rt solutions-practice test papers xiii vxy 1 to 6 sol

PRACTICE TEST

SOLUTIONS

TARGET IIT JEE 2012

MATHEMATICS

CONTENTS

PRACTICE TEST�1 ........................................................... 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

3

1

3

1

3

1

3

3

3

13

3

13

3

1

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) = 0

hence 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 = ± 1

Hence 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. = 2

1× 8 × 3 = 12

R = 4

abc=

124855

A

B C

b = 53

44

c = 5

aR =

625

]

5. Given,

2BA

sin

2BA

cos·

2BA

cos

2BA

sin =

151

7878

baba

2BA

tan

2BA

tan

(Using Napier's analogy) . Ans.]

6. Given,

=

2

2

)(

)(

= 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

q4

p2

= c4

b2

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 , 2

3, 3,

2

7, .......... 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 = 2

m6 = 3m

f(3m) = 9m2 � 18m2 + m2 + 4m + 2 = �8m2 + 4m + 2 = �8

4

1

2

mm2 =

2

5 � 8

2

4

1m

f(3m) is maximum if m = 4

1.

(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 =

8

522 =

8

1642 =

4

51

4

51,

4

51m

4

51

4

510

largest integral value of 4

236.21 =

4

236.3

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 = �2

1. 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. an = 53 + (n � 1)(�2) = 55 � 2n

As, an < 0 55 � 2n < 0 n > 2

55

number of terms = 27

S27 = 2

27[2 × 53 + 26 (� 2)] =

2

27[106 � 52] = 54

2

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

2

x

4tan =

2

x

4cot . Ans.]

15. yx y xlog·y 10 = log10y ........(1)

and 4y xy ylog·y 10 = 4 · log10 x ........(2)

On putting, log10x = y

1 log10y from (1) in (2), we get

ylog·y 10 = y

4 · log10 y

y

4y 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 21�1 + p 0 p � 1

1p

2�2 + p 0 p 41

Now, let f(x) = x2 �2 (p + 7) x + 5

f(1) 0 p � 44p

f(3) 0 p 314

So, p [� 4, �1]

a = � 4; b = � 1

a + b = � 5 | a + b | = 5. Ans.]

PAGE # 6

19. Let y = px4x3

4x3px2

2

(p + 4y) x2 + 3(1 � y)x � (4 + py) = 0

As 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 R

So, 9 + 16p > 0 and (4p2 + 46)2 � 4(9 + 16p)2 0

or 4 (p2 + 8p + 16) (p2 � 8p + 7) 0

or (p + 4)2 (p2 � 8p + 7) 0

or 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 < 7

So, the number of possible integral values of p are 5 (i.e., p = 2, 3, 4, 5, 6). Ans.]

20.

n

1k1k1kkk

k

2323

6 =

n

1k1k1k

1k

kk

k

23

3

23

3

= 22

2

1

1

23

3

23

3

= 33

3

2

2

23

3

23

3

= := :

Sn = 1n1n

1n

23

3

1

3

nnSLim

= 3 � 1 = 2. Ans.]

PAGE # 7

PRACTICE TEST # 2

Syllabus : Straight line and Circle.1. Only one is correct : 3 marks each.





[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, 1b

y

a

x 222222 ba

ab

ba

ay

ba

bx

2p = 22 ba

ab

4p2 = 22

22

ba

ba

8p2 = 22

22

ba

ba2

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) from

point (� 2, 3) will be 22 )8()6( = 10

So, number of required lines = zero. Ans. ]

PAGE # 8

5.O

(0, � 1)

0,2

5x

y

2

5,0

y = 2 (3, 2)

0,

2

9

(1, 0)

x � y = 5/2x � y = 1

From above graph, 3 < a < 2

9. Ans.]

6. Coordinates of G =

9

8,1

1 : 2

GO H�1

323

11 3

43

, ,� � �Now AG : GD = 2 : 1

3

1h2 = 1,

3

10k2 =

9

8

1 89

,�G

A(1, 10)

B CD(h, k)

2

1 (h, k) =

3

11,1 ]

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 � 1

x � 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) = �

32

12 (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,

5

3

= 22

3

LR

RL

=

41

81

=

5

8 required

PAGE # 9

Area of quadrilateral PQOR = 2 × area of POQ = 2 × 2

1 × 1 × 2 = 2

required ratio = 2

58 =

5

4

(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 = 4

3

equation of PQ is 3x � 4y + 5 = 0 ......(ii)Solving (i) and (ii)

Q =

5

4,

5

3

Let S = (x1, y1) x1 + 1 = 1 � 5

3 x1 =

5

3 Also y1 + 2 = 0 +

5

4 y1 =

5

6 S

5

6,

5

3

Equation of family of circle x2 + y2 � 1 + (x + 2y � 1) = 0.

It passes through

5

6,

5

3

1

5

151

25

36

25

9 = 0

= 5

1 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 = 5

1 =

5

1

QR = 25

11 =

5

4

For max. area perpendicular distance from A to line QR should be maximum and it will be equal to

= 1 + 5

1 =

5

15

Area of AQR = 5

15

5

4

2

1 = 15

5

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) = 3

1 (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) = 3

4 (x � 3) 4x = 3y

PAGE # 10

Co-ordinates of orthocentre (H)

9

8,

3

2

(B) Co-ordinates of centroid =

3

yyy,

3

xxx 321321 = (3, 1)

� ��2 1

O G C

G = 3

OC2

2

OG3C co-ordinates of circumcentre

C=

298

3,

232

9

C =

18

19,

6

25

P(1,1)

R(3,4)

Q

�H

E

k2 = 2

1819

625

= 18

47

value of 18k = (47)(C) Equation of line joining orthocentre and & centroid

G(3, 1) & H

9

8,

3

2 (y � 1) =

332

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

abcd

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 = cba

cxbxax 321

B(3, 6)

A(3, 2)C(6, 2)

y

xO

3

4 5

= 435

643335

= 9 center = (9, 8)

ya = cba

cybyay 321

= 8 radius ra = s. tan

2

A

(x � 9)2 + (y � 8)2 = 62 = 6 × 1

= 6Since ABC is a right angle triangle is hence

Circumcenter =

2

26,

2

63 =

4,2

9

PAGE # 11

The In-circle will be smallest circle which touches the sides

x1 = cba

cxbxax 321

=

435

643335

= 4

y1 = cba

cybyay 321

=

12

426325 = 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)

7

5,

7

5 ]

15. Let the equation of chord be

cos

1x =

sin

y = 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 3

4

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

4

1(x1

2 + y12 + x1y1 + y2

2 + 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 | = 2

represents 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 22

Ctan 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 < 5

6.......(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 < 3

2......(2)

(1) (2)

c

3

2,0 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) = 0

or, x2 + y2 + ( � 2) x � ( + 2)y + 2 = 0 .......(2)As, (1) and (2) are orthogonal so, using condition of orthogonality, we get

22

224

2

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 = 4

4c3

Equation of line PQ is,

c2

2

5x +

c

2

35 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 = 5c4

1c

,

10c3

1ck

Required locus is 9x � 13y + 25xy = 0

So, 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.





[STRAIGHT OBJECTIVE TYPE]1. Let f(x) = sin�1 x � tan�1 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 cos�1 x and 0 cos�1 (�x) Using A.M. G.M., We get

0

22

xcosxcosxcos.xcos

1111

f (x)

4,0

2

. Ans.]

3. Clearly,

9

8f =

39

8sincos 1

=

9

11sincos 1

=

9

11

2coscos 1

=

1813

coscos 1 =

18

13. Ans.]

4. For 2xx to be defined, 2xx 0 0 1x

Also,2

2

2

1x

4

1xx

so, 0 2xx 2

1 0 sin�1

6xx 2

. Ans. ]

PAGE # 15

5. We have sin�16x = x36sin2

1

sin (sin�16x) =

x36sin2

sin 1

6x = � x36sincos 1

Squaring both the sides, we get

36x2 = 1 � 108x2 144x2 = 1 x = ± 12

1

But x = 12

1 (Rejected)

Hence, x = 12

1. Ans.]

6. We have

1

x

11loglog)x(f

41

313

1

x

11log

4

13

1

> 0 log3 1

x

11

4

1

3

x

11

4

1

41x

1 > 2

41

x < 2

1 x <

16

1

0 < x < 16

1 (As, x 0)

Hence, domain of f(x) =

16

1,0 . 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) =

21

x1

x2sin2 = 2 +

21

x1

x2sin

As, sin�1 2x1

x2

2,

2

21

x1

x2sin = � 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, 1

Sum = � 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 < 5

110 1 2 3

x-axis

f(x) = � x � 2ax � a � 22

(1) (2) a Hence, no real a exists. Ans.]


11. Clearly, 1xgxgf 2 =

1

1x

12

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

� cos�1 )x(sincos 1 + cos�1 )x(cossin 1

= xcossincosxcos2

coscos2

1111

=

2

.

(B) For example : Let f(x) = x + 2 and g(x) = x + 3

Now, )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 R

But f(x) and g(x) are not inverse of each-other.(C) We have f(x) = sin�1 (sin ax)

Period = 2a

2

(Given)

| a | = 4 a = � 4, 4

Hence, sum of possible values of a = � 4 + 4 = 0. ]

PAGE # 17

(D) As, cot�1 (cot 6) = 6 � and tan�1 (tan 6) = 6 � 2 cot�1(cot 6) + tan�1 (tan 6) = 12 � 3 . ]

13. onto functions = !3!2

1·

!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 27

9C2 = 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) Tr + 1 = 20Cr (� 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 tan�1x + 2 tan�1x + 2 tan�1x = 6 tan�1x.(B) f(x) = 2 tan�1x � 2 tan�1x + 2 tan�1x = 2 tan�1x.(C) f(x) = � 2 tan�1x + 2 tan�1x + 2 tan�1x � = 2tan�1x(D) f(x) = � � 2 tan�1x � 2 tan�1x + + 2 tan�1x = � 2 tan�1 x.

[MATCH THE COLUMN]16.

(A)dx

dy = 3x2 + 2(a + 2) x + 3a

Put 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) tan�1 (2 tan x) + tan�1 (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 = 6

1

(C) Coefficient of x7 in 11

2

bx

1ax

= 11C5 5

6

b

a and coefficient of x�7 in

11

2bx

1ax

= 11C6 6

5

b

a.

ab = 1. Ans.

PAGE # 18

(D) We have

x3

xx3

x1

x3

xx3

xtan 1 =

x

6tan 1

2x

9x

x6

x6

x6

x2 � 2x

9 = 0 (x2)2 = 9 x2 = 3. Ans.]

[INTEGER TYPE / SUBJECTIVE]

17.

10

1k

1

)1k(k1

k1ktan

10

1k

1 )1k(tan � tan�1(k)

T1 = tan�12 � tan�11T2 = tan�13 � tan�12

T10 = tan�111 � tan�110

S = tan�111 � tan�11 = tan�112

10 = cot�1

10

12

cot

10

12cot 1

= 10

12 =

5

6 =

b

a a + b = 11 Ans. ]

18. Let a = cos�1x and b = sin�1y

So, 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 = 1

Case-I : When b = a,

Now, equation (1) becomes

a2 + a � 1 = 0 a = 2

51

But a [0, ] and b

2,

2 a = b =

2

15

(x, y) =

2

15sin,

2

15cos

Case-II :When a + b = 1,

Now, equation (1) becomes

a2 + (1 � a) = 1 a(a � 1) = 0

PAGE # 19

So, a = 0 or a = 1

If 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,

2

15cos , (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.





[STRAIGHT OBJECTIVE TYPE]

1.1xsinxsin

12

is always defined x R.

And sin�1 1x

12 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 = 0

Using parametric form of C1C2

13

33y

13

22x

= ± 132 . P(2,3)

C1

C2

2x�3y+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

10cotx

2

5 2 .

decagonregularofArea

pentagonregularofArea =

10cotx

25

5cotx5

2

2

=

18cot

36cot2 =

5

2.

[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) = 0

sin2x [5 sin2x � 2] = 0 sin x = 0 or sin2x = 5

2

sin x = 0 give x = and sin x = ± 5

2 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)

90º

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 = 1xtanxtan

1xtanxtan24

24

(y � 1) tan4x + (y � 1) tan2 x + (y + 1) = 0

Case I : y 1 tan x R D 0

(y � 1)2 � 4 (y � 1) (y + 1) 0 3

5 y < 1

Case II : y = 1 It didn't satisfy above equation

35

y < 1. Ans.]


Sol. As, tan = 43

71

1

71

1

sin = 53

Area = OA · OB sin = 9

B

A

C

y=x

y=17

xx

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°) =

2

3,

2

3

coordinate of A is =

50

1·50,

50

750 =

2

1,

2

7

Mid point of diagonal m =

2

2,

2

5, then line OC is y = x

52

.

0 = 2, b = 5 a + b = 7.

Also, AB = 10 (using distance formula). Ans.]


11. We have (y � 3) = m (x � 2) are other sides. Then, 3

tanm1

1m

.

m = � (2 + 3 ) or ( 3 � 2).

Find two values of m and the equation of sides.

A(2,3)

CB60° 60°

x�y+3=0

slope = m

Also, Area of triangle = 3

2 = cosec 60°. Ans.]

12.

(A) tan�1(x2) = cot�1

2x

1 is obviously true for all x R � {0}.

(B) cos�1

2

2

x1

x1 [0, ) for all x R cos�1

2

2

x1

x1 .

(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 �

2

n n I.

(D) is incorrect (f is not defined for any x. Domain is ) ]

13. Here, exponent of 2 is 2

1 and exponent of 3 is

5

1 and L.C.M. of 2 and 5 is 10.

So, only those terms will be rational in which power of both 2 and 5

1

3 are multiples of 10.

Here, index of

5

1

32 is 10, therefore only first and last terms in the expansion of 10

5

1

32

will be rational.

PAGE # 23

Now, sum of rational terms = t1 + t11 = 10C0 0

5

110

32

+ 10C0

10

5

10

32

= 25 + 32 = 32 + 9 = 41. Ans.]

14.

(A) f(x) = )x2sin1(

x2sin·x2cos2

)x2cos1(

x2cos·x2sin22

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(cosx2sin

xcosxsin)xsinx(cos442

2222

=

xcosxsin4

xcosxsin22

22

= 4

1

h(x) = 4

1 C is also correct.

(D) l(x) =

2x

cos2

)xcos1(xtan

2

= 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 argument

we have 2x + y = 3x � yx = 2y

y = 2

x.

Hence put y = 2

x 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........

2

3

2

2

2

1

Let P = ........2

3

2

2

2

132 ..........(1)

2

P = ........

2

2

2

132 ..........(2)

��

2

P = ........

2

1

2

1

2

132 (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

8

7

bbb

aaa

321

321 8

7

)d5)(d5(5

)d5)(d5(5

22

11

87

d25

d2522

21

25 · 8 � 8 · (d2 + 1)2 = 25 · 7 � 7

22d

25 = 8 ( 22d + 2d2 + 1) � 7 2

2d

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

sin�1x cos�1x = 18

2 sin�1x

xsin2

1 =

18

2

(sin�1x)2 � xsin.2

1 +

18

2

6xsin

3xsin 11

= 0

x1 = 3

sin

= 2

3; x2 = sin

6

=

2

1

x12 + x2

2 = 14

4

2

1

2

322

. Ans.]

19.y

x3 /4/2

� /4

/4

O

y=|cot 2x|

y=tan (tan x)� /4�1

Number of solution = 1 N = 1 Ans.

13 cosec x + 13 sec x = 4 2

PAGE # 26

2

1·

21

23

·2

1 cosec x +

2

1·

21

23

·2

1 sec x = 2

sin (60° � 45°) cosec x + cos (45° � 30°) sec x = 2

cos x sin 15° + sin x cos 15° = 2sin x cos x

sin 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 # 5

Syllabus : Calculus1. Only one is correct : 3 marks each.





[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

0

dxxf3dx1)x(f .

1�1x

y

3

�1

=

0

1

1

0

33

25

dxxx2dx2xx . Ans.]

Q.2

[Sol. ln c + ln x + ln y = y

x.

dx

dy

y

1

x

1 = 2y

dxdy

xy =

dx

dy

y

x

y

12

dx

dy =

1yx

yx

1yx

yx

dxdy

so (z) = )1z(z

1z

. 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(3

4 =

34

sq. units. Ans.]

PAGE # 28

Q.4[Sol. Let y = vx

dxdy

= v + xdxdv

v + x dx

dv = 2

222

vx2

x·vx 2v1

v2

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 4A42

. Ans.]

Q.6

[Sol. (x3y3 � xy)dy = �2dx

x3y3dy � yxdy = �2dx

�2dx + yx dy = x3y3dy

3

23y

x

y

dy

dx·

x

2

Let 2x

1 = t dy

dt

dy

dx·

x

23

dy

dt+ yt = y3

2/ydy·y 2ee.F.I

cdye·ye.t 2/y32/y 22 ce

2y

·e2e.t 2/y2

2/y2/y 222

2/y

22/y

2

2

e

c)2y(et

t = (y2 � 2) + c · 2/y2

e

1 = (xy)2 + cx2 · 2/y2e � 2x2

at x = 1, y = 0

1 = c � 2 c = 3

1 = x2 (y2 + 3·2/y2

e � 2) ]

PAGE # 29

Q.7

[Sol. Let r (x) = )4x(

x2x)2x)(1x( 2

12

2

x2x)4x(

)2x)(1x(2Lim)x(r Lim

2xix

]

Q.8[Sol. [2x � 1] is discontinuous at three points

x = 2

5,

2

3, �2

f(x) may be continuous if f(x) = ax3 + x2 + 1 = 0 at x = 2

5,

2

3, �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 < e�x < 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

sin·m·2Limsin

m N and n = 1 or 2. Ans.]

Q.11

[Sol. f(g(x)) = (sgn x)3 � (sgn x) =

0x00x,0

y

x1O�1

x � x3

g(f(x)) = sgn (x3 � x) =

1,0,1xat.0x1,1

1x0,10x1,1

1x,1

.

y

x1O�1x'

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 ] �

11

4

1=

4

1

�1�2 1 2 3 4

A2A1

y

xA2 =

3

2

234

xx4

x

=

4

81 � 27 � 9 =

4

9

A3 =

4

3

234

xx4

x

= 64 � 64 + 16 �

4

9 =

4

55.

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,e

0xif,e)x(f x

x

; f ' (0�) = h

)0(f)h0(fLim

0h

= 1

h

1eLim

h

0h

Hence20x

2x

cos1

2x

cos1cos1

Lim

1

x.2

2x

cos1

nm

2

[Using 2

1cos1Lim

20x

]

2x2

2x

cos1Lim qp

2

0x

[13th, 05-09-2010, P-1]

2x2

4x

sin4Lim qp

4

0x

PAGE # 31

2x24

x4Lim qp4

4

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

3s2sLim

2

1s

= 1s

)3s)(1s(Lim

1s

= )3s(Lim1s

= 4 (D)

II By substituting x = s into the equation of the secant line, and cancelling by s � 1 again, we get

y = 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 2

x12

0 < cot x 2 + 3

Hence, number of points of discontinuity is 3

i.e. x = cot�13, cot�12 and cot�11 = 4

. Ans.

(D)49/itf

Hence two solutions. ]

PAGE # 32

PART-C

Q.1

[Sol.704/itf B � A = )3(cot3)2(cot2 11 �

3

1cot

3

1

2

1cot

2

1 11[13th, 05-08-2007]

= 2(cot�12 + cot�13) + cot�13 �

2

1cot

6

1

3

1cot

2

1cot

3

1 111

= 2

+ cot�13 �

2tan6

1

41

= 4

+ cot�13 �

6

1

3tan4

3 1

= 8

+ cot�13 +

6

1tan�13 =

8

+ cot�13 �

6

1

3cot2

1=

8

+

12

+ cot�13 �

6

1cot�13

= 3cot6

5

24

5 1

hence a = 5; b = 24; c = 5; d = 6a + b + c + d = 40 Ans. ]

Q.2

[Sol. f (x) = 9x12x41a3an

9x12x422

2

l

22

2

)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 = 0

The 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) + = 22

16 � 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) = h

0)h1(fLim

0h

exists; but f ' (1) = A =

h

)h1(fLim

0h

now f ' (x) = h

)x(f)hx(fLim

0h

=

h

)x(fxh

1·xf

Lim0h

using functional relationship

f ' (x) =

h

)x(fx

)xh(1f)xh(1

)x(f

;

f ' (x) =

h

)x(f)xh(1

)x(f

)xh(·x

xh

1fLim

20h =

h

1xh

1Lim

x

1

t

)t1(fLim

0h20t

f ' (x) = x

)x(f

x

A2 where A = f ' (1)

let f (x) = y

yx

1

dx

dy = 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 # 6

Syllabus : 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.




Time : 120 min. Marks : 115

PART-AQ.1

[Sol. Let k�i�a

; k�i�3b

and k�j�2i�4c

Volume of parallelopiped = cba

= |240301

|

= 1 (0 � 2) � 0 + (6) = 6 � 2 = 4. Ans.]

Q.2[Sol. For | z � 1| maximum, z = � 4

centroid () = 3

z =

3

14� = �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

= 6

5. 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, 0z

0z

1

2

= 3

i

e

0z

0z

1

2

= 3

i

e

= 1

| z2 � 0 | = | z1 � 0 |O(0)

A(z )1

Re(z)

B(z )2

60º

Im(z)

Also,

1

2

z

zarg =

3

z1oz2 =

3

Triangle is equilateral. Ans.]

Q.6

[Sol. As, 1n2z

1 = (cos (2n + 1) � i sin (2n + 1))

1n2z

i4 = 4i (cos (2n + 1) � i sin (2n + 1))

1n2z

i4Re = 4 sin (2n + 1)

9

0n1n2z

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 = 230102k�j�i�

= )6(k�)4(j�)3(i� = k�6j�4i�3

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

410

y

310

x

= 1

Area of triangle ABC =

9

100

36

100

36

100

16

100

16

100

9

100

2

1

= 50 936

1

3616

1

169

1

=

643

1693650

=

72

6150 =

36

6125. Ans.]

Q.8[Sol. Given | z � 1 | = 2 Im(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)

3

1,0

3

1,0

Also, L3 : x = 0

Hence, area = 2

3

11

21

= 3

1. Ans.]

PAGE # 36

Q.9

[Sol. Equation of line L is 4

0z

0

2y

3

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 2z4

iz4

=

2i

i

e44

)e2()i4((As | z | = 2 z = 2ei)

= )sini()2cos1(

)sini(cosi2

=

cos·sini2sini2

)sini(cosi222

= )sini(cossin

sinicos

= cosec (� , � 1] [1, ) Option (B) is correct.]

Paragraph for question nos. 11 to 13[Sol. We have

L1 :

4

2z

2

6y

3

7x

and L2 :

3

4z

1

3y

2

5x

A(7�3 , 6+2 , 2+4 )

B(2µ+5, µ+3, 3µ+4)

L2

L1

(7,6,2)

(5,3,4)

Now,2

µ232 =

2µ23

=

1µ342

3 + 2µ � 2 = 3 + 2 � µ + 3µ = 5 ........(1)and 3 + 2µ � 2 = � 4 + 8 � 6µ 5 � 8µ = 2 ........(2) On solving (1) and (2), we get

= 2, µ = 1

So, 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

7

c

17

b

2

a

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)

= OBOAOP6

1 = |

74710101321

|61

= 426

1 = 7. Ans.]

PAGE # 37

Paragraph for question nos. 14 to 16

[Sol.6zz

zz.arg

12

13

A = 30°

C(z )3

B(z )2A(z )1

30°(i) Circumcenter is origin and centroid is 3

zzz 321 .

As centroid divides orthocenter and circumcentre in the ratio 2 : 1 (internally).

3

)02()z1( =

3

zzz 321 z = z1 + z2 + z3 H(z) O(0)

2 1G

(ii) Clearly, Asin

a =

Bsinb

(using Sine law)

a = | z2 � z3 |, b = | z1 � z3 |

Also, sin A = 2

1, B =

23

21

zz

zzarg

| z2 � z3 | =

23

21

31

zzzz

argsin2

|zz| | z2 � z3 | =

23

2131 zz

zzargcosec|zz|

2

1

(iii) If HBTC be parallelogram then midpoint of HT and BC should be same.

2

zzzz 321 =

2

zz 32

z = � z1 | z � z1 | = 2 | z1 | = 2 (circumradius)

Ra

R21

2aR2Asin

a,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) = 22

So, a = max. | z � (�2 + 3i) | = 9 + 22

and b = min. | z � (�2 + 3i) | = 9 � 22Hence, a + b = 18

Also, (a � b) = 24 and ab = 4

)ba()ba( 2

= 4

32324 =

4

292 = 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 2

7z

1

6y

2

5x

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 16

3z22y

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

rqp·rqp

= 1 + 16 + 64 + q·p2

= 81 (As , q·p

= 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

=

k�2j�2i�

|122232303

|

= 3. Ans.

(C) As, (, , ) lies on plane x + 2y + z = 4 + 2 + = 4 ......(1)

Now, 0vj�j�

0vj�·j�j�v·j�

vj�v·j�

k�j�i�j� k�i� = 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

b

y

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 = 4

a a = 4x ; y =

4

b b = 4y and z =

4

c c = 4z

Also,

c

1

b

1

a

1

1

22

= 2 4

1 = 222 c

1

b

1

a

1

So, 4

16 = 22 z

1

y

1

x

1

Hence, k = 4. Ans.]

PART-CQ.1[Sol. As, O, A, B and D are concyclic,

so cos 60º = AD

3

AD

BD

AD = 6 and OD = 5

Re(z)

Im(z)

D(0, 5) B(z )2

A(z )3O(0, 0)

60º

/3

C (z )1

Hence, | z3 | = OA = 2536 = 11Hence | z3 |

2 = 11. Ans.]

Q.2

[Sol. The normal vector of plane is parallel to vector = 311122

k�j�i�

= k�4j�7i�5

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 = 120011k�j�i�

= k�2j�i�

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 = 3

2 z =

3

2

Hence

2i,3

2z ]

Q.5

[Sol. Here, |v|1|u|

Also, v·u

= 0 vu

Now, vu4vu

= vvuu4vu

= vv·vvv·uuv·uvu·u4

= 4 u0v

= v4u

vu4vu·v4u

= 2

v4u

= 1 + (4)2 + v·u8

= 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) = 4

3 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(

2

1 = 32 (square unit.)

= 32 2 = 12 Ans.]

rt solutions-practice test papers xiii vxy 1 to 6 sol

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