fixed value of a minimum number of points of discontinuity = 2. ]
Q.9[Sol. 0 < ex < 2 and 0 < ex < 2
x (ln2, ln2)
f(x) =
}0{)2n,2n(x,2
0x,ll ]
Q.10
[Sol.
n
m
0x x
xcos1sinLim
=
n
m
0x x
xcos1Limsin =
n
2
0x x2x
sinm2Limsin
m N and n = 1 or 2. Ans.]
Q.11
[Sol. f(g(x)) = (sgn x)3 (sgn x) =
0x00x,0
y
x1O1
x x3
g(f(x)) = sgn (x3 x) =
1,0,1xat.0x1,1
1x0,10x1,1
1x,1
.
y
x1O1x'
y'
]
-
PAGE # 30
Q.12[Sol. y = (x) (x 1) (x 2) = x(x2 3x + 2)
y dx = dxx2x3x 23 = 234
xx4
x
A1 =
2
1
234
xx4
x
= [4 8 + 4 ]
1141
= 41
12 1 2 3 4
A2A1
y
x
A2 =
3
2
234
xx4
x
= 481
27 9 = 49
A3 =
4
3
234
xx4
x
= 64 64 + 16 49
= 455
.
so one value in (3, 4) and another in ( 2, 1). Ans.][ curve is symmetric about x = 1.]
Q.13
[Sol. f(x) is not differentiable at x = 1, 0, 3. 1
y
x33 1/31O
Ans.]
PART-BQ.1[Sol.
(A)
0xif,e0xif,e)x(f
x
x
; f ' (0) = h
)0(f)h0(fLim0h
= 1h
1eLimh
0h
Hence 20x2x
cos1
2x
cos1cos1Lim
1
x.22x
cos1
nm
2
[Using 21cos1Lim 20x
]
2x2
2x
cos1Lim qp
2
0x
[13th, 05-09-2010, P-1]
2x2
4x
sin4Lim qp
4
0x
-
PAGE # 31
2x24
x4Lim qp44
0x
for limit to exist n = 4
12
2p8
28 + p = 2 p = 7 q = 4Hence 2q + p = 8 7 = 1 Ans.
(B) I By definition f '(1) is the limit of the slope of the secant line when s 1. [29-01-2006, 12&13]
Thus f '(1) = 1s
3s2sLim2
1s
= 1s)3s)(1s(Lim
1s
= )3s(Lim
1s
= 4 (D)
II By substituting x = s into the equation of the secant line, and cancelling by s 1 again, we gety = s2 + 2s 1. This is f (s), and its derivative is f '(s) = 2s + 2, so f ' (1) = 4.
(C) We know that
[cot x] =
Ixcotif,xcot
Ixcotif,xcot[12th, 25-07-2010 P-1]
f(x) = xcotxtan =
Ixcotif,0
Ixcotif,1
0]x[cotxtan1]x[cotxtan
xcot]x[cot,so,Ixcotwhen:Note
so, points of discontinuity are those points where cot x I
Now 2x
12
0 < cot x 2 + 3
Hence, number of points of discontinuity is 3
i.e. x = cot13, cot12 and cot11 = 4
. Ans.
(D)49/itf
Hence two solutions. ]
-
PAGE # 32
PART-CQ.1
[Sol.704/itf B A = )3(cot3)2(cot2 11
31
cot31
21
cot21 11 [13th, 05-08-2007]
= 2(cot12 + cot13) + cot13
21
cot61
31
cot21
cot31 111
= 2
+ cot13
2tan
61
41
= 4
+ cot13 61
3tan4
3 1
= 8
+ cot13 + 61
tan13 = 8
+ cot13 61
3cot2
1= 8
+ 12
+ cot13 61
cot13
= 3cot65
245 1
hence a = 5; b = 24; c = 5; d = 6a + b + c + d = 40 Ans. ]
Q.2
[Sol. f (x) = 9x12x41a3an9x12x4
22
2
l
222
)3x2(1a3an)3x2(
l
f(x) will be discontiuous when|a2 3a + 1| + (2x 3)2 = 0 or 1
0
0 |a2 3a + 1| 1 1 a2 3a + 1 1 (a2 3a + 2) 0 & a2 3a 0 (a 1) (a 2) 0 a (a 3) 0 a + (, 1] [2, ) a [0, 3] a [0, 1] [2, 3]Hence, integral values of 'a' for which f(x) will be discontinuous at atleast one real x are.
0, 1, 2 & 3.Q.3[Sol. For f (x) to be continuous
2x3 18x + = 6x + 10 2x3 24x + 10 = 0The above equation should have exactly two roots one root is repeating f ' (x) = 0 has that root f ' (x) = 6x2 24 = 6 (x2 4)
x2 4 = 0 x = 2 2(2)3 18(2) + = 6(2) + 10
16 + 36 + = 2= 2 20
-
PAGE # 33
= 22 at x = 2
2(2)3 18(2) + = 2216 36 + = 22 = 22 + 20 = 42
sum = 42 22 = 20 ]
Q.4
[Sol.1048/de x)y(
y)x()xy( fff [13th, 16-12-2007]
put x = 1, y = 1, f (1) = f (1) + f (1)hence f (1) = 0
Also f ' (1) = h0)h1(fLim
0h
exists; but f ' (1) = A = h)h1(fLim
0h
now f ' (x) = h)x(f)hx(fLim
0h
=
h
)x(fx
h1xfLim
0h
using functional relationship
f ' (x) =
h
)x(fx
)xh(1f)xh(1
)x(f
;
f ' (x) =
h
)x(f)xh(1)x(f
)xh(xx
h1fLim 20h = h
1x
h1Lim
x
1t
)t1(fLim0h20t
f ' (x) = x
)x(fx
A2 where A = f ' (1)
let f (x) = y
yx
1dxdy
= 2x
A which is a linear differential equation with I.F. = xe
dxx
1
x y = dxx
A hence xy = A ln x + C ....(1)
if x = 1, y = 0 C = 0 {Using (1)}
y = x
xnA l
if x = e, y = e
1
e
1 =
e
A A = 1
hence y = f (x) = x
xnl Ans. ]
-
PAGE # 34
PRACTICE TEST # 6Syllabus : Vector, 3-D and Complex Number1. Only one is correct 1 to 5 : 3 marks each.
Only one is correct 6 to 10 : 4 marks each.2. One or more than one is/are correct : 4 marks each.3. Matrix Match : 3 marks for each row.4. Integer type : 5 marks each.Time : 120 min. Marks : 115
PART-AQ.1[Sol. Let kia ; ki3b
and kj2i4c
Volume of parallelopiped = cba = |240301
|
= 1 (0 2) 0 + (6) = 6 2 = 4. Ans.]
Q.2[Sol. For | z 1| maximum, z = 4
centroid () = 3z
= 314
= 1(1,0)(4,0) (4,0) Re(z)
O
Im(z)
Re () = 1 Ans. ]
Q.3[Sol. Given, |c||b||a| = 1
Let ba
=
Now, 0cba3
bca3
2
bca3
3 + 1 + 2 3 cos = 1 cos = 2
3
= 65
. Ans.]Q.4 .
[Sol. Given, z
z1 is real
z
z1 =
z
z1
(z z2) = 2zz O
Re(z)
Im(z)
x=1/2
y=0 0)zz)(zz()zz( 0)1zz()zz(
Either z = z (z 0) or z + z = 1 x = 21
Ans. ]
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PAGE # 35
Q.5
[Sol. Given, 0z0z
1
2
= 3i
e
0z0z
1
2
= 3i
e
= 1
| z2 0 | = | z1 0 | O(0)
A(z )1
Re(z)
B(z )260
Im(z)
Also,
1
2z
zarg = 3
z1oz2 = 3
Triangle is equilateral. Ans.]
Q.6
[Sol. As, 1n2z1 = (cos (2n + 1) i sin (2n + 1))
1n2z
i4 = 4i (cos (2n + 1) i sin (2n + 1))
1n2z
i4Re = 4 sin (2n + 1)
9
0n1n2
z
i4Re = 4
9
0n)1n2(sin
= 4 (sin + sin 3 + sin 5 + ....... + sin 19)
= 4
sin)10(sin)10(sin
= 4
3sin)30(sin2
= cosec 3. Ans.]
Q.7
[Sol. The normal vector of plane P is parallel to vector = 230102kji
= )6(k)4(j)3(i = k6j4i3
Equation of plane P is 3 (x 0) 4 (y 1) + 6 (z + 1) = 0 3x 4y + 6z + 10 = 0
3x 4y + 6z = 10 or 610
z
410y
310x
= 1
Area of triangle ABC =
9100
36100
36100
16100
16100
9100
21
= 50 9361
36161
1691
=
6431693650
=
726150
=
366125
. Ans.]
Q.8[Sol. Given | z 1 | = 2 I
m(z)
(x 1)2 + y2 = 4y2 3y2 = (x 1)2 3 y = (x 1) L1 : x 3 y 1 = 0 and L2 : x + 3 y 1 = 0. O Re(z)
Im(z)
(0, 0) (1, 0)
31
,0
3
1,0
Also, L3 : x = 0
Hence, area = 2
311
21
=
31
. Ans.]
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PAGE # 36
Q.9
[Sol. Equation of line L is 4
0z0
2y3
1x
= (say)
Any point on it is (3 + 1, 2, 4)Now, above point will satisfy x y + z = 13, so (3 + 1) 2 + 4 = 13 7 = 14 = 2So, co-ordinates of Q is (7, 2, 8). Ans.]
Q.10
[Sol. Consider 2z4iz4
=
2i
i
e44)e2()i4( (As | z | = 2 z = 2ei)
= )sini()2cos1()sini(cosi2
=
cossini2sini2)sini(cosi2
22
= )sini(cossinsinicos
= cosec ( , 1] [1, ) Option (B) is correct.]
Paragraph for question nos. 11 to 13[Sol. We have
L1 :
42z
26y
37x
and L2 :
34z
13y
25x
A(73 , 6+2 , 2+4 )
B(2+5, +3, 3+4)
L2
L1(7,6,2)
(5,3,4)Now, 2
232 = 2
23
= 1
342
3 + 2 2 = 3 + 2 + 3 = 5 ........(1)and 3 + 2 2 = 4 + 8 6 5 8 = 2 ........(2) On solving (1) and (2), we get
= 2, = 1So, A (1, 10, 10) and B (7, 4, 7)
(i) AB = 222 10710417 = 93636 = 81 = 9. Ans.
(ii) Let equation of plane parallel to L1 and containing L2 bea(x 5) + b(y 3) + c(z 4) = 0 .......(3)
Now, 2a + b + 3c = 0 and 3a + b + 4c = 0
7c
17b
2a
From equation (3), we get 2(x 5) 17(y 3) + 7(z 4) = 0
or 2x + 17y 7z = 33. Ans.(iii) Volume of tetrahedron OPAB (where O is origin)
= OBOAOP61
= |747
10101321
|61
= 4261
= 7. Ans.]
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PAGE # 37
Paragraph for question nos. 14 to 16
[Sol.6zz
zz.arg
12
13
A = 30C(z )3
B(z )2A(z )1 30(i) Circumcenter is origin and centroid is 3zzz 321
.
As centroid divides orthocenter and circumcentre in the ratio 2 : 1 (internally).
3)02()z1(
= 3zzz 321 z = z1 + z2 + z3 H(z) O(0)
2 1G
(ii) Clearly, Asina
= Bsinb (using Sine law)
a = | z2 z3 |, b = | z1 z3 |
Also, sin A = 21
, B =
23
21zz
zzarg
| z2 z3 | =
23
21
31
zz
zzargsin2
|zz| | z2 z3 | =
23
2131
zz
zzargcosec|zz|
21
(iii) If HBTC be parallelogram then midpoint of HT and BC should be same.
2zzzz 321
= 2zz 32
z = z1 | z z1 | = 2 | z1 | = 2 (circumradius)
Ra
R212aR2
Asina
,As
Also | z2 z3 | = circumradius | z z1 | = 2| z2 z3|. ]
Q.17[Sol. As, locus of P() satisfying
| 4 | + | + 4 | = 16 is an ellipse with foci (4, 0) and (4, 0) and e = 21
.
So, a conic with foci (4, 0) and (4, 0) and eh = 2 will be a hyperbola and its equation is| z 4 | | z + 4 | = 4 Ans. ]Q.18
[Sol. Given, cba = 15 |p212p1432
| = 15
p24p234p2 2 = 15 6p7p2 2 = 15 2p2 7p + 6 = 15 or 2p2 7p + 6 = 15
p = 1, 29
or 2p2 7p + 21 = 0 has non-real roots. Ans.]
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PAGE # 38
Q.19[ Sol. Centre of circle is (4, 5).
Also, radius = 940)5()4( 22 B
A
C(4, 5)P(2,3)
Distance of centre (4, 5) from (2, 3) = 22So, a = max. | z (2 + 3i) | = 9 + 22and b = min. | z (2 + 3i) | = 9 22Hence, a + b = 18
Also, (a b) = 24 and ab = 4)ba()ba( 2
= 432324
= 4292
= 73. Ans. ]
Q.20[Sol. Image of A(1, 2, 3) in the plane x + y + z = 12 is (5, 6, 7)
Equation of BC is 27z
16y
25x
A(1,2,3) C(3,5,9)
B(7,0,19)
(5,6,7)
B is ( 7, 0, 19)
Now, equation of AB is 163z
22y
81x
Equation of plane containing the incident and reflected ray is 1628632
3z2y1x
= 0.
i.e., 3x 4 + z + 2 = 0. Ans.]
PART-BQ.1[Sol.(A) We have 2rqp rqprqp
= 1 + 16 + 64 + qp2 = 81 (As , qp = 0) rqp
= 9. Ans.
(B) Let A0( 3, 6, 3), B0(0, 6, 0), 2,3,2c
and 1,2,2d
Then ABmin = dconBAofProjection 00
=
dc
dcBA 00
=
k2j2i
|122232303
|
= 3. Ans.
(C) As, (, , ) lies on plane x + 2y + z = 4 + 2 + = 4 ......(1)Now, 0vjj 0vjjjvj vjvj kjij ki = 0, which is possible when = = 0.So, from equation (1), we get 2 = 4 = 2. Ans.
(D) Let equation of variable plane be c
z
by
a
x
1 = 0
-
PAGE # 39
Now, A (a, 0, 0) ; B (0, b, 0) ; C (0, 0, c)
Centroid of tetrahedron OABC =
4c
,
4b
,
4a
So, x = 4a
a = 4x ; y = 4b
b = 4y and z = 4c
c = 4z
Also,
c1
b1
a
11
22
= 2 41
= 222 c
1b1
a
1
So, 416
= 22 z
1y1
x
1
Hence, k = 4. Ans.]
PART-CQ.1[Sol. As, O, A, B and D are concyclic,
so cos 60 = AD3
ADBD
AD = 6 and OD = 5
Re(z)
Im(z)
D(0, 5) B(z )2
A(z )3O(0, 0)
60
/3C (z )1
Hence, | z3 | = OA = 2536 = 11Hence | z3 |2 = 11. Ans.]
Q.2
[Sol. The normal vector of plane is parallel to vector = 311122
kji
= k4j7i5
Equation of plane is 5x 7y 4z = 0 ......(1)So, distance of plane in equation (1) from P 102,0,104
=
)4()7()5(1024)0(71045
22= 103
1012 = 4. Ans.]
Q.3
[Sol. The normal vector of plane 1, is = 120011kji
= k2ji
Equation of plane 1 is, 1 (x 2) 1 (y 3) + 2 (z 4) = 0 or 1 : x y + 2z 7 = 0.Also, 2 : x y + 2z 19 = 0
So, d = 6|197| = 6
12 = 2 6
Hence, d2 = 24. Ans.]
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PAGE # 40
Q.4[Sol. We have z2 + 2 | z |2 = 2
Put z = x + iy, we get (x2 y2) + 2i xy + 2(x2 + y2) = 2 On equating real and imaginary parts, we get3x2 + y2 = 2 .....(1)2xy = 0 .....(2)
Case I: x = 0, so y2 = 2 y = 2 z = 2 i
Case II: y = 0 3x2 = 2 x = 32
z = 3
2
Hence
2i,32
z ]
Q.5[Sol. Here, |v|1|u|
Also, vu = 0 vu Now, vu4vu = vvuu4vu
= vvvvvuuvuvuu4 = 4 u0v = v4u
vu4vuv4u = 2v4u = 1 + (4)2 + vu8 = 17 (As, vu ). Ans.]Q.6
[Sol.C(5,0)
B(z ) = 2 + i 32
A(1,0)
y
x O(0,0)
D
Clearly A(z1) is the point of intersection of arg(z 2 + i) = 43
and arg (z + 3 i) = 3
z1 = 1
Also, B(z2) is the point on arg (z + 3 i) = 3
such that | z2 5 | is minimum, so z2 = 3i2 .also, C(z3) be the centre of the circle | z 5 | = 3, so z3 = 5.Hence, area of ABC = )BD()AC(2
1
= 3)4(21
= 32 (square unit.)
= 32 2 = 12 Ans.]
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