NDA & NA (II) 2017 Exam: Mathematics Practice Questions - Set 01

Mathematics practice questions help the aspirants in understanding the nature of NDA & NA (II) 2017 Exam that will be held on 10 September 2017.

Created On: Jul 11, 2017 17:05 IST For the benefit of NDA & NA (II) 2017 Exam aspirants, Jagran Josh is providing Mathematics practice questions. The questions along with answers and detailed explanations help the aspirants in understanding the nature of the questions in the exam. The questions are given below.

1. If  A = {1,2,3} and B = {3,4}, then (A ∪ B) × (A ∩ B) is

(a) {3,3}
(b) {(1,3),(2,3),(3,3),(1,4),(2,4),(3,4)}
(c)  {(1,3),(2,3),(3,3)}

(d)  {(1,3),(2,3),(3,3),(4,3)}

2. If R be a relation on N × N defined by (a, b) R (c, d) if and only if ad = bc; then R is

(a) an equivalence relation
(b) symmetric and transitive but not reflexive
(c) reflexive and transitive but not symmetric
(d) reflexive and symmetric but not transitive

3. What is the maximum value of sin 3θ cos 2θ + cos 3θ sin 2θ?
(a) 1
(b) 2
(c) 4
(d) 10
4.  If     A + B + C = 3π/2, then the value of cos 2A + cos 2B + cos 2C is equal to
(a) 1 – 4cos A cos B cos C
(b) 4 sin A sin B sin C
(c) 1 + 2 cos A cos B cos C
(d) 1 – 4 sin A sin B sin C
5. If f(x) is differrerntiable everywhere, then which one of the following is correct?
(a) ⎹ f⎹ is differentiable everywhere
(b) ⎹f⎹2 is differentiable everywhere
(c) f⎹f⎹  is not differentiable at same points
(d) None of the above.
6. After subtending an angle of 1000ᵒ from its initial position, the revolving line will be situated in which one of the following quadrants?
7.  What is the gradient of the coplanar and non- intersecting line which lies midway between the lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0?
(a) 2/3
(b) -2/3
(c) 3/2
(d) -3/2
8. The number of circle having radius 5 and passing through the points (-2, 0) and (4, 0) is
(a) one
(b) two
(c) four
(d) infinite
9. If G is the centroid of ΔABC, then GA + GB + GC is equal to
(a) 0
(b) 3 GA
(c) 3 GB
(d) 3 GC
10. If C is the mid-point of AB and P is any point not lying on AB, then
(a) PA + PB = 2PC
(b) PA + PB = PC
(c) PA + PB = - PC
(d) PA + PB =  -2PC

1. (d) {(1,3),(2,3),(3,3),(4,3)}
(A∪B) = {1, 2, 3, 4} and (A∩B) = {3}
∴ (A ∪ B) × (A ∩ B) = {(1,3),(2,3),(3,3),(4,3)}
2. (a) an equivalence relation
R is a relation defined on N × N by (a,b)R(c.d) if and only if ad = bc, then R is
(i) Reflexive    Since, for any a , b∊ N
=>     ab = ba      =>   (a, b) R (a, b)
(ii) Symmetric    Let (a, b) R (c, d) => ad = bc  => bc = ad
=>  cb = da (∵   commutativity of natural numbers)
=>     (c, d) R (a, b)
(iii) Transitive     Let (a, b) R (c, d)and (c, d) R (e, f)
=>     ad = bc  and  cf = de
=>     ade = bce  =>  acf = bce
=>     af = be                      (cancellation law)
=>      (a, b) R (e, f)
Hence, from above observations, we conclude that R is an equivalence relation on N × N.
3.(a) 1
Let f(θ) = sin 3θ . cos 2θ + cos 3θ . sin 2θ
= sin (3θ + 2θ)
=sin 5θ    [∵ sin (A+B) = sinA. cosB +cosA. sinB)
We know that,
-1 ≤ sin 5θ ≤ 1 => -1 ≤ f(θ) ≤ 1
So the maximum value is 1.
4. (d) 1 – 4 sin A sin B sin C
Given that, A + B + C = 3π/2
∴         cos 2A + cos 2B + cos 2C = (cos 2A + cos 2B )+ cos 2C
= 2cos (A + B) cos (A – B)+ cos 2C
= -2sin C cos (A – B) + 1 – 2sin2 C
= 1 – 2sin C [cos (A – B) + sin C]
= 1 – 2sin C [cos (A – B) – cos (A +B)]
= 1 – 2sin C 2sin A sin B
= 1 – 4sin A sin B sin C.
5. (c) f⎹f⎹  is not differentiable at same points
If f(x) is differentiable everywhere, then f⎹f⎹  is not differentiable at same points.
Since , the product of two functions in which one is differentiable and other is not differentiable, then the resultant is not differentiable.
∵ 1000ᵒ = 2 X 360ᵒ+280
∴ From above it is clear that the revolving line will be in the fourth quadrant.
7. (d) -3/2
The given lines are 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0  and these lines are parallel to each other, therefore their slope is -3/2.
The required line will also be parallel to these lines, so the gradient of the required line will be – 3/2.
8. (b) two
Two centre of each lying on the perpendicular bisector of the join of the two points.
9. (a) 0
We have, GB + GC = (1 +1) GD, where D is mid-point of BC.
∴    GA + GB + GC = GA + 2GD =GA – GA = 0
(G divides AD in the ratio 2 : 1,∴  2GD = - GA)
10. (a) PA + PB = 2PC
Since, C is the mid-point of AB
∴ AC = CB
Now, in ΔAPC,
PA + AC = PC    ….(i)
And in ΔBPC,
PC + CB = PB
==>    PC = PB – CB    ….(ii)
In adding Eqs. (i) and (ii), we get
2PC = PA + PB + AC – CB
==>    2PC = PA + PB

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