NDA I - Mathematics Question Paper 2019 - NDA MCQ

Substituting n = 1, 2, 3, 4, 5, on option (d), we obtain a₅ = (−1)⁵⁻¹5⁵⁺¹ = 5⁶ = 15625.

Therefore, the required terms are 25, – 125, 625, – 3125, and 15625.

Given X = {1, 2, 3, 4} and R = [(1, 1), (2, 2), (3, 3), (1, 2), (2,1), (2, 3), (3, 2)}
Reflexive Property states that for every real number x, x=x.
Here (1,1) ∈ R, (2,2) ∈ R, (3,3) ∈ R, but (4,4) ∉ R. So, R is not Reflexive.

Symmetric Property states that for all real numbers x and y, if x=y, then y=x.
(1, 2) ∈ R, (2, 1) ∈ R and, (2, 3) ∈ R, (3,2) ∈ R

So, R is symmetric.

Transitive Property states that for all real
numbers x, y and z, if x=y and y=z, then x=z.

Here (1, 2) ∈ R, (2, 1) ∈ R and (1, 1) ∈ R,

Similarly, (2, 3) ∈ R, (3, 2) ∈ R, (2, 2) ∈ R

But for (1, 2) ∈ R, (2, 3) ∈ R, (1, 3) ∉ R

So, R is not transitive.

Thus, R is neither reflexive nor transitive, but symmetric.

So, R is not transitive.
Hence, R is neither symmetric nor  transitive, but reflexive.

By drawing venn diagram

This statement is possible

The second statement is impossible.

Hence matrix A is an Involutory matrix. 

By comparison both sides, we get

Given

Now according to options there are four roots i.e. 1, –1, ω, ω² But 1 and –1 are not satisfying the given equations. Hence ω, ω² are the required roots.

A straight line can be drawn by joining two points.

Hence the number of possible straight lines is

Given

Here sum is negative & product is positive this is possible only when both the roots are negative. i.e. a < 0, b < 0.

Let A be a set, then the of all the possible subsets of is called the power set of A and is denoted by P(A).

⇒ 50 + 28 + x + 12 − x + 21 + x
⇒ 111 + x (total) Minimum value when x=0 will be 111 + 0 = 111

⇒ 50 + 28 + x + 12 − x + 21 + x
⇒ 111 + x (total)
The maximum value for x=12 111 + x = 0
111 + 12 = 0
The maximum value = 123

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

Similarly, for A = A, A⁻¹ = A

A is a square matrix of order 

Determinant |A| = 4 Since, we know that  |A′ | = |A|

So, there is no value change when determinant to transpose, since both are equal.
Transpose |A′ | = 4

From 6 programmers and 4 typists, an office wants to recruit 5 people.

The number of ways this can be done so as to recruit at least one typist= 252 – 6 = 246

Given expression is 

Number of terms = 4 + 1 = 5

Given:

Therefore, n = 9

The numbers 1,5 and 25 can be three terms (not necessarily consecutive) of one or many numbers of Arithmetic and Geometric Progressions.
To check the Arithmetic Progression: Let us assume that given number series 1, 5 and 25 are I^th, J^th and Q^th terms in the AP series whose distance is equal to d.
Therefore,

Here x is a natural number.

So, the given three numbers can be terms in one or many numbers of

Arithmetic Progressions.

A is a square matrix of order n > 1. det (–A) = (−1)ⁿ det A is correct because, when 1 is substituted with I and if we multiplied the RHS then eventually equals to LHS

where −I is the matrix with −1 on the diagonal and 0 elsewhere. 

Given det A = 4 and det B = 3 det(2AB)

= 2³ det A det B 

= 8 × 4 × 3

= 96

Given complex number, 

We know that, Principal argument of z is