mathematics code - d...nda 21st april 2019 mathematics code - d 1. in a school, all the students...

NDA 21st April 2019

MATHEMATICS CODE - D

1. In a school, all the students play at least one of three indoor games-chess, carrom and table tennis. 60 playchess, 50 play table tennis, 48 play carrom, 12 play chess and carrom, 12 play chess and carrom, 15 play carrom and table tennis, 20 play table tennis and chess.

What can be the minimum number of students in the school?

(A) 123

(B) 111

(C) 95

(D) 63

Solution: (B)

P (A )=60, P (B )=50,P (C )=48

P (A∩B )=20,P ( B∩C )=15, P (C∩ A )=12

∴P ( A∪B∪C )=P ( A )+P (B )+P (C )−P ( A∩B )−P (B∩C )−P (C∩ A )+P (A ∩B∩C )

¿60+50+48−20−15−12+P ( A∩B∩C )

¿111+P ( A∩B∩C )

∵P ( A ∩B∩C )≥0

∴P ( A∪B∪C )≥111

∴ Minimum number of students in the school ¿111

2. In a school, all the students play at least one of three indoor games-chess, carrom and table tennis. 60 playchess, 50 play table tennis, 48 play carrom, 12 play chess and carrom, 12 play chess and carrom, 15 play carrom and table tennis, 20 play table tennis and chess.

What can be the maximum number of students in the school?

(A) 111

(B) 123

(C) 125

(D) 135

Solution: (B)

P (A )=60, P (B )=50,P (C )=48

P (A∩B )=20,P ( B∩C )=15, P (C∩ A )=12

NDA 21st April 2019 MATHS∴P ( A∪B∪C )=P ( A )+P (B )+P (C )−P ( A∩B )−P (B∩C )−P (C∩ A )+P (A ∩B∩C )

¿60+50+48−20−15−12+P ( A∩B∩C )

¿111+P ( A∩B∩C )

∵P ( A ∩B∩C )≤12

∴P ( A∪B∪C )≤111+12=123

∴ Maximum number of students in the school ¿123

3. If A is an identity matrix of order 3, then its inverse ( A−1 )

(A) Is equal to null matrix

(B) is equal to A

(C) is equal to 3A

(D) does not exist

Solution: (B)

∵ I−1=I

∴ A−1=A

4. A is a square matrix of order 3 such that its determinant is 4. What is the determinant its transpose?

(A) 64

(B) 36

(C) 32

(D) 4

Solution: (D)

∵|A '|=|A|

∴|A '|=4

5. From 6 programmers and 4 typists, an office wants to recruit 5 people. What is the number of ways this can be done so as to recruit at least one typist?

(A) 209

(B) 210

(C) 246

(D) 242

Solution: (C)

Required number of ways ¿ Total number of ways – (number of ways without typist)

¿ C510

− C 56

NDA 21st April 2019 MATHS

¿10×9×8×7×6

120−6=246

6. What is the number of terms in the expansion of [ (2x−3 y )2

(2 x+3 y )2 ]

2?

(A) 4

(B) 5

(C) 8

(D) 16

Solution: (B)

[ (2x−3 y )2

(2 x+3 y )2 ]

2

¿ [ (4 x2−9 y2 )

2 ]2

=(4 x2−9 y2 )

4

∴ Number of terms ¿4+1=5

7. In the expansion of (1+ax )n , the first three terms are respectively 1,12 x and 64 x2 . What isn equal

to?

(A) 6

(B) 9

(C) 10

(D) 12

Solution: (B)

C0n

=1, C 1ax=12x , C2nn a2 x2

=64 x2

⇒na=12,n (n−1 )

2a2

=64

⇒n (n−1 )

2×

144

n2=64

⇒ n−1n

=128144

=89

∴n=9

8. The numbers 1, 5 and 25 can be three terms (not necessarily consecutive) of

(A) only one AP

(B) more than one but finite number of APs

(C) infinite number of APs

(D) finite number of GPs

NDA 21st April 2019 MATHSSolution: (C)

Let for AP, t1=1,tp=5, tq=25,c .d=d

∴1+( p−1 )d=5⇒ (p−1 ) d=4 …(i)

1+ (q−1 ) d=25⇒ (q−1 ) d=24 …(ii)

∴ from (i) and (ii)

q−1p−1

=6⇒q−1=6 p−6

⇒6 p−q=5

∴ there are infinite AP possible.

In same ways, we can show that infinite GP are possible.

9. The sum of ( p+q )th and ( p−q )th terms of an AP is equal to

(A) (2 p )th terms

(B) (2q)th terms

(C) Twice thee pth term

(D) Twice the q th term

Solution: (C)

Let first terms of AP be ‘a’ and common difference be ‘d’, the

T p+q+T p−q=a+( p+q−1)d+a+ ( p−q−1 )d

¿2a+2 (p−1 ) d

¿2 [a+( p−1 ) d ]

¿2T p

NDA 21st April 2019 MATHS10. If A is a square matrix of order n>1 , then which one of the following is correct?

(A) det (−A )=det A

(B) det (−A )=(−1 )ndet A

(C) det (−A )=−det A

(D) det (−A )=ndet A

Solution: (B)

As we know if A is a square matrix of order n, then ¿kA∨¿k n.|A| (where k is a scalar)

Hence |A−|¿ (−1 )n|A|

11. What is the least value of 25 cose c2 x+36 sec2 x ?

(A) 1

(B) 11

(C) 120

(D) 121

Solution: (D)

Given expression Δ=25 cosec2 x+36 sec2 x

⇒Δ=25(1+cot2 x )+36 (1+ tan2 x )

Δ=61+25cot2 x+36 tan2 x

For minimum value of 25 cot2×+36 tan2 x

applying AM−GM inequality

25cot2 x+36 tan2 x2

≥ (25cot2 x . 36 tan2 x )1/2

⇒25cot2 x+36 tan2 x ≥60

⇒61+25cot2 x+36 tan2 x ≥121

Hence minimum value of

Δ=121

12. Let A and B be (3×3 ) matrices with det A=4 and det B=3.

What is det (2 AB ) equal to?

(A) 96

(B) 72

(C) 48

(D) 36

NDA 21st April 2019 MATHSSolution: (A)

Given A and B are 3×3 matrices with ¿A∨¿ 4 and ¿B∨¿3

Now ¿2 AB∨¿23∨A∨¿B∨¿8.4 .3=96

13. Let A and B be (3×3 ) matrices with det A=4 and det B=3.

What is det (3 A B−1 ) equal to?

(A) 12

(B) 18

(C) 36

(D) 48

Solution: (C)

Given A and B are 3×3 matrices with ¿A∨¿ 4 and ¿B∨¿3

Now |3 A B−1|=33|A||B−1|=27.|A|.1

|B|=27.4 .

13=36

14. A complex number is given by z=1+2 i

1−(1−i )2

What is the modulus of z?

(A) 4

(B) 2

(C) 1

(D) 12

Solution: (C)

Given complex number

z=1+2i

1−(1−i)2

⇒ z=1+2i

1−(1−1−2 i)=

1+2 i1+2 i

=1

⇒ z=1+0 i

Clearly ¿ z∨¿1 and arg (z)=0

15. A complex number is given by z=1+2 i

1−(1−i )2

What is the principal argument of z?

(A) 0


(B) π4

(C) π2

(D) π

Solution: (A)

Given complex number

z=1+2i

1−(1−i)2

⇒ z=1+2i

1−(1−1−2 i)=

1+2 i1+2 i

=1

⇒ z=1+0 i

Clearly ¿ z∨¿1 and arg (z)=0

16. If the roots of the equation x2+ px+q=0 are tan 19 ° and tan 26° , then which one of the

following is correct?

(A) q−p=1

(B) p−q=1

(C) p+q=2

(D) p+q=3

Solution: (A)

Given tan 19 ° and tan 26 ° are roots of x2+ px+q=0

⇒ tan19°+ tan26 °=−p and tan 19 ° . tan26 °=q

⇒ tan(19 °+26°)=tan 19 °+tan 26 °

1− tan 19° , tan 26 °

⇒1=−p1−q

⇒q−p=1

17. What is the fourth term of an AP of n terms whose sum is n (n+1 ) ?

(A) 6

(B) 8

(C) 12

(D) 20

Solution: (B)

Given Sn=n (n+1 ) , we known T n=Sn−Sn−1

NDA 21st April 2019 MATHST n=n(n+1)−(n−1)n

⇒T n=2n

⇒T 4=8

18. What is (1+tan α tan β )2+ (tan α−tan β )2−sec2 α sec2 β equal to?

(A) 0

(B) 1

(C) 2

(D) 4

Solution: (A)

Given trigonometric expression can be written as

1+tan2α tan2β+2 tan α tan β+ tan2α+ tan2 β−2 tanα tan β−sec2α . sec2β

¿(1+ tan2α)(1+tan 2β)−sec2α sec2β

¿sec2α sec2β−sec2α sec2β

¿0

19. If p=cosec θ−cot θ and q=(cosecθ+cot θ )−1 then which one of the following is correct?

(A) pq=1

(B) p=q

(C) p+q=1

(D) p+q=0

Solution: (B)

As we known cose c2θ−cot2θ=1

⇒(cosecθ+cotθ)(cosecθ−cot θ)=1

⇒ 1q∙ p=1

⇒ p=q


20. If the angle of a triangle ABC are in the ratio 1:2 :3, then the corresponding sides are in the ratio

(A) 1:2 :3,

(B) 3 :2:1,

(C) 1: √3 :2

(D) 1: √3 :√2

Solution: (C)

Given angles A, B and C of a Δ ABC are in the Ratio 1:2 :3 ⇒ A=30, B=60 and C=90

From sine rule a

sin A=

bsin B

=c

sinC

⇒ a1/2

=b

√3/2=

c1

⇒a :b : c=1:√3 :2

21. Consider the following statements:

(1) For an equation of a line,

x=cos θ+ y sin θ=p , in normal form, the length of the perpendicular from the point (α ,β ) to the line is |α cos θ+β sin θ+p|.

(2) The length of the perpendicular from the point (α , β ) to the line xa+

yb=1 is

|aα+bβ−ab

√a2+b2 |

Which of the above statements is is/are correct?

(A) 1 only

(B) 2 only

(C) both 1 and 2

(D) Neither 1 nor 2

Solution: (D)

Length of perpendicular from (α ,β ) to the line

x cosθ+ y sinθ−p=0 is |α cosθ+β sin θ−p|

Hence statement (i) is false

Length of perpendicular from (α ,β ) to the line

xa+

yb−1=0 or bx+ay−ab=0 is ¿|αb+ βa−ab

√b2+a2 |

Hence statement (ii) is also false.

NDA 21st April 2019 MATHS22. A circle is drawn on the chord of a circle x2

+ y2=a2 as diameter. The chord lies on the line

x+ y=a . What is the equation of the circle?

(A) x2+ y2

−ax−ay+a2=0

(B) x2+ y2

−ax−ay=0

(C) x2+ y2

+ax+ay=0

(D) x2+ y2

+ax+ay−2a2=0

Solution: (B)

Given equation of circle is x2+ y2

=a2 and equation of chord x+ y=a

For point of interesection eliminate y, we get

x2+(a−x)2=a2

⇒ x=0, a

Hence intersection point A (0,a) and B (0,a)

Now circle with AB as diamter is

(x−0)(x−a)+( y−a)( y−0)=0

⇒ x2+ y2

−ax−ay=0

23. The sum of the focal distance of a point on an ellipse is constant and equal to the

(A) length of minor axis

(B) length of major axis

(C) length if latus rectum

(D) sum of the lengths of semi-major and semi-minor axes

Solution: (B)

We know ps+ ps1=2a i.e sum of the focal distances of a point an ellipse is constant and equal to length

of major axis.

NDA 21st April 2019 MATHS24. The equation 2x2

−3 y2−6=0 represents

(A) a circle

(B) a parabola

(C) an ellipse

(D) a hyperbola

Solution: (D)

The given equation can be written as

x2

3−

y2

2=1 which is a rectangular hyperbola

25. The two parabolas y2=4ax and x2

=4 ay intersect

(A) at two points on the line y=x

(B) only at the origin

(C) at three points one of which lies on y+x=0

(D) only at (4 a ,4a )

Solution: (A)

Clearly parabola y2=4ax and x2

=4 ay intersect at (0,0) and (4 a ,4a ) which also lies on the line y=x .

26. The points (1,3 ) and (5,1 ) are two opposite vertices of a rectangle. The other two vertices lie on the line y=2 x+c . What is the value of c ?

(A) 2

(B) −2

(C) 4

(D) −4

Solution: (D)

In rectangle diagonals bisect each other, hence mid point of BD must lie on y=2 x+c

⇒ 2=6+c

⇒c=−4

27. If the lines 3 y+4 x=1, y=x+5 and 5 y+bx=3 are concurrent, then what is the value of b ?

NDA 21st April 2019 MATHS(A) 1

(B) 3

(C) 6

(D) 12

Solution: (C)

Since given lines are concurrent

|b 5 −31 −1 54 3 −1|=0

⇒b=6

28. What is the equation of the straight line which is perpendicular to y=x and passes through (3,2 )

(A) x− y=5

(B) x+ y=5

(C) x+ y=1

(D) x− y=1

Solution: (B)

Line passing through (3,2 ) and perpendicular to x− y=0 is given by y−2=−1 ( x−3 )

⇒ x+ y=5

29. The straight lines x+ y−4=0, 3x+ y−4=0 and x+3 y−4=0 , from a triangle, which is

(A) isosceles

(B) right-angled

(C) equilateral

(D) scalene

Solution: (A)

Let the slope of the lines be m1=−1,m2=−3 and m3=−13

Angle between lines L1 and L2 is tan θ1=|−1+31+3 |=1

2

Angle between lines L1 and L3 is tan θ2=|−1+13

1+13

|=12


Angle between lines L2 and L3 is tan θ2=|−3+13

1+1 |=43

Hence triangle is isosceles

30. The circle x2+ y2

+4 x−7 y+12=0 , cuts an intercept on y-axis equal to

(A) 1

(B) 3

(C) 4

(D) 7

Solution: (A)

y -intercept ¿2√ f 2−c

In given equation of circle f =−72

and c=12

y -intercept ¿2√ 494

−12=1

31. What is the nth term of the sequence 25−125, 625,−3125, .....?

(A) (−5 )2n−1

(B) (−1 )2n5n+ 1

(C) (−1 )2n−15n+1

(D) (−1 )n−15n+ 1

Solution: (D)

Its GP with t1=25,r=−5

∴t n=ar n−1=25 (−5 )n−1

¿ (−5 )2 (−5 )n−1

¿ (−5 )n+1=(−1 )n+ 1 (5 )n+1

¿ (−1 )n−1 (5 )n+ 1

32. Suppose X=[1,2, 3, 4 ] and R is a relation on X . If 2,1

R={(1,1) , (2,2 ) (3,3 ) (1,2 ) (¿, (2,3 ) (3,2 ) }

Then which one of the following is correct?

(A) R is reflexive and symmetric, but not transitive

(B) R is symmetric and transitive, but not reflexive

(C) R is reflexive and transitive, but not symmetric

NDA 21st April 2019 MATHS(D) R is neither reflexive nor transitive, but symmetric

Solution: (D)

X={1,2,3, 4 }

R= {(1,1 ) , (2,2 ) , (3,3 ) (1,2 ) , (2,1 ) (2,3 ) , (3,2 ) }

(i ) Reflexivity

R is reflexive if xRx for all x∈ X

But (4,4 )∉R⇒R is not reflexive

(ii ) Transitivity

R is transitive if xRy , yRz⇒ xRz ∀ x , y , z∈ R

But (1,2 ) , (2,3 )∈ R and (1,3 )∉R

⇒R is not transitive

(iii ) Symmetry

R is symmetric if xRy⇒ yRx ∀ x , y∈R which is true for given relation.

33. A relation R is defined on the set N of natural numbers as xRy⇒ x2−4 xy+3 y2

=0 . Then which one of the following is correct?

(A) R is reflexive and symmetric, but not transitive

(B) R is reflexive and transitive, but not symmetric

(C) R is reflexive, but neither symmetric nor transitive

(D) R is reflective, but neither symmetric nor transitive

Solution: (D)

R :N → N ,xRy⇒ x2− yxy+3 y2

=0

⇒ ( x− y ) ( x−3 y )=0

(i ) Reflexive

xRx⇐ x2− yxx+3 x2

=0⇐0=0 Identify

⇒R is reflexive

(ii ) Symmetric

For any a∈N (3a ,a )∈R [ (3a )2−4 (3 a ) (a )+3a2=0 ]

But (a ,3a )∉R [a2−4 (a ) (3a )+3 (3a )2=16 a2≠0 ]

⇒R is not symmetric

(iii ) Transitive

For any a∈N (9a ,3a )∈R , (3a , R )∈R

But (9a ,a )∉R

⇒R is not transitive


34. If A= [x∈Z : x3−1=0 ] and B=[ x∈Z : x2+x+1=0 ] where Z is set of complex number, then what is A ∩B equal to?

(A) Null set

(B) {−1+√3 i2

,−1−√3 i

2 }(C) {−1+√3 i

4,−1−√3 i

4 }(D) {1+√3 i

2,1−√3 i

2 }Solution: (B)

x3=1⇒ x=1,ω,ω2

x2+x+1=0⇒ x=ω,ω2

∴ A={1,ω,ω2 } ,B={ω ,ω2 }

∴ A∩B= {ω,ω2 }

¿{−1+√3 i2

,−1−√3 i

2 }

35. Consider the following statements for two non-empty sets A and B

1. (A∩B)∪(A∩ B)∪( A∩B)=A∪B

2. (A∪)( A ∩B)=A∪B

Which of the above statements is/are correct?

(A) 1 only

(B) 2 only

(C) both 1 and 2

(D) Neither 1 no 2

Solution: (A)

1. From Venn Diagram

A∪B=( A∩B )∪ ( A∩B )∪ ( A∩B )

NDA 21st April 2019 MATHS2. AU ( A∩ B )

¿ ( A∪ A )∩ (A∪ B ) Distributive law

¿U ∩ (A∪B ) Complement law

¿A∪ B

≠ A∪B

36. Let X be a non-empty set and let A, B, C be subsets of X. consider the following statements:

1. A⊂C⇒(A∩B)⊂ (C ∩B ) (A∪B)⊂(C∪B)

2. (A∩B)⊂ (C∩B ) for all sets B⇒ A⊂C

3. (A∪B)⊂ (C∪B ) for all sets B⇒ A⊂C

Which of the above statements are correct?

(A) 1 and 2 only

(B) 2 and 3 only

(C) 1 and 3 only

(D) 1, 2 and 3

Solution: (D)

1. A⊂C⇒ (A∩B )⊂ (C∩B ) , ( A∪B )⊂ (C∪B )

Let x∈ A∩B⇒ x∈ A , x∈B

⇒ x∈C , x∈B (∵ A⊂C )

⇒ x∈C∩B

Let y∈ A∪B⇒ y∈ A or y∈B

⇒ y∈C or y∈B (∵ A⊂C )

⇒ y∈C∪B

2. (A ∩B )⊂ (C∩B ) for all sets B

Let B=⋃ (Universal set)

⇒ A∩B=A∩U=A⊂ (C∩U )=C

3. ( A∪B )⊂ (C∪B ) for all sets B

Let B=ϕ (Empty set)

⇒ A∪B=A∪ϕ=A⊂ (C∪ϕ )=C

37. If B=[3 2 02 4 01 1 0 ] , then what is adjoint of B equal to?


(A) [0 0 00 0 0

−2 −1 8](B) [

0 0 −20 0 −10 0 8 ]

(C) [0 0 20 0 10 0 0]

(d) It does not exist

Solution: (A)

Co-factor matrix of B ,C=[0 0 −20 0 −10 0 8 ]

∴ adj B=C '=[

0 0 00 0 0

−2 −1 8]38. What are the roots of the equation ¿ x2

−x−6∨¿ x+2?

(A) −2,1,4

(B) 0,2,4

(C) 0,1,4

(D) −2,2,4

Solution: (D)

Roots will exist if x+2≥0

∴|x2−x−6|=x+2

⇒|( x+2 ) ( x−3 )|=x+2

⇒ ( x+2 )|x−3|−( x+2 )=0

⇒ ( x+2 ) (|x−3|−1 )=0

∴ x+2=0 or |x−3|=1

∴ x=−2 or x=4,2

∴ x=−2,2,4

39. If A=[0 11 0] , then the matrix A is a/an

(A) Singular matrix

(B) Involutory matrix

NDA 21st April 2019 MATHS(C) Nilpotent matrix

(D) Idempotent matrix

Solution: (B)

A2=[0 1

1 0 ][0 11 0]

⇒ A2=[1 0

0 1]=I

Hence matrix A is involuntary.

40. If [x −3i 1y 1 i0 2i −i ]=6+11 i , then what are the values of x and y respectively?

(A) −3,4

(B) 3,4

(C) 3,−4

(D) −3,−4

Solution: (A)

Given |x −3 i 1y 1 i0 2 i −i|=6+11 i

|x −3 i 1y 1 i0 2 i −i|=x (−i− (2 i ) (i ) )− y ( (−3 i ) (−i )− (2 i ) )+0

¿ x (−i+2 )− y (−3−2 i)

¿ (2x+3 y )+(2 y−x )i=6+11 i (∴given )

⇒2 x+3 y=6 …(i)

2 y−x=11 …(ii)

Solving (i) and (ii),

we get x=−3 and y=4.

41. The common roots of the equations z3+2 z2

+2 z+1=0 and z2017+z2018

+1=0 are

(A) −1,ω

(B) 1, ω2

(C) −1, ω2

(D) ω, ω2

Solution: (D)

NDA 21st April 2019 MATHSGiven equation z3

+2 z2+2 z+1=0

⇒ ( z+1 ) ( z2− z+1 )+2 z (z+1 )=0

⇒ ( z+1 ) ( z2+z+1 )=0

⇒ z=−1,ω,ω2

Clearly ω,ω2 satisfy z2017+z2018

+1=0.

42. If C (20, n+2 )=C (20, n−2 ) , then what is n equal to?

(A) 8

(B) 10

(C) 12

(D) 16

Solution: (B)

Given Cn+2= ¿20 Cn−2

20

¿

⇒20=n+2+n−2

⇒2n=20

⇒n=10

43. There are 10 points in a plane. No three of these points are in a straight line. What is the total number of straight lines which can be formed by joining the points?

(A) 90

(B) 45

(C) 40

(D) 30

Solution: (B)

Total number of straight lines which can be formed by joining given 10 points ¿ C210

¿10×9

2

¿45

44. The equation px2+qx+r=0 (where p ,q , r , all are positive) has distinct real roots a and b .

Which one of the following is correct?

(A) a>0,b>0

(B) a<0,b<0

(C) a>0,b<0

NDA 21st April 2019 MATHS(D) a<0,b>0

Solution: (B)

Given

px2+qx+r=0 (where p ,q , r all are positive) a and b are distinct real roots

a+b=−qp

⇒a+b<0 …(i)

ab=rp

⇒ab>0 …(ii)

from (i) and (ii) we get

a<0,b<0.

45. If A= {λ , {λ , μ } } , then the power set of A is

(A) {ϕ , {ϕ } , {λ }, {λ , μ } }

(B) {ϕ , {λ } , {{λ , μ }} , {λ , {λ ,μ }}}

(C) {ϕ , {λ } , {λ , μ } , {λ , {λ , μ }}}

(D) { {λ } , {λ ,μ } , {λ , {λ ,μ }}}

Solution: (B)

Given A= {λ , {λ ,μ }}

∴ Power set of A= {ϕ , {λ } , {{λ ,μ }} , {λ , {λ ,μ }}}

46. What is the value of sin 34o cos236o

−sin 56o sin 124o

cos28o cos88o+cos178o sin 208o ?

(A) −2

(B) −1

(C) 2

(D) 1

Solution: (A)

Given sin 34o cos236o

−sin 56o sin 124o

cos28o cos88o+cos178o sin 208o

⇒−sin 34o cos56o−sin 56o cos34o

cos 28o sin 2o+cos2o sin 28o

⇒−sin (34o

+56o )

sin (28o+2o )

NDA 21st April 2019 MATHS⇒−2

47. tan 54o can be expressed as

(A) sin 9o

+cos 9o

sin 9o−cos 9o

(B) sin 9o

−cos 9o

sin 9o+cos 9o

(C) cos9o

+sin 9o

cos9o−sin 9o

(D) sin 36o

cos 36o

Solution: (C)

tan (54o )=tan (45o+9o )

¿1+ tan 9o

1−tan 9o

∴ tan (54o )= cos9o+sin 9o

cos9o−sin 9o

Consider the following for the next 03 (three) items:

If p=X cos θ−Y sin θ , q=X sin θ+Y cos θ and p2+4 pq+q2

=A X 2+BY 2 , 0≤θ≤

π2

.

48. What is the value of θ?

(A) π2

(B) π3

(C) π4

(D) π6

Solution: (C)

Given,

p=X cosθ−Y sin θ

q=X sin θ+Y cosθ

p2+4 pq+q2

=A X 2+BY 2 …(i)

p2+4 pq+q2

=X2cos2θ+Y 2sin2θ−2XY cosθ sin θ+x2 sin2θ+Y 2cos2θ

+2 XY sinθ cosθ+4 [ ( X cos θ−Y sin θ ) (X sin θ+Y cosθ ) ]

NDA 21st April 2019 MATHS¿X2+Y 2+4 [ X2sin θ cosθ+XY cos2θ−XY sin2θ−Y 2sin θ cosθ ]

¿X2+Y 2+4 [ (X2−Y 2 ) sin 2θ2

+XY cos2θ]¿X2

+Y 2+2 (X2

−Y 2 ) sin 2θ+4 XY cos 2θ …(ii)

From (i) and (ii) we get

cos2θ=0

⇒θ=π4

∴X2+Y 2

+2 (X2−Y 2 )=A X2

+BY 2

3 X2−Y 2

=A X2+BY 2

⇒ A=3 ;B=−1

49. What is the value of A ?

(A) 4

(B) 3

(C) 2

(D) 1

Solution: (B)

Given,


q=X sin θ+Y cosθ

p2+4 pq+q2

=A X 2+BY 2 …(i)

p2+4 pq+q2



¿X2+Y 2+4 [ X2sin θ cosθ+XY cos2θ−XY sin2θ−Y 2sin θ cosθ ]

¿X2+Y 2+4 [ (X2−Y 2 ) sin 2θ2

+XY cos2θ]¿X2

+Y 2+2 (X2



cos2θ=0

⇒θ=π4

∴X2+Y 2

+2 (X2−Y 2 )=A X2

+BY 2

3 X2−Y 2

=A X2+BY 2

⇒ A=3 ;B=−1


50. What is the value of B?

(A) −1

(B) 0

(C) 1

(D) 2

Solution: (A)

Given,


q=X sin θ+Y cosθ

p2+4 pq+q2

=A X 2+BY 2 …(i)

p2+4 pq+q2



¿X2+Y 2+4 [ X2sin θ cosθ+XY cos2θ−XY sin2θ−Y 2sin θ cosθ ]

¿X2+Y 2+4 [ (X2−Y 2 ) sin 2θ2

+XY cos2θ]¿X2

+Y 2+2 (X2



cos2θ=0

⇒θ=π4

∴X2+Y 2

+2 (X2−Y 2 )=A X2

+BY 2

3 X2−Y 2

=A X2+BY 2

⇒ A=3 ;B=−1

Consider the following for the next 02 (two) items:

It is given that cos (θ−α )=a , cos (θ−β )=b .

51. What is cos (α−β ) equal to?

(A) ab+√1−a2√1−b2

(B) ab−√1−a2√1−b2

(C) a√1−b2−b√1−a2

(D) a√1−b2+b√1−a2

Solution: (A)

cos (θ−α )=a , cos (θ−β )=b

NDA 21st April 2019 MATHScos (α−β )=cos (θ−β−(θ−α ) )

¿cos (θ−β )cos (θ−α )+sin (θ−β )sin (θ−α )

sin (θ−α )=√1−a2 , sin (θ−β )=√1−b2

∴cos (α−β )=ab+√1−a2√1−b2

52. What is sin2 (α−β )+2ab cos (α+β ) equal to?

(A) a2+b2

(B) a2−b2

(C) b2−a2

(D) – (a2+b2 )

Solution: (A)

sin2 (α−β )+2abcos (α−β )

¿1−cos2 ( α−β )+2abcos (α−β )

¿1−cos (α−β ) [cos ( α−β )−2ab ]

¿1−cos (α−β ) [√1−a2√1−b2−ab ]

¿1−(√1−a2√1−b2+ab )(√1−a2√1−b2

−ab )

¿1−[ (1−a2 ) (1−b2 )−a2b2 ]

¿1−[1−a2−b2+a2b2−a2b2 ]

¿a2+b2

∴sin2 (α−β )+2abcos (α−β )=a2+b2

53. If sin α+cos α=p , then what is cos2 (2α ) equal to?

(A) p2

(B) p2−1

(C) p2 (2−p2 )

(D) p2+1

Solution: (C)

Given

sinα+cosα=p

(sinα+cosα )2=p2

1+sin 2α=p2

sin 2α=p2−1

NDA 21st April 2019 MATHScos2 (2α )=1−sin22α

¿1−( p2−1 )

2

¿1−( p4−2 p2

+1 )

¿2 p2−p4

∴cos2 (2α )=p2 (2−p2 )

54. What is the value of sin−1 45

+sec−1 54−

π2?

(A) π4

(B) π2

(C) π

(D) 0

Solution: (D)

Given expression sin−1 45

+sec−1 54−

π2

¿sin−1 45+cos−1 4

5−

π2

¿π2−

π2

¿0

55. If sin−1 2 p1+ p2−cos−11−q2

1+q2 =tan−1 2 x1−x2 , then what is x equal to?

(A) p+q

1+ pq

(B) p−q1+ pq

(C) pq

1+ pq

(D) p+q

1−pq

Solution: (B)

Given equation is

sin−1 2 p1+ p2−cos−11−22

1+22 =tan−1 2x1−x2

⇒2 tan−1 p−2 tan−1q=2 tan−1 x


⇒ tan−1 x= tan−1 p−q1+pq

⇒ x=p−q1+ pq

56. If tan θ=12

and tan φ=13, then what is the value of (θ+φ )?

(A) 0

(B) π6

(C) π4

(D) π2

Solution: (C)

Given tan θ=12

and tan ϕ=13

Now tan (θ+ϕ )=tan θ+ tanϕ

1−tanθ tanϕ

¿

12+

13

1−12∙13

=1

⇒θ+ϕ=π4

57. If cos A=34, then what is the value of sin( A

2 )sin( 3 A2 )?

(A) 58

(B) 516

(C) 524

(D) 732

Solution: (B)

Given expression sinA2

. sin3 A2

¿12 (2sin

A2

. sin 3A2 )


¿12

[cosA−cos 2 A ]

¿12

[cosA−2cos2 A+1 ]

¿12 [ 34−2∙

916

+1]= 516

58. What is the value of tan 75o+cot75o?

(A) 2

(B) 4

(C) 2√3

(D) 4 √3

Solution: (B)

¿ tan 75o+cot75o

¿2+√3+2−√3

¿4

59. What is the value of cos 46o cos 47o cos 48o cos 49o cos 50o…cos 135o?

(A) −1

(B) 0

(C) 1

(D) Greater than 1

Solution: (B)

Clearly in the given product are value is cos90o which is zero, hence given product is zero.

60. If sin 2θ=cos 3θ , where 0<θ<π2, then what is sin θ equal to?

(A) √5+14

(B) √5−14

(C) √5+116

(D) √5−116

Solution: (B)

Given equation sin 2θ=cos3θ

NDA 21st April 2019 MATHS⇒sin 2θ=sin (90o

−3θ )

⇒2θ=90o−3θ

⇒θ=18o

⇒sin θ=sin 18o=

√5−14

61. For r>0, f (r ) is the ratio of perimeter to area of a circle of radius r . Then f (1 )+ f (2 ) is equal to

(A) 1

(B) 2

(C) 3

(D) 4

Solution: (C)

Given f (r )=2 πr

π r2

⇒ f (r )=2r

Now f (1 )+ f (2 )=21+

22=3

62. If f ( x )=31+x , then f ( x ) f ( y ) f ( z ) is equal to

(A) f ( x+ y+z )

(B) f ( x+ y+z+1 )

(C) f ( x+ y+z+2 )

(D) f ( x+ y+z+3 )

Solution: (C)

Given f ( x )=31+x , hence

f ( x ) , f ( y ) , f (z )=31+ x ∙31+ y ∙31+ z

¿33+ x+ y+z

¿ f ( x+ y+z+2 )

63. The number of real roots for the equation x2+9|x|+20=0 is

(A) Zero

(B) One

(C) Two

(D) Three

NDA 21st April 2019 MATHSSolution: (A)

In the given equation x2+9|x|+20=0, we can see x2≥0 and 9|x|≥0, hence

x2+9|x|+20>0∀ x∈R . Given equation has no solution.

64. If f ( x )=sin (cos x ) , then f ' ( x ) is equal to

(A) cos (cos x )

(B) sin (−sin x )

(C) (sin x ) cos (cos x )

(D) (−sin x )cos (cos x )

Solution: (D)

Given f ( x )=sin (cos x )

⇒ f ' ( x )=cos (cos x ) ∙ (−sin x ) (applying chain rule)

⇒ f ' ( x )=−sin x ∙cos (cos x )

65. The domain of the function f ( x )=√(2−x ) ( x−3 ) is

(A) (0, ∞ )

(B) [0, ∞ )

(C) [2, 3 ]

(D) (2,3 )

Solution: (C)

For the existence of the given function

(2−x ) ( x−3 ) ≥0

⇒ ( x−2 ) ( x−3 )≤0

⇒ x∈ [2,3 ]

66. The solution of the differential equation dydx

=cos ( y−x )+1 is

(A) ex [sec ( y−x )−tan ( y−x ) ]=c

(B) ex [sec ( y−x )+ tan ( y−x ) ]=c

(C) ex sec ( y−x ) tan ( y−x )=c

NDA 21st April 2019 MATHS(D) ex

=c sec ( y−x ) tan ( y−x )

Solution: (A)

Given differential equation dydx

=cos ( y−x )+1

⇒ dy−dxdx

=cos ( y−x )

⇒d ( y−x )

cos ( y−x )=dx

⇒∫ sec ( y−x ) d ( y−x )=∫ dx

¿ ln ( sec ( y−x )+ tan ( y−x ) )=x+c

⇒ sec ( y−x )+ tan ( y−x )=c . ex

⇒ex (sec ( y−x )−tan ( y−x ) )=c

67. ∫0

π2

|sin x−cos x|dx is equal to

(A) 0

(B) 2 (√2−1 )

(C) 2√2

(D) 2 (√2+1 )

Solution: (B)

Let I=∫0

π2

|sin x−cos x|dx

¿∫0

π4

|sin x−cos x|dx+∫π4

π2

|sin x−cos x|dx

¿∫0

π4

|cos x−sin x|dx+∫π4

π2

|sin x−cos x|dx

¿ (sin x+cos x )|0π4+ (−cos x+sin x )|π

4

π2

¿2 (√2−1 )

68. If y=acos 2 x+b sin 2 x , then

(A) d2 yd x2+ y=0


(B) d2 yd x2+2 y=0

(C) d2 yd x2−4 y=0

(D) d2 yd x2+4 y=0

Solution: (D)

Given equation of family of curve

y=acos2 x+b sin 2 x

⇒ dydx

=−2a . sin 2 x+2bcos2 x

⇒ d2 yd x2=−4 acos2 x−4 b sin2 x

⇒ d2 yd x2=−4. y

⇒ d2 yd x2 +4 y=0

69. A given quantity of metal is to be cast into a half cylinder (i.e., with a rectangular base and semicircular ends). If the total surface area is to be minimum, then the ratio of the height of the half cylinder to the diameter of the semicircular ends is

(A) π : (π+2 )

(B) (π+2 ): π

(C) 1:1

(D) None of the above

Solution: (A)

Let the radius of cylinder be r and height be h , then

Volume of half cylinder V=12π r2h …(i)

Surface area of half cylinder

S=2 r .h+πrh+π r2

⇒ S=4Vπr

+π .2Vπr

+π r2 (from (i) rh=

2Vπr

⇒ dsdr

=−4V

π r2−

2V

r 2+2πr=0

⇒2πr=4V

π r2+

2V

r2⇒ h

2r=

ππ+2


⇒2πr=4.π r2h2π r2 +

2r2 .

12π r2h

70. ∫0

π2

esin x cos x dx is equal to

(A) e+1

(B) e−1

(C) e+2

(D) e

Solution: (B)

Let I=∫0

π2

esin x .cos x dx put sin x=t ⇒cos x dx=dt

⇒ I=∫0

1

et dt

⇒ I=e t|01

⇒ I=e−1

71. If f ( x )=x−2x+2

, x ≠−2 , then what is f −1 ( x ) equal to?

(A) 4 ( x+2 )

x−2

(B) x+2

4 ( x−2 )

(C) x+2x−2

(D) 2 (1+x )

1−x

Solution: (D)

Given f ( x )=x−2x+2

Replacing x by f −1 ( x ) , we get

f ( f−1 (x ) )=f −1 (x )−2

f −1 ( x )+2

⇒ x f−1 (x )+2 x= f−1 ( x )−2

⇒ f−1 ( x )=2 (x+1 )

1−x


72. What is ∫ ln (x2 )dx equal to?

(A) 2x ln ( x )−2 x+c

(B) 2x+c

(C) 2x ln ( x )+c

(D) 2 ln ( x )

x−2x+c

Solution: (A)

Let I=∫ ln x2dx

⇒ I=2∫ 1. ln x .dx (applying integration by parts)

⇒ I=2( ln x . x−∫ 1x. xdx )

⇒ I=2 ( x ln x−x )+c

73. The minimum distance from the point (4, 2 ) to y2=8 x is equal to

(A) √2

(B) 2√2

(C) 2

(D) 3√2

Solution: (B)

The minimum distance will occur along the normal to the parabola passing through (4,2 ) .

Normal to parabola at point Q (am2 ,−2am) where a=2 is

y=mx−2.2 .m−2.m3…. (i) this normal passes through (4,2 ) , hence

2=4m−4m−2m3

⇒m3=−1

⇒m=−1. Hence point θ (2, (−1 )2 ,−2 ∙2∙ (−1 ) )

⇒θ (2,4 )

NDA 21st April 2019 MATHSHence PQ=2√2

74. The differential equation of the system of circles touching the y-axis at the origin is

(A) x2+ y2

−2 xydydx

=0

(B) x2+ y2

+2 xydydx

=0

(C) x2− y2

+2xydydx

=0

(D) x2− y2

−2 xydydx

=0

Solution: (C)

Equation of family of circle touching y -axis at origin is ( x−α )2+ ( y−0 )2=α2

⇒ x2+ y2

−2αx=0

⇒ x+y2

x−2α=0 ...(i)

Differentiating once, we get

1+

x .2 ydydx

− y2

x2=0

⇒ x2− y2

+2xydydx

=0

75. Consider the following in respect of the differential equation: d2 yd x2+2( dydx )

2

+9 y=x

1. The degree of the differential equation is 1.

2. The order of the differential equation is 2.


(A) 1 only

(B) 2 only

(C) Both 1 and 2

(D) Neither 1 nor 2

NDA 21st April 2019 MATHSSolution: (C)

In the given D.E. highest order differential coefficient is d2 yd x2 , hence order of D.E. is 2. In D.E.

power of d2 yd x2 is 1, hence degree of D.E. is 1.

76. If A ,B ,C are three events, then what is the probability that at least two of these events occur together?

(A) P (A∩B )+P ( B∩C )+P (C∩ A )

(B) P ( A∩B )+P (B∩C )+P (C∩ A )−P ( A∩B∩C )

(C) P ( A∩B )+P (B∩C )+P (C∩ A )−2P ( A∩B∩C )

(D) P (A∩B )+P ( B∩C )+P (C∩ A )−3P ( A∩B∩C )

Solution: (C)

Probability of simultaneous occurrence of at least two events

¿P (A∩B )+P (B∩C )+P (A∩C )−2P ( A∩B∩C )

77. If two variables X and Y are independent, then what is the correlation coefficient between them?

(A) 1

(B) −1

(C) 0

(D) None of the above

Solution: (C)

Since X and Y are independent, their correlation coefficient must be equal to zero.

78. Two independent events A and B are such that P (A∪B )=23

and P (A∩B )=16. If

P (B )<P ( A ) , then what is P (B ) equal to?

(A) 14


(B) 13

(C) 12

(D) 16

Solution: (B)

Since A and B are independent events P (A∩B )=P ( A ) . P (B )

⇒P ( A ) . P (B )=16

….(i)

Now P ( A∪B )=23

⇒P ( A )+P (B )−P ( A∩B )=23

⇒P ( A )+P (B )−16=

23

⇒P (A )+P (B )=56

…(ii)

From (i) and (ii)

P (B )=13,12

⇒P (B )=13

(givenP (B )<P ( A ) )

79. The mean of 100 observations is 50 and the standard deviation is 10. If 5 is subtracted from each observation and then it is divided by 4, then what will be the new mean and the new standard deviation respectively?

(A) 45,5

(B) 11 .25,1 .25

(C) 11 .25,2 .5

(D) 12.5, 2.5

Solution: (C)

N=100 x=50 σ=10

x '=∑ x i

'

N=∑

x i'−54

N

¿14 (∑ x i

'

N−5)=1

4(50−5 )=11.25


σ '=σ ( x i−5

4 )=14σ (x i−5 )=

14σ (xi )

¿104

=2.5

80. If two fair dice are rolled then what is the conditional probability that the first dice lands on 6 given that the sum of numbers on the dice is 8?

(A) 13

(B) 14

(C) 15

(D) 16

Solution: (C)

Total cases (2,6 ) (3,5 ) (4,4 ) (5,3 ) (6,2 )

Favourable case (6,2 )

Hence required probability ¿15

81. Two symmetric dice flipped with each dice having two sides painted red, two painted black, one painted yellow and the other painted white. What is the probability that both land on the same colour?

(A) 318

(B) 29

(C) 518

(D) 13

Solution: (C)

Sides of dice R ,R , B ,B ,Y ,W

Total cases ¿6×6=36

Favourable cases ¿ C1 ∙2 C1+ C1

2 ∙ C1+ C1 ∙ C1+ C 1∙ C1111122

¿10

Hence required probability ¿1036

=518

NDA 21st April 2019 MATHS82. There are n socks in a drawer, of which 3 socks are red. If 2 of the socks are chosen randomly

and the probability that both selected socks are red is 12

then what is the value of n ?

(A) 3

(B) 4

(C) 5

(D) 6

Solution: (B)

Given out of n socks 3 socks are red. Now is 2 socks are selected then probability that both the

socks are red ¿C2

3

C2n

Given C2

3

C2n =

12

⇒ C2n

=6

⇒n=4

NDA 21st April 2019 MATHS83. Two cards are chosen at random from a deck of 52 playing cards. What is the probability that both of them have the same value?

(A) 117

(B) 317

(C) 517

(D) 717

Solution: (A)

Total number of ways of selecting two cards from 52 cards ¿ C252

Favourable cases ¿13 ∙ C24

Hence required probability ¿13. C2

4

C252

¿117

84. In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes is respectively given by

(A) 83,169

(B) 83,43

(C) 43,43

(D) 43,69

Solution: (B)

Here n=8, p=26=

13, q=

23

Mean ¿np=8.13=

83

Variance ¿npq=8 ∙13∙23=

169

Standard deviation ¿43

85. A and B are two events such that A and B are mutually exclusive. If P (A )=0 ⋅5 andP (B )=0 ⋅6, then what is the value of P (A|B )?


(A) 15

(B) 16

(C) 25

(D) 13

Solution: (B)

Given A and B are mutually exclusive

⇒P ( A ∩B )=0

P (A )=0.5⇒P ( A )=0.5

P (B )=0.6⇒P (B )=0.6

P( AB )= P (A ∩B )

P ( B )=

1−P ( A∪ B )

P ( B )

⇒1−[P ( A )+P (B ) ]

P (B )

¿1−0.90.6

¿16

∴P( AB )=1

6

86. Consider the following statements:

1. The algebraic sum of deviations of a set of values from their arithmetic mean is always zero.

2. Arithmetic mean > Median > Mode for a symmetric distribution.


(A) 1 only

(B) 2 only

(C) Both 1 and 2

(D) Neither 1 nor 2

Solution: (A)

1. x=∑ x i

N

⇒∑ (x i− x )=∑ x i−∑ x

⇒N x−N x=0


2.

For symmetric distributions

Mean ¿ Median ¿ Mode.

87. Let the correlation coefficient between X and Y be 0 .6 . Random variables Z and W

are defined as Z=X+5 and W=Y3. What is the correlation coefficient between Z and W ?

(A) 0 .1

(B) 0 .2

(C) 0 .36

(D) 0 .6

Solution: (D)

Correlation coefficient

r (Z ,W )=r(X+5,Y3 )

¿ r (X ,Y3 ) Correlation coefficient is independent of origin

¿ r ( X ,Y ) Correlation coefficient is independent of scale

¿0.6

88. If all the natural numbers between 1 and 20 are multiplied by 3, then what is the variance of the resulting series?

(A) 99 .75

(B) 199 .75

(C) 299 .25

(D) 399 .25

Solution: (C)

Observation: 3,6,9, 12,…,60


σ2=∑ x i

2

N−(∑ x i

N )2

¿

9∙20∙21 ∙41

620

−(3 ∙20 ∙21

220 )

2

¿63 .41

2−(632 )

2

¿63.19

4=

11974

=299.25

89. What is the probability that an interior point in a circle is closer to the center than to the circumference?

(A) 14

(B) 12

(C) 34

(D) It cannot be determined

Solution: (A)

Let radius of circle be r

P (E )=

π ( r2 )2

π r2

P (E )=14

90. If A and B are two events, then what is the probability of occurrence of either event A or event B?

(A) P (A )+P ( B )

(B) P ( A∪B )

NDA 21st April 2019 MATHS(C) P (A∩B )

(D) P (A ) P ( B )

Solution: (B)

Probability of occurrence of either event A or event B is P (A∪B )

91. The centroid of the triangle with vertices A (2,−3, 3 ) , B (5, −3,−4 ) and C (2,−3, −2 ) is the point

(A) (−3,3,−1 )

(B) (3, −3,−1 )

(C) (3,1,−3 )

(D) (−3,−1,−3 )

Solution: (B)

Given vertices of triangle are A (2,−3,3 ) , (5,−3,−4 ) and C (2,−3,−2 )

Centroid ¿( x1+x2+x3

3,y1+ y2+ y3

3,z1+z2+z3

3 )¿( 2+5+2

3,−3−3−3

3,3−4−2

3 )¿ (3,−3,−1 )

92. What is the radius of the sphere x2+ y2

+z2−6 x+8 y−10 z+1=0?

(A) 5

(B) 2

(C) 7

(D) 3

Solution: (C)

Given equation of sphere is x2+ y2

+z2−6 x+8 y−10 z+1=0

⇒ (x2−6 x+9 )−9+( y2

+8 y+16 )−16+( z2−10 z+25 )−25+1=0

⇒ ( x−3 )2+( y+4 )2+( z−5 )2=49

∴ Radius of given sphere ¿7

93. The equation of the plane passing through the intersection of the planes2x+ y+2 z=9,4 x−5 y−4 z=1 and the point (3,2,1 ) is

(A) 10 x−2 y+2 z=28

(B) 10 x+2 y+2 z=28

(C) 10 x+2 y−2 z=28

NDA 21st April 2019 MATHS(D) 10 x−2 y−2 z=24

Solution: (A)

Family of plane passing through line of intersection of planes 2x+ y+2 z=9 and 4 x−5 y−4 z=1 is(2x+ y+2 z−9 )+ λ (4 x−5 y−4 z−1 )=0 …(i)

This plane passes through (3,2, 1 )

⇒ λ=13

Hence equation of required plane is

(2x+ y+2 z−9 )+13

(4 x−5 y−4 z−1 )=0

⇒10 x−2 y+2 z=28

94. The distance between the parallel planes 4 x−2 y+4 z+9=0 and 8 x−4 y+8 z+21=0 is

(A) 14

(B) 12

(C) 32

(D) 74

Solution: (A)

Given equations of plane are

4 x−2 y+4 z+9=0⇒ 8 x−4 y+8 z+18=0 and 8 x−4 y+8 z+21=0

∴ Distance between parallel planes ¿|d1−d2|

√x2+ y2

+z2

¿|18−21|

√64+16+64=

3√144

¿14

95. What are the direction cosines of z-axis?

(A) ¿1,1, 1>¿

(B) ¿1,0, 0>¿

(C) ¿0,1, 0>¿

(D) ¿0,0,1>¿

Solution: (D)

NDA 21st April 2019 MATHSClearly, direction cosines of z -axis are cos90o ,cos 90o and cos0o

⇒ ⟨ 0,0, 1 ⟩

96. If a=i−2 j+5 k and b=2 i+ j−3 k then what is ( b−a ) ∙ (3 a+b ) equal to?

(A) 106

(B) −106

(C) 53

(D) −53

Solution: (B)

Given a=i−2 j+5 k and b=2 i+ j−3 k

Now, b−a=(2−1 ) i+ (1+2 ) j+(−3−5 ) k

¿ i+3 j−8 k

3 a+b=3 ( i−2 j+5 k )+ (2 i+ j−3 k )

¿ (3+2 ) i+ (−6+1 ) j+(15−3 ) k

¿5 i−5 j+12 k

∴ ( b−a ) ∙ (3 a+b )=( i+3 j−8 k ) ∙ (5 i−5 j+12 k )

¿5−15−96

¿−106

97. If the position vectors of points A and B are 3 i−2 j+k and 2 i+4 j−3 k respectively, then what is the length of AB ?

(A) √14

(B) √29

(C) √43

(D) √53

Solution: (D)

Given position vector of A and B are

OA=3 i−2 j+ k

OB=2 i+4 j−3 k

∴ AB=OB−OA

¿ (2 i+4 j−3 k )−(3 i−2 j+k )

−i+6 j+4 k

∴ length of AB=√ (−1 )2+(6 )

2+(−4 )

2

NDA 21st April 2019 MATHS¿√1+36+16

¿√53

98. If in a right – angled triangle ABC , hypotenuse AC=p , then what isAB . AC+BC . BA+CA . CB equal to?

(A) p2

(B) 2 p2

(C) p2

2

(D) p

Solution: (A)

Given AC=P ,

Now AB . AC+BC . BA+CA . CB

¿ AC . ( AB+BC )+BC . BA

¿ AC . AC+0 ( BC⊥ BA )

¿|AC|2

¿ p2

99. The sine of the angle between vectors a=2 i−6 j−3 k and b=4 i+3 j−k is

(A) 1

√26

(B) 5

√26

(C) 526

(D) 126

Solution: (B)

a . b= (2 i−6 j−3 k )∙ (4 i+3 j−k )

|a||b|cosθ=−7


cosθ=−7

7×√26=

−1

√26

sinθ=√1−cos2θ

¿5

√26

∴sin θ=5

√26

100. What is the value of λ for which the vectors 3 i+4 j−k and −2 i+λ j+10 k are perpendicular?

(A) 1

(B) 2

(C) 3

(D) 4

Solution: (D)

For perpendicular vectors dot product is zero

⇒ (3 i+4 j−k )∙ (−2 i+λ j+10 k )=0

⇒−6+4 λ−10=0

⇒ λ=4

101. What is the derivative of sec2 ( tan−1 x ) with respect to x ?

(A) 2x

(B) x2+1

(C) x+1

(D) x2

Solution: (A)

y=sec2 ( tan−1 x )

Let tan−1 x=θ

x=tanθ

y=sec2θ=1+ tan2θ

⇒ y=1+x2

⇒ dydx

=2x

102. If f ( x )=log10 (1+x ) , then what is 4 f (4 )=5 f (1 )−log102 equal to?

(A) 0

NDA 21st April 2019 MATHS(B) 1

(C) 2

(D) 4

Solution: (D)

Given f ( x )=log10 (1+x )

Now 4 f (4 )+5 f (1 )−log102

¿4 log105+5 log10

2−log10

2

¿4 [ log105+ log10

2 ]

¿4 log1010

¿4

103. A function f defined by f ( x )=ln (√x2+1−x ) is

(A) an even function

(B) an odd function

(C) Both even and odd function

(D) Neither even nor odd function

Solution: (B)

Given

f ( x )=ln (√x2+1−x )

⇒ f (−x )= ln (√x2+1+x )

¿ ln( x2+1−x2

√ x2+1−x )

¿ ln( 1

√ x2+1−x )

¿−ln (√ x2+1−x)

f (−x )=−f ( x )

∴ f ( x ) is odd function

104. The domain of the function f defined by f ( x )=logx 10 is

(A) x>10

(B) x>0 excluding x=10

(C) x≥10

(D) x>0 excluding x=1

NDA 21st April 2019 MATHSSolution: (D)

f ( x )=logx10

¿log10

10

log10x

f ( x )=1

log10x

x>0 and log10x ≠0

⇒ x>0 excluding x=1

105. limx→0

1−cos34 x

x2 is equal to

(A) 0

(B) 12

(C) 24

(D) 36

Solution: (C)

limx→0

1−cos34 x

x2 ( 00 form)limx→0

−3cos24 x × (−sin 4 x ) ×4

2 x

limx→0

12cos24 x sin 4 x

4 x2

limx→ 0

24 cos24 xsin 4 x4 x

¿24

106. What is the general solution of the differential equation dydx

+xy=0?

(A) x2+ y2

=c

(B) x2− y2

=c

(C) x2+ y2

=cxy

(D) x+ y=c

Solution: (A)


Given dydx

+xy=0

⇒ dydx

=−xy

⇒ y dy+xdx=0

⇒ y2

2+

x2

2+c=0

⇒ x2+ y2

=c

107. The value of k which makes f (x)={sin x x≠0k x=0

continuous at x=0, is

(A) 2

(B) 1

(C) −1

(D) 0

Solution: (D)

Given f ( x )={sin x ,∧x≠0k ,∧x=0

Given f ( x ) is continuous at x=0

limx→ 0

f ( x )=f (0 )

⇒ limx→ 0

sin x=k

⇒k=0

108. What is the minimum value of a2 x+b2 y where xy=c2?

(A) abc

(B) 2abc

(C) 3abc

(D) 4 abc

Solution: (B)

Given xy=c2

For two terms a2 x and b2 y

A.M. ≥ G.M

⇒ a2 x+b2 y2

≥√a2 x .b2 y

⇒a2 x+b2 y≥2√a2b2c2

NDA 21st April 2019 MATHS⇒a2 x+b2 y≥2abc

109. What is ∫ exln ( a) dx equal to?

(A) ax

ln (a )+c

(B) ex

ln (a )+c

(C) ex

ln (ae )+c

(D) aex

ln (a )+c

Solution: (A)

∫e xln ( a) dx=∫e ln (a )x

dx

¿∫ axdx

¿ax

ln (a )+c

110. What is the area of one of the loops between the curve y=c sin x and x-axis?

(A) c

(B) 2c

(C) 3c

(D) 4 c

Solution: (B)

Required area

¿∫0

π

c sin x dx

¿−c (cos x )|0π

¿−c (−1−1 )

¿2 c


111. If sin θ+cosθ=√2cosθ , then what is (cosθ−sin θ ) equal to?

(A) −√2cosθ

(B) −√2sin θ

(C) √2sin θ

(D) 2sinθ

Solution: (C)

Given sin θ+cosθ=√2cosθ

⇒sinθ=(√2−1 )cosθ ...(i)

Now, cosθ−sin θ=sin θ

√2−1−sin θ

¿ sin θ .[ 1− (√2−1 )

√2−1 ]¿

(2−√2 )

√2−1sin θ

cosθ−sin θ=√2sin θ

112. In a circle of diameter 44 cm, the length of a chord is 22cm. What is the length of minor arc of the chord?

(A) 48421

cm

(B) 24221

cm

(C) 12121

cm

(D) 447

cm

Solution: (A)

Given diameter of circle ¿44 cm ⇒ radius ¿22 cm

Length of chord ¿22 cm

NDA 21st April 2019 MATHSLet AB be the given chord

From figure, it is clear that ∆OAB is equilateral triangle

∴ length of arc AB (minor )=16

(Perimeter of circle)

¿16

(πD )

¿16×

227

×44

¿48421

cm

113. If sin θ=−12

and tan θ=1

√3, then in which quadrant does θ lie?

(A) First

(B) Second

(C) Third

(D) Fourth

Solution: (C)

sin θ=−12

and tan θ=1

√3

Since sin θ is negative and tan θ is positive.

Hence θ lies in third quadrant.

114. How many three-digit even numbers can be formed using the digits 1,2,3, 4 and 5 when repetition of digits is not allowed?

(A) 36

(B) 30

(C) 24

(D) 12

Solution: (C)

2,42,42,4↑3↑4↑2

∴ Total number of 3 digits even numbers

¿3×4×2=24

115. The angle of elevation of a tower of height h from a point A due South of it is x and from a point B due East of A is y . If AB=z , then which one of the following is correct?

NDA 21st April 2019 MATHS(A) h2 (cot2 y−cot2 x )=z2

(B) z2 (cot2 y−cot2 x )=h2

(C) h2 (tan2 y−tan2 x )=z2

(D) z2 ( tan2 y−tan 2 x )=h2

Solution: (A)

tan x=hx

∴X=hcot x ,Y=hcot y

In ∆OAB

OB2−O A2

=A B2

⇒h2 cot2 y−h2 cot2 x=Z2

⇒h2 (cot2 y−cot2 x )=Z2

116. From a deck of cards, cards are taken out with replacement. What is the probability that the fourteenth card taken out is an ace?

(A) 151

(B) 451

(C) 152

(D) 113

Solution: (D)

Required probability ¿ (1 )13 .452

¿113

NDA 21st April 2019 MATHS117. If A and B are two events such that P (A )=0 .5, P (B )=0 .6 and P (A∩B )=0 .4, then what is P ( ´A∪B ) equal to

(A) 0 .9

(B) 0 .7

(C) 0 .5

(D) 0 .3

Solution: (D)

P ( ´A∪B )=1−P (A∪B )

¿1−(P ( A )+P (B )−P (A ∩B ) )

¿1−(0.5+0.6−0.4 )

¿1−0.7

¿0.3

118. A problem is given to three students A ,B and C whose probabilities of solving the problem are12,34

and 14

respectively. What is the probability that the problem will be solved if they all solve the

problem independently?

(A) 2932

(B) 2732

(C) 2532

(D) 2332

Solution: (A)

E=¿ event that problem will be solved

∴P ( E )=1−P (E1 )

¿1−(1−12 )(1− 3

4 )(1−14 )

¿1−12×

14×

34

¿1−332

¿2932

NDA 21st April 2019 MATHS119. A pair of fair dice is rolled. What is the probability that the second dice lands on a higher value than does the first?

(A) 14

(B) 16

(C) 512

(D) 518

Solution: (C)

n (S )=6×6=36

n ( E )=12

(36−6 )=15

∴P ( E )=n ( E )

n (S )=

1536

¿512

120. A fair coin is tossed and an unbiased dice is rolled together. What is the probability of getting a 2 or4 or 6 along with head?

(A) 12

(B) 13

(C) 14

(D) 16

Solution: (C)

n (S )=2×6=12

n ( E )=1×3=3

∴P ( E )=n ( E )

n (S )=

312

=14

mathematics code - d...nda 21st april 2019 mathematics code - d 1. in a school, all the students...

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