practice test papers xiii vxy 1 to 6 sol

PRACTICE TESTSOLUTIONS

TARGET IIT JEE 2012

MATHEMATICS

CONTENTS

PRACTICE TEST1 ........................................................... Page 2




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

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

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.

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.]

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.]

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. ]

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.]


[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

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

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

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)

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.]

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. ]

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.]


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. ]

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.]

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.]

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.]

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.]

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.]

practice test papers xiii vxy 1 to 6 sol

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