ASPIRE SCHOLARSHIP Exam Solutions - … Class... · The question paper consists of 45 multiple choice questions. (Single correct Answer) Each right answer carries 4 marks and −

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Exam Solutions

11th Class Test Name: P (Set–I)

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1. z1 and z2 are two complex numbers such that 1 2

1 2

z 2z

2 z z

is unimodular, while z2 is not unimodular.

Find 1|z | .

(a) 4 (b) 2 (c) – 2 (d) None of these

Ans. (b)

Solution:

Here 1 2

1 2

z 2z1

2 z z

1 2

1 2

|z 2z |

|2 z z |

= 1 (By Prop. VI)

1 2 1 2|z 2z | |2 z z |

2 21 2 1 2|z 2z | |2 z z |

1 2 1 2 1 2 1 2(z 2z )(z 2z ) (2 z z )(2 z z ) (By Prop. IV)

1 2 1 2 1 2 1 2(z 2z )(z 2z ) (2 z z )(2 z z )

1 1 1 2 2 1 2 2 1 2 1 2 1 1 2 2z z 2z z 2z z 4z z 4 2z z 2z z z z z z

2 2 2 21 2 1 2|z | 4|z | 4 |z | |z |

2 2 2 21 1 2 2|z | |z | |z | 4|z | 4 0

2 21 2(|z | 4)(1 |z | ) 0

But 2|z | 1 (given)

21|z | 4

1|z | 2.

2. Let z1 = 10 + 6i, z2 = 4 + 6i. If z is a complex number such that the argument of (z – z1)/(z – z2) is

π/4 then prove that |z 7 9i| 3 2 .

(a) 3 2 (b) 2 3 (c) 7 9 2 (d) 9 7 2

Ans. (a)

Solution:

We have

arg 1

2

z z π

z z 4




arg (z – z1) – arg (z – z2) = π/4

Let z = x + iy

arg ((x – 10) + i (y – 6)) – arg ((x – 4) + i (y – 6)) =π

4

1 1y 6 y 6 πtan tan

x 10 x 4 4

1

y 6 y 6πx 10 x 4tan

(y 6)(y 6) 41(x 10)(x 4)

2 2

6(y 6) πtan

4x y 14x 12y 76

2 2

6(y 6)1

x y 14x 12y 76

x2 + y2 – 14x – 18y + 112 = 0

(x – 7)2 + (y – 9)2 = 18

(x – 7)2 + (y – 9)2 = 2

3 2

|(x 7) i(y 9)| 3 2

|z 7 9i| 3 2

3. If r r r

π πx cos i sin

3 3

, then x1x2x3….upto infinity =

(a) i (b) –i (c) 1 (d) None of these

Ans. (a)

Solution:

We have

r r r

π πx cos i sin

3 3

1 2 3 2 3 2 3

π π π π π πx x x ... cos .... i sin ....

3 33 3 3 3

=

π π

3 3cos i sin1 1

1 13 3




= π π

cos i sin2 2

= 0 + i

= i

4. If α, β, γ are roots of x3 – 3x2 + 3x + 7 = 0 (and ω is imaginary cube roots of unity), then the value of

α 1 β 1 γ 1

β 1 γ 1 α 1

is.

(a) 2ω (b) 3ω (c) –3ω (d) 23ω

Ans. (d)

Solution:

We have x3 – 3x2 + 3x + 7 = 0

(x – 1)3 + 8 = 0

(x – 1)3 = (–2)3

3

x 11

2

1/3x 1

12

= 21, ω, ω (cube roots of unity)

x = – 1, 1 – 22ω, 1 2ω

Here 2α 1, β 1 2ω, γ 1 2ω

2α 1 2, β 1 2ω, γ 1 2ω

Then 2

2

α 1 β 1 γ 1 2 2ω 2ω

β 1 γ 1 α 1 2ω 22ω

= 21 1ω

ω ω

= 2 2 2ω ω ω

= 23ω

5. If 8π 8π

α cos i sin11 11

then Re 2 3 4 5α α α α α is

(a) 1/2 (b) –1/2 (c) 0 (d) None of these

Ans. (b) Here

8πi11

8π 8πα cos i sin e

11 11




6. Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n

sides. If n 1 nT T 21, then n equals

(a) 5 (b) 7 (c) 6 (d) 4

Ans. (b)

Solution:

n+1C3 – nC3 = 21

nC2 + nC3 – nC3 = 21

n 72 2

21 2 7 6C 21 C

2 2

n = 7

7. Two numbers are chosen from 1, 3, 5, 7,…., 147, 149 and 151 and multiplied together in all possible

ways. The number of ways which will give us the product a multiple of 5, is

(a) 75 (b) 1020 (c) 95 (d) 1030

Ans. (b)

Solution:

In the given numbers 1, 3, 5, 7, …, 147, 148, 151 the numbers which are multiple of 5 are 5, 15, 25, 35,

…, 145, which is an arithmetic sequence

Tn = a + (n – 1) d

145 = 5 + (n – 1) 10

n = 15

and if total number of terms in the given sequence is m, then

151 = 1 + (m – 1) 2

m = 76

So, the number of ways in which product is a multiple of 6

= (both two numbers from 5, 15, 25, …., 145)

or (one number from 5, 15, 25, …, 145 and one from remaining numbers)

= 15C2 + 15C1 76 – 15C1

= 15C2 + 15C1 61C1

= 15 14

15 612

= 105 + 915

= 1020.




8. The remainder obtained, when 1! + 2! + 3! + … + 175! Is divided by 15 is

(a) 5 (b) 0 (c) 3 (d) 8

Ans. (c)

Solution:

5, 6, 7, 8,..., 175 each multiple of 15 and 1 2 3 4 33

15 33 2

30

3

Hence, required remainder = 3.

9. When simplified, the expression 5

47 52 n4 3

n 1

C C

equals

(a) 47C5 (b) 49C4 (c) 52C5 (d) 52C4

Ans. (d)

Solution:

547 52 n

4 3n 1

C C

= 47C4 + 51C3 + 50C3 + 49C3 + 48C3+ 47C3

= 51C3 + 50C3 + 49C3 + 48C3 + (47C3 + 47C4)

= 52C4

(Using Pascal rule successively)

10. If 100

x iy 1 i 3 , then (x, y) =

(a) (299, 299 3 ) (b) (299, – 299 3 ) (c) (–299, 299 3 ) (d) None of these

Ans. (c)

Solution:

If x + iy= 100

1 i 3 [Given]

=

100

1 i 32

2

1 i 3

ω2

100 100( 2) .ω

1002 3 33.(ω ) ,ω 3[ ω 1]

1002 .ω




100 1 i 32 .

2

992 1 i 3

99 992 i2 3)

Hence (x, y)= 99 99( 2 ,2 3)

11. Given that n is the odd, the number of ways in which three numbers in AP can be selected from

1, 2, 3, 4, …, n is

(a)

2n 1

2

(b)

2

n 1

4

(c)

2

n 1

2

(d)

2

n 1

4

Ans. (d)

Solution:

Let n = 2m + 1

If a, b, c are in AP

then 2b = a + c

i.e., sum of two numbers is even.

Here, m + 1 is odd and m is even number.

Sum of two numbers is even, then both numbers are even or odd.

Required number of ways = mC2 + m+1C2

= m m 1 m m 1 m

2 2

= m2

= 2

n 1

2

=

2n 1

4

12. The maximum number of different permutations of 4 letters of the word EARTHQUAKE is

(a) 1045 (b) 2190 (c) 4380 (d) 2348

Ans. (b)

Solution:

Here 2E, 2A, 1R, 1T, 1H, 1Q, 1U, 1K




Maximum number of permutations = coefficient of x4 in

22

6 x x4 1 x 1

1 2

= coefficient of x4 in 6(1 + x)6 [(1 + x)2 + 1]2

= coefficient of x4 in 6{(1 + x)10 + 2(1 + x)8 + (1+ x)6}

= 6. {10C4 + 2.8C4 + 6C4}

= 6. 10.9.8.7 8.7.6.5

2. 151.2.3.4 1.2.3.4

= 2190 .

13. The value of 1515

2 r15

r 1 r 1

Cr

C

is equal to:

(a) 1085 (b) 560 (c) 680 (d) 1240

Ans. (c)

Solution:

1515 15 15

2 2r15

r 1 r 1 r 1r 1

C 15 r 1 15 16 15 16 31 15 16r r r 16 r 16 17 680.

C r 2 6 6

14. If z 0, then 100

x 0

[arg|z|]dx

is (where |z|denotes modulus and [.] denotes greatest integer function

(a) 0 (b) not defined (c) 100 (d) none of these

Ans. (a)

Solution:

100

X 0

arg z dx

arg of z is defined but arg of |z| is 0

So,100

X 0

arg z dx

=0

15. If positive numbers a–1, b–1, c–1 are in AP, then the product of roots of the equation

2 101 101 101x kx 2b a c 0, k R has

(a) > 0 (b) < 0 (c) = 0 (d) undefined

Ans. (b)

Solution:

a–1, b–1, c–1 are in AP




a, b, c are in HP

Now for any three numbers a101, b101, c101

AM > GM

101 101

101 101a c( ac) b

2

( ac b )

101 101 101a c 2b 0

101 101 1012b a c 0 …(i)

Product of roots = 101 101 1012b a c

01

[from Eq. (i)]

Product of roots < 0

16. If 2 n 2 41 λ λ ... λ 1 λ 1 λ 1 λ 8 161 λ 1 λ , then the value of n is (where n N)

(a) 32 (b) 16 (c) 31 (d) 15

Ans. (c)

Solution:

LHS =

n 1 n 11 1 λ 1 λ

1 λ 1 λ

and RHS = 2 4 8 161 λ 1 λ 1 λ 1 λ 1 λ

=

2 4 8 161 λ 1 λ 1 λ 1 λ 1 λ 1 λ

1 λ

=

321 λ

1 λ

n 1 321 λ 1 λ

1 λ 1 λ

n 1 321 λ 1 λ

n + 1 = 32 n = 31

17. If log3 2, log3(2x – 5) and log3 x 72

2

are in A.P., then the value of x is

(a) 3 (b) 1 (c) 4 (d) 5

Ans. (a)

Solution:

Since given numbers are in A.P.




x x3 3 3

x 2 x

72 log (2 5) log 2 log 2

2

7(2 5) 2 2

2

(22x – 10 . 2x + 25 = 2. 2x – 7

22x – 12 . 2x + 32 = 0

y2 – 12y + 32 = 0, where y = 2x

(y – 8) (y – 4) = 0

y = 8 or 4

2x = 23 or 22

x = 3 or 2.

But x = 2 gives 2x – 5 = 4 – 5 = – 1

x = 2 is not possible. Hence x = 3.

18. Given that α, γare roots of the equation ax2 – 4x + 1 = 0, and β, δ the roots of the equation

bx2 – 6x + 1 = 0, the values of a and b such that α, β, γ and δare in H.P. are

(a) a = 3, b = 8 (b) a = – 3, b = 8 (c) a = 3, b = – 8 (d) none of these

Ans. (a)

Solution:

1 1 1 1α,β, γ andδare inH.P. , , , are in A.P.

α β γ δ

Le td be the common difference of this A.P.

Now, , are roots of Ax2 – 4x + 1 = 0

4α γ 1 1 1 1A 4 or 4 i.e., 2d 4

1αγ α γ α αA

1or d 2

α

, are roots of Bx2– 6x + 1 = 0




6β δ 1 1 1 1B 6 or d 3d 6

1βδ β δ α αB

1or 2d 3

α

From (1) and (ii), on solving, we get

1 1 1 1 11,d 1. 1, 2, 3, 4

α α β γ δ

1 1Since, A A 3. Also, B, B 8.

αγ βδ

Hence, A = 3 and B = 8.

19. For any three positive real numbers a, b and c, 9 (25a2 + b2) + 25(c2 – 3ac) = 15b(3a+ c). Then

(a) b, c and a are in G.P. (b) b, c and a are in A.P. (c) a, b and c are in A.P. (d) a, b and c are in G.P.

Ans. (b)

Solution:

225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0

(15a)2 + (3b)2 + (5c)2 – (15a) (3b)– (3b) (5c) – (15a) (5c) = 0

1

2[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)]2 = 0

15a = 3b, 3b = 5c, 5c = 15a

5a = b, 3b = 5c, c = 3a

a b c

1 5 3

a = , b 5 , c 3 a, c, b are in A.P or b, c, a are in A.P

20. The numbers 2 sin 2x 1 4 2 sin 2x3 , 14, 3 form first three terms of an AP, its fifth term is equal to

(a) – 25 (b) – 12 (c) 40 (d) 53

Ans. (d)

Solution:

Since, 2 sin 2x 1 4 2 sin 2x3 , 14, 3

are in AP.

2 sin 2x 4

2sin 2x

1 128 3 . 3 .

3 3

Put 2 sin 2x3 t




t 81

283 t

84t = t2 + 243

t2 – 84t + 243 = 0

(t – 81) (t – 3) = 0

t = 3, 81

2 sin 2x 1 43 3 , 3

2 sin 2x = 1, 4

sin 2x = 1

, 22

( sin 2x 2)

sin 1

2x2

First term = 1 13 1

Second term = 14

Third term = 27

Here, common difference = 13

Fifth term = 1+ 4 13 = 53.

21. If three positive real numbers a, b, c are in AP with abc = 4, then minimum value of b is

(a) 4 (b) 3 (c) 2 (d) 1/2

Ans. (a)

Solution:

1/3a b c

abc3

1/32b b

643

b 4

Minimum value of b is 4.

22. If the sum of the first ten terms of the series 2 2 2 2

23 2 1 4 161 2 3 4 4 ...., is

5 5 5 5 5

m, then m is

equal to

(a)102 (b) 101 (c) 100 (d) 99

Ans. (b)

Solution:




2 2 2 2

2

1 16[8 12 16 20 .... upto 10 terms] m

5 5

22 3 2

2

4 16[2 3 ... 11 ] m

5 5

2 2 2 21[ 1 2 .... 11 1 ] m

5

1 11 12 231 m

5 6

22 23 1 5m

m = 101

23. A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinate

axes at points P and Q. As L varies, the absolute minimum value of OP + OQ is (O is origin)

(a) 10 (b) 18 (c) 16 (d) 12

Ans. (b)

Solution:

The equation of the line L be y – 2 = m(x – 8), m < 0 coordinates of P and Q are P2

8 , 0m

and

Q(0, 2 – 8m).

So, OP + OQ = 8 – 2 2 2

2 8m 10 8( m) 10 2 8 m 18m ( m) ( m)

So, absolute minimum value of OP + OQ = 18

24. Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx + 1 equals

(a)

2

|m n|

m n

(b)

2

|m n| (c)

1

|m n| (d)

1

|m n|

Ans. (d)

Solution:

If p1 and p2 be the distance between parallel sides and θ be the angle between adjacent sides, then




Required area = p1p2 cosec θ

Where, 1 22 2

1 1p , p

(1 m ) (1 n )

(distance between lines)

and |m n|

tan θ|1 mn|

Required area

= 2 2

2 2

(1 m ) (1 n )1.

|m n|(1 m ) (1 n )

= 1

|m n|

25. A ray of light coming along the line 3x + 4y – 5 = 0 gets reflected from the line ax + by – 1 = 0 and goes

along the line 5x – 12y – 10 = 0, then

(a) 64 112

a , b115 15

(b) 64 8

a , b115 115

(c) 64 8

a , b115 115

(d) 64 8

a , b115 115

Ans. (c)

Solution:

Equation of bisectors of the given lines are

2 2 22

3x 4y 5 5x 12y 10

3 4 5 12

(39x + 52y – 65) = (25x – 60y – 50)

14x + 112y – 15 = 0 or 64x – 8y – 115 = 0

or 14 112

x y 1 015 15




or 64 8

x y 1 0115 115

14 112

a , b15 15

or 64 8

a , b115 115

26. If m2 2 1 2

1!9! 3!7! 5!5! n! , then orthocentre of the triangle having sides x – y + 1 = 0, x + y + 3 = 0

and 2x + 5y – 2 = 0 is

(a) (2m – 2n, m – n) (b) (2m – 2n, n – m) (c) (2m – n, m + n) (d) (2m – n, m – n)

Ans. (a)

Solution:

m2 2 1 2

1!9! 3!7! 5!5! n!

m1 2 10! 2 10! 10! 2

10! 1!9! 3!7! 5!5! n!

m

10 10 10 10 101 3 5 7 9

1 2{2 C C C C C }

10! n!

m

10 11 2(2)

10! n!

m = 9 and n = 10

Hence, x – y + 1 = 0 and x + y + 3 = 0 are perpendicular to each other, then orthocentre is the point of

intersection which is (–2, – 1)

–2 = 2m – 2n

and –1 = m –n

Point is (2m – 2n, m – n).

27. A straight line through origin O meets the line 3y = 10 – 4x and 8x + 6y + 5 = 0 at points A and B

respectively. Then O divides the segment AB in the ratio

(a) 2: 3 (b) 1 : 2 (c) 3 : 4 (d) 4: 1

Ans. (d)

Solution:




1

2

| 10|POA 45

5OB P 1

2

2

28. The value of (21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + ….. + (21C10 – 10C10) is

(a) 221 – 211 (b) 221– 210 (c) 220 – 29 (d) 220 – 210

Ans. (d)

Solution:

(21C1 +21C2 +21C3 + …. + 21C10) – (10C1 + 10C2 + 10C3 + …..10C10) = S1 – S2

S1 = 21C1 + 21C2 + 21C3 + ….. 21C10

21 21 21 21 21 21 21 211 1 2 20 0 1 2 20 21

1 1S ( C C ... C ) ( C C C .... C C 2)

2 2

201S 2 1

S2 = (10C1 + 10C2 + 10C3 + ….10C10) = 210 – 1

Therefore, S1 – S2 = 220 – 210

29. (–6, 0), (0, 6) and (–7, 7) are the vertices of ABC. The incircle of the triangle has the equation

(a) x2 + y2 – 9x – 9y + 36 = 0 (b) x2 + y2 + 9x – 9y + 36 = 0

(c) x2 + y2 + 9x + 9y – 36 = 0 (d) x2 + y2 + 18x – 18y + 36 = 0

Ans. (b)

Solution:

The triangle is isosceles and therefore the median through C is the bisector of C. The equation of the

angle bisector can be taken as y = – x and l = (–a, a), where a is positive.




Equation of AC is y – 0 = – 7(x+ 6) or 7x + y + 42 = 0 and equation of AB is x – y + 6 = 0.

The length of the perpendicular from l to AB and AC are equal

7a a 42 a a 6

50 2

giving the positive value a =

9

2

Centre is 9 9

,2 2

and radius =3

2

The equation of the circle is 2 2

9 9 9x y

2 2 2

x2 + y2 + 9x – 9y + 36 = 0

30. Let a1, a2, a3,…., an, ….be in A.P. If a3 + a7 + a11 + a15 = 72, then the sum of its first 17 terms is equal to

(a) 153 (b) 306 (c) 612 (d) 204

Ans. (b)

Solution:

a1 + 2d + a1 + 6d + a1 + 10d + a1 + 14 = 72

4(a1 + 8d) = 72

a1 + 8d = 18

17 1

17S 2a 16d

2

17S 17 18 306.

31. One of the diameter of the circle x2+ y2 –12x + 4y + 6 = 0 is given by

(a) x + y = 0 (b) x + 3y = 0 (c) x = y (d) 3x + 2y = 0

Ans. (b)

Solution:

Diameter always passes through the centre of circle.

32. If y + 3x = 0 is the equation of a chord of the circle, x2 + y2 – 30x = 0, then the equation of the circle with

this chord as diameter is

(a) x2 + y2 – 3x – 9y = 0 (b) x2 + y2 + 3x + 9y = 0 (c) x2 +y2 – 3x + 9y = 0 (d) x2 + y2 + 3x – 9y = 0

Ans. (c)

Solution:

2 2x y 30x λ y 3x 0

centre 3λ 30 λ

,2 2

centre lies on y + 3x = 0

λ 9

circles is x2 + y2 – 3x + 9y = 0




33. If the sum of the binomial coefficients in the expansion of n

1x

x

is 64, then the term independent of x

is equal to

(a) 10 (b) 20 (c) 40 (d) 60

Ans. (b)

Solution:

n

11 64

1

n 62 2

n = 6

General term r

n rnr 1 r

1T C x

x

= 6 6 2rrC x

For independent of x,

Let 6 – 2r = 0

r = 3

6 03 1 3T C x

= 6C3

= 20

34. The number of irrational terms in the expansion of (21/5 + 31/10)55 is

(a) 47 (b) 56 (c) 50 (d) 48

Ans. (c)

Solution:

Here 55 r r

55 5 10r 1 rT C (2) 3

r cannot be equal to 0, 10, 20, 30, 40, 50

As in total there are 56 terms.

no of terms which are irrational

are = 56 – 6 = 50 terms

35. If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle

S, whose centre is at (– 3,2), then the radius of S is -

(a) 5 2 (b) 5 3 (c) 5 (d) 10

Ans. (b)

Solution:




x2 + y2 – 4x + 6y – 12 = 0 ..(1)

centre C1(2, – 3)

radius r1 = 5

& centre of circle (2) is C2 (–3, 2)

2 2

1 2C C 2 3 3 2 25 25 5 2

radius (r2) of circle (2) is

2

2 2 22 2 1 1 2r C A C A C C 5 5 2

= 25 50 75 5 3

36. The sum of the last ten coefficients in the expansion of (1 + x)19, when expanded in ascending powers of

x is

(a) 218 (b) 219 (c) 218 – 19C10 (d) none of these

Ans. (a)

Solution:

Sum of the last ten coefficients in the expansion of (1 + x)19

= (sum of all coefficients) – (sum of first ten coefficients)

= 219 – (19C0 + 19C1 + 19C2 + …. + 19C9)

= 19 12

2 (19C0 + 19C1 + 19C2 + …. + 19C9 + 19C10 + 19C11 + …. + 19C19)

= 19 1912 . 2

2

= 19 1812 1 2

2

37. The term independent of x in the binomial expansion of 8

5 21 11 3x 2x is :

x x

(a) 496 (b) – 496 (c) 400 (d) – 400

Ans. (c)

Solution:

r

8 r5 8 2

r

1 11 3x . C 2x

x x

= r r r1

8 r 8 r 8 r8 2 8 2 5 8 2x

r r r

1 1 1C 2x C 2x 3x C 2x

x x x




= r

r r r8 r8 8 r 16 3r 8 8 r 15 3r 8 21 3r

r r r

1C 2 1 x C 2 1 x 3 C 2 1 x

x

for independent term

16 – 3r = 0, 15 – 3r = 0, 21 – 3r = 0

r = 5, r = 7 in III term

in II term

8 3 8

5 7C 2 1 3. C . 2

= 448 – 6 × 8 = 448 – 48 = 400

38. Let N denote the set of natural numbers and A = {n2 : n N } and B = {n3/2 :n N}.

Which one of the following is correct?

(a) A B N

(b) The complement of (AB) is an infinite set

(c) A B must be a finite set.

(d) A B must be a proper subset of {m6: m N}.

Ans. (d)

Solution:

2 3A {n : n N} and B {n :n N}

A = {1, 4, 9, 16, 25, 49, 64, 81, …}

B = {1, 8, 27, 64, 125, …}

A B {1, 64, ...}

Hence, A B must be a proper subset of {m6 : m N}.

39. Let A = {x : x 9, x N }. If B = {a, b, c} be the subset of A where (a + b + c) is a multiple of 3.

What is the largest possible number of subsets like B?

(a) 12 (b) 21 (c) 27 (d) 30

Ans. (d)

Solution:

A = {x : x 9, x N} = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Total possible multiple of 3 are

3,6,9,12,15,18,21,24,27

But 3 and 27 are not possible.

6 1 + 2 + 3

9 2 + 3 + 4, 5 + 3 + 1, 6 + 2 + 1

12 9 + 2 + 1,8 + 3 + 1,7 + 1 + 4,7 + 2 + 3, 6 + 5 + 1,5 + 4 + 3,6 + 4 + 2,

15 9 + 4 + 2, 9 + 5 + 1,8 + 6 + 1,8 + 5 + 2, 7 + 6 + 2,7 + 5 + 3, 6 + 5 + 4, 8 + 4 + 3,

18 9 + 8 + 1,9 + 7 + 2,9 + 6 + 3,9 + 5 + 4, 7 + 6 + 5, 8 + 7 + 3,

21 9 + 8 + 4, 9 + 7 + 5, 8 + 7 + 6

24 9 + 8 + 7

Hence, total number of largest possible subsets are 30.

40. A relation R is defined over the set of non-negative integers as xRy x2 + y2 = 36. What is R?

(a) {(0, 6)}




(b) {(0, 6), ( 11,5), (3,3 3)}

(c) {(6, 0), (0, 6)}

(d) {( 11, 5), (2, 4 2), (5, 11), (4 2,2)} Ans. (c)

Solution:

Since, R is defined over the set of non-negative integers, then R = {(6, 0), (0, 6)} because the options (b)

and (c) does not have integers in all pairs. Also in options (a) and (c), the most appropriate option is (c).

41. Let P = {θ: sin θ cos θ 2 cos θ} and Q = {θ : sin θ cos θ 2 sin θ} be two sets. Then

(a) Q P (b) P Q (c) P Q and Q P (d) P = Q

Ans. (d)

Solution:

sinθ 2 1 cos θ

tan θ 2 1

2 1 sin θ cos θ

tan θ 2 1

42. If 0 x 2π, then the number of real values of x, which satisfy the equation

cos x + cos 2x + cos 3x + cos 4x = 0, is:

(a) 3 (b) 5 (c) 7 (d) 9

Ans. (c)

Solution:

cos x + cos 2x + cos 3x + cos 4x = 0

4 1 4xcos x x . sin

2 20

xsin

2

[By using series formula]

5x

cos . sin 2x 02

But x

sin 02

sin 2x = 0 5x

cos 02

2x nπ 5x π

2n 12 2

nπ

x , n I2

π

x 2n 1 , n I5

π 3π

x 0, , π, , 2π2 2

π 3π 7π 9π

x , , π, ,5 5 5 5

But x = 0, 2π does not satisfy ( sin x

02 )




Solutions are π 3π π 3π 7π 9π

x , π, , , , ,2 2 5 5 5 5

7 solutions

43. If 5(tan2 x – cos2x) = 2 cos 2x + 9, then the value of cos4x is:

(a) 3

5

(b)

1

3 (c)

2

9 (d)

7

9

Ans. (d)

Solution:

5(tan2x – cos2x) = 2cos 2x + 9 2

2

2 2

1 1 t5 t 2 9

1 t 1 t

5(t4 + t2 – 1) = 2 – 2t2 + 9 + 9t2

5t4 – 2t2 – 16 =0

5t4 – 10t2+ 8t2 – 16 = 0

5t2(t2 – 2) + 8(t2 – 2) =0

(5t2 + 8) (t2 – 2) = 0

tan2x = 2

cos 2x = 1 2 1

1 2 3

cos 4x = 2 cos2 2x – 1 = – 7

9

44. The number of solutions of the equation cos x2 |sinx| in [ 2π, 2π] is

(a) 1 (b) 2 (c) 3 (d) 4

Ans. (d)

Solution:

We have,

cos x2 |sin x|

It is true only for |sin x| 1

sin x 1

π

x 2nπ ,2

x [ 2π,2π]

π π 3π 3π

x , , ,2 2 2 2

i.e., no. of solutions = 4.

45. The real roots of the equation cos7x + sin4 x = 1 in the interval π, π are

(a) π

,02

(b) π π

, 0,2 2

(c) π

,02

(d) π π

0, ,4 2

Ans. (b)

Solution:




7 2cos x cos x …(i)

and 4 2sin x sin x …(ii)

Adding Eqs. (i) and (ii), then

cos7 x + sin4 x 1

But given cos7 x + sin4 x = 1

Equality holds only, if

cos7 x = cos2 x and sin4 x = sin2 x

Both are satisfied by x = π

,02

.

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ASPIRE SCHOLARSHIP Exam Solutions - … Class... · The question paper consists of 45 multiple choice questions. (Single correct Answer) Each right answer carries 4 marks and −