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Master SolutionPremier Institute for the preparation of IIT-JEE /

Date 01 05 2011 Class : 10 + 1 M.M:171 Time: 1:30 hrs

Test Code 4001 TEST : Mathematical Tools (Master solution)

SECTION-A (ONE OPTION CORRECT TYPE QUESTIONS)

The value ofo2o4

o42

20cos20sin

20cos20sin

++

isQ.1

(a.) 1 (b.) 2 (c.)

2

1

(d.) None of these

Ans. (a)

o2o4

o4o2

20cos20sin

20cos20sin

++

=o2o22

o2o2o2

20cos]20cos1[20sin

)20sin1(20cos20sin

++

120cos20sin1

20cos20sin1o2o2

2o2

==

sinP2P 20 + sinP2 P70Q.2

(a.) 0 (b.) 1 (c.) 2 (d.) sinP2Px

Ans. (b)

Let p = sinP2P 20 + sinP2P 70= sinP2P20 + sinP2P(90 - 20)= 1

The value of8

7sin

8

5sin

8

3sin

8sin 2222

+

+

+

isQ.3

(a.) 1 (b.) 2 (c.)

8

11

(d.)

8

12

Ans. (b)= sinP2P

8

+ sinP2

8sin

8

5sin

8

3 22 7++

=

+

+

+

8

3sin

8

3sin

8sin

8sin 2222

=

+

8

3sin

8sin2 22

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Master SolutionPremier Institute for the preparation of IIT-JEE /

=

+

82sin

8sin2 22

=

+

8cos

8sin2 22 = 2

The value of cos 1 cos 2 cos 3.. cos 179 is equal toQ.4

(a.)

2

1

(b.) 0 (c.)

2

1

(d.) 2

Ans. (b)

cos1 cos2 cos3 cos 179= 0 {as cos 90 = 0

If tan A =2

1and tan B =

3

1, then the value of A + B is

Q.5

(a.)

4

(b.) 0 (c.) (d.)

4

Ans. (a/d)

Tan (A + B) =Btan.Atan1

BtanAtan

+

Put tan A =2

1; tan

3

1B =

1)BAtan( =+A + B = /4The maximum and minimum values of 6 sin x cos x + 4 cos 2x are:Q.6

(a.) 5, 5 (b.) 4, 4 (c.) 7,7 (d.) 132,132

Sol: (a)

6 sin x cos x + 4 cos 2x = 3sin2x + 4 cos 2x

Max = 22 43 + 5

The greatest value of sinx. cosx isQ.7

(a.) 1 (b.) 2 (c.)

2

1

(d.) 1

Sol: (c)

y = sin x cos x = x2sin2

1 greatest value =

2

1

The third term of a G. P. is 4. The product of the first five terms isQ.8

(a.) 4P3P (b.) 4P5P (c.) 4P4P (d.) 4P2P

Sol: (b)

TB3B = arP2

P = 4

Product of first five terms

= a ar arP2P arP3P arP4

P aP5PrP10P = (arP2P)P5P = 4P5P

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Which term of the sequence 72, 70, 68, 66,.. is 40?Q.9

(a.) 17 (b.) 15 (c.) 19 (d.) 16

Sol: (a)

72, 70, 68, 66 ., 40 A. P

a = 72, d = 2, TB

nB

= 40 40 = 72 + (n 1) (2)32 = 2 (n 1) n = 17

The 10PthP and 18PthP terms of an A. P. are 41 and 73 respectively. Find 26PthP term.Q.10

(a.) 100 (b.) 105 (c.) 110 (d.) 108

Sol: (b)

aB10B = 41 and aB18B = 73

a + 9d = 41 .(1) and a + 17 d = 73 ..(2)

Solving eq (1) & (2), we get

8d = 32 d = 4 a + 9 4 = 41 a = 41 36 = 5, a = 5

aB26B = a + 25 d = 5 + 25 4 = 5 + 100

aB2BB6B = 105

One root of the equation 5xP2P + 13x + K = 0 is the reciprocal of the other, ifQ.11

(a.) K = 0 (b.) K = 5 (c.)K =

6

1

(d.) 6

Ans. (b)5xP2P + 13x + K = 0

Let one root be Then other root be

1

Product of roots ==

1

5

K

K = 5For what value of p the difference of the roots of the equation xP2P px + 8 = 0 is 2?Q.12

(a.) 2 (b.) 4 (c.) 6 (d.) 8

Ans. (c)xP2P px + 8 = 0

Let its root be &

+ = p = 8( - ) = 2

222 p2 =++

16p222 =+

4222 =+

48216p2 =p = 6

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If , are the roots of the equation xP2P p(x + 1) c = 0, then ( + 1) ( + 1) =Q.13

(a.) c (b.) c -1 (c.) 1 c (d.) None of these

Ans. (c)xP2P p(x + 1) c = 0xP2P px (p + c) = 0

)cp( +=

.1)1()1( +++=++= p c + p + 1= 1 - c

The roots of the equation 2P2xP 10. 2PxP + 16 = 0 areQ.14

(a.) 2, 8 (b.) 1, 3 (c.) 1, 8 (d.) 2, 3

Ans. (b)

2P2xP 10 2PxP + 16 = 0Let 2PxP = y

yP2P 10y + 16 = 0yP2P 8y 2y + 16 = 0

y(y 8) - 2(y 8) = 0(y 2) (y 8) = 0y = 2, 8

2PxP = 2P1 P x = 12PxP = 2P3P x = 3

The angle between the lines 2x y +3 = 0 and x + 2y + 3 = 0 isQ.15

(a.) 90 (b.) 60 (c.) 45 (d.) 30

Sol: (a)

2x y + 0

x + 2y + 3 = 0

mB1B = + 2, mB2B =2

1

0

2/5

m.m1

mmtan

12

12 =+

=

= /2

The equation of the line which has Y-intercept 4 units and is parallel to the line 2x 3y 7 = 0. Will cuts

the X-axis.

Q.16

(a.) (-6, 0) (b.) (-5, 0) (c.) (5, 0) (d.) (4, 0)Ans. (a)

2x 2y 7 = 0Line parallel to above line2x 3y + k = 0 (1)y- intercept = 4y-intercept is obtained by putting x = 0

y =3

k

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Master SolutionPremier Institute for the preparation of IIT-JEE /

12k;43

k==

Eqn (1) become ; 2x 2y + 12 = 0It will cuts the x ax is it y = 02x = -12x = -6 pt, (-6, 0)The inclination of the straight line passing thro the point (-3, 6) and the midpoint of the line joining the

point (4, -5) and (-2, 9) is

Q.17

(a.)

4

(b.)

6

(c.)

3

(d.)

4

3

Ans. (d)

xB1B=2

24 yB1B=

2

59

xB1B = 1 yB1B = 2xB2B = -3 yB2B = 6

113

26m ==

tan =4

3

The ratio in which the line segment joining (2, 3) and (4, 1) is divided by the line joining (1, 2) and ( 4, 3)

is

Q.18

(a.) 1 : 2 (b.) 1 :3 (c.) 2 : 3 (d.) None of these

Sol: (b)

eq. of CD is( ) ( )112

121 xx

xx

yyyy

=

(4,1

y 2 = ( 1x1423

) y 2 = ( )1x

31

(x, y)

(4,3)

(2,3)

m n

P

D

C(xB1B, yB1B)

(1,2)

A B

(xB2B, yB2B)

3(y 2) = (x 1) .(1)

nm

n2m4

nm

2n4m

++

=++

P(x,y) is x =

nm

n3m

nm

3n1my

++=

++

=

Put in eq. (1), we get

++= ++1

nmn2m4

12

nmn3m3 + += + + nm

nmn2m4nm

n2m2n3m3

nm31

nm3 +=

+

3m + 3n = 3m + n 3m 3m = n 3n

6

2

n

m= m : n = 1 : 36m = - 2n

Q.19 The vertices of a triangle ABC are (1, 2), (0, 1) and (2, 0) respectively. The equation of the median

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through (1, 2) is

(b.) (c.) (d.) None of these(a.) 5x 4y 3 = 0 x + 2y 2 = 0 4x + y 1 = 0

Sol: (a)

++=

2

01,

2

20D

=

2

1,1D

B

A CD(0,1) (1,1/2)

(1,2)

(2,0)

)xx(xx

yy1

12

12

eq of BD is y yB1B =

( )1x11

2

12

2

1y

= ( )1x

22

14

2

1y

=

( 1x4

5

2

1y2=

( ) 5x51y22 =)

5x 5 4y +2 = 05x52y4 =

5x 4y 3 = 0

If 7 times the 7 PthP term of an A.P is equal to 11 times its 11PthP term. Then 18PthP term of the A.P. isQ.20

(a.) Zero (b.) 7 (c.) 13 (d.) 9

Ans. (a)

7(a + 6d) = 11(a + 10d)

7a + 42d = 11a + 10d

4a + 68d = 0

4(a + 17d) = 0

A + 17d = 0

18PthP term = a + 17d

= 0

SECTION-B (ONE OR MORE THAN ONE CORRECT TYPE QUESTIONS)5

12Q.1If cot = , is in quadrant III then

13

5

13

12cos=

(b.)(a.)sin =

1312cos =

13cossin 7=+(c.) (d.)

Ans. (a,b)

cot =5

12

5=

13

5

12

12

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The fourth, seventh and last term of a G. P. are 10, 80 and 2560 respectively. Which of the following is/are

correct.

Q.2

Ratio of 5PthP term of series to the 4 PthP term is 2.(a.)

(b.) The common ratio of the series is 2.

(c.) Total number of terms of the series is 12

(d.) The fifth term of the series is 20

Sol: (a, b, c, d)

aB4B = 10, aB7B = 80, aBnB = 2560

arP3P = 10 ..(1)

arP6P = 80 ..(2)

Divide (2) by (1), we get

10

80

ar

ar3

6

= rP3P = 8 r = 2

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a(2)P3P =10 a 8 = 10 a =4

5

8

10 =4

5a =

24

5 aB2B = ar = aB2B =

2

5

2560)2(4

5 1n=

Also arPn1P = 2560

5(2)Pn1P = 2560 4

(2)Pn1P = 512 4

(2)Pn1P = (2)P11P n 1 = 11 n =12

aB5B = arP4P

= ( )424

5 16

4

5= = aB5B =20

Q.3

4

1If sin = , then which of the following is/are correct?

8

7

15

1

15

4(c.)(b.) (d.)(a.)

4

3 Tan = Sec =Cos 2 =Sin 2 =

Sol: (b, c, d)

sin =4

1

sin 2 = 2 sin. cos = 2 4

15

4

1

sin2 =8

15 .(1) A B

C

1

4

15

4P2P = 1P2P + ABP2Pcos 2 = 1 2 sinP2

P

= 1 2 2

4

1

16

1

8

18 AB116 = = 1 2 =

AB15 =

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cos 2 =8

7

tan =15

1

Q.4 If 3 sin + 4 cos = A sin ( + ), then which of the following is/are correct?

(d.)(a.) A = 5 (b.) A = 7 (c.)

3

4tan =

4

3tan =

Sol: (a, c)

3sin + 4 cos = A sin ( + )22 baA += 22 43 + 169 + 25= = = = 5

As [ ] + cos4sin3

=

+ cos

5

4sin

5

35

5

4sin,

5

3=cos =

3

5

5

4 tan =

tan =3

4

Q.5 The equation of straight line is given as 3x 4y + 10 = 0 . Then

4

3

4

3

(a.) (b.)Its slope is Its slope is

3

10

(c.) (d.) It will pass through (6, 7)Its x-intercept is

Ans. (b, c, d)3x 4y + 10 = 0 (1)4y = 3x + 10

y =4

10x

4

3+

slope =4

3

to find x-intercept put y = 03x = -10

x =3

10

Since pt (6,7) satisfy eq(1) so the eqn pass through (6,7)

Q.6 The right bisector of segment joining (3, 4) and (-1, 2) has

(a.) Slope 2 (b.) Slope 2

(c.) Equation 2x + y = 5 (d.) Equation 2x + y + 5 = 0

Ans. (a, c)A (3, 4) B = (-1, 2)

2

24,

2

13 +Pt. from which bisector pass =

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= (1, 3)& slope of right bisector = -2

Eqn = y 3 = -2(x -1)y 3 =-2x + 2

2x + y -3 2 = 0

2x + y 5 = 0Q.7 It and are the roots of 4xP2P + 3x + 7 = 0, then

7

3

7

3

16

47

16

47

(a.) (b.) (c.) (d.) + = + = P2P + P2P = P2P + P2P =Sol: (d)

4xP2P + 3x + 7 = 0

4

3=+

4

7=

+=+ 2)( 222

=4

72

16

9

2

7

16

9=

16

47=22 +

Q.8 A person standing on the bank of a river, observes that the angle of elevation of the top of a tree on the

opposite bank of the river is 60 and when he retires 40 m away from the tree the angle of elevation

becomes 30. Then

(a.) width of the river is 20 m (b.) width of the river is 60 m

(c.) (d.)320 3/20Height of the tree is Height of the tree is

Ans. (a, c)

6040 x

30y

o60tanx

y= x3y =

o30tan40x

y=+

3

40xy+= 40x)x3(3 +=

3x = x + 40 x = 20

320y =

Width of the river = 20 m

3 Height of tree = 20

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Q.9 The sequence 9, 12, 15, 18, is an A.P. Then

(a.) Its general term is 3n + 6 (b.) Its general term is 3n 6

(c.) Its 16PthP term is 54 (d.) Its 10PthP term is 36

Ans. (a,c, d)

9, 12, 15, 18 .

a = 9d = 3general term = a +(n -1) d

= 9 + (n -1)3= 3n + 6

10PthP term 9 +(10 1)9 + 27 = 36

16PthP term = 9 + 15 3= 54

Q.10 A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3.Then

31

34(a.) (b.)Its y-intercept is Its y-intercept is

Its equation is x 3y + 4 = 0 Its equation is x 3y = 4(c.) (d.)

Ans. (b,c)

3

1 Slope of required line =

3

1Eqn of line = y 2 = (x 2)

3y 6 = x 2 x 3y + 4 = 0

3

4

Its y-intercept =SECTION-C (SUBJECTIVE TYPE QUESTIONS)Q.1

=

+

2tan

2tan

3 sin = 5 sin , then

Sol: [4]

= sin5sin3

3

5

sin

sin=

35

35

sinsin

sinsin

+=+

2sin.

2cos2

2cos.

2sin2

+

+

= 4

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2.tan

2tan

+

= 4

Q.2 Find the sum of the given series up to 2 decimal places

..........321

161

81

41

211 +++

Sol: [0.66 or 0.67]

As it is infinite G.P. series; usingr1

aS= =

2

11

1=

3

2= 0.67

Q.3 Find the slope of the equation of the perpendicular bisector of the line segment joining the points A(2, 3)

and B( 6, 5)

Sol: [0.5 ]

Sol: 0.5

12

12

xx

yy

4

826

35

mB1B = 2= =Slope of AB =

Using mB1B mB2B = 1

-2 mB2B = 1

mB2B =2

1 mB2B = 0.5

Q.4 If one root of the equation (x 1) (7 x) = m is three times the other, then m is equals to

Ans. [5](x 1) (7 x) = m

Let one root = Other root = 37x xP2P -7 + x = mxP2P 8x + 7 + m = 0

4 = 8 = 27 + m = 3P2Pm + 7 = 12m = 5

Q.5 The x-intercept of equation of a line perpendicular to x 2y + 3 = 0 and passing through the pt(3, -2) is

Ans. [2]

Slope of required line = -2

Eqn of line y + 2 = -2(x 3)y + 2 = -2x + 62x + y 4 = 0

For its x-intercept put y = 0x = 2

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SECTION-D (MATCH THE COLUMN TYPE QUESTIONS)(Items of Column- I can match one or more than one Items of column- II with individual marking scheme)

Q.1 In the following options symbols have their usual meaning. Match the column I with column-II

Column-I Column- II

r1a

(i.) Sum of n terms in arithmetic progression (a.)

( )r1

r1a n

(ii.) nPthP term of arithmetic progression (b.)

(iii.) nPthP term of geometric progression (c.) a + (n 1)d

arPn1P(iv.) Sum of infinite series of geometric progression (d.)

( )[ ]d1na22

n+

(e.)

Sol: (i) (e), (ii) (c), (iii) (d), (iv) (a)

Column-I Column- IIQ.2

(a.) 0(i.) If sinB1B + sinB2B + sinB3B = 3 then1

cos B1B + cos B2B + cos B3B =

(b.) 1(ii.) tan(180 + ) . tan(90 + ) =

(c.) 2(iii.) cotA + tan (180 + A) + tan(90 + A) +

tan(360 - A) =

(iv.) (d.) 3

o

o

o

o

70cot

20tan

36tan

54cot+ =

(e.) -1

Ans. (i a), (ii e), (iii a), (iv c)(i) sin B1B + sinB2B + sinB3B = 3

2

So, B1B = B2B = B3B =

So, cosB1B + cosB2B + cosB3B

= 0

(ii) tan(180 + ) . tan(90 + )

=tan.[-cot ] = -1

(iii) cotA + tan(180 + A) + tan(90 + A) +tan(360 - A)

= cotA + tanA + [-cotA) tanA

= 0

o

o

o

o

70cot

20tan

36tan

54cot+(iv)

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=oo

o

o

oo

2090cot(

20tan

36tan

)3690cot(

+

= 1 +1 = 2