NDA Mock Test - 1 - Defence MCQ

Correct Answer :- d

Explanation : (1-w)(1-w²)(1+w⁴)(1+w⁸)......(1)

w⁴ = w³ * w

w⁴ = 1*w = w.......(2)

w⁸ = w³ * w³ * w²

= w².........(3)

Put the values of (2) and (3) in eq (1)

(1-w)(1-w²)(1+w)(1+w²)

(1-w)(1+w)(1-w²)(1+w²)

=> (1-w²) (1-w)

From eq(2) w⁴ = w

1 - w - w² + w³ 

As we know that 1 + w + w^2 = 0

w² + w = -1……..(4)

Putting in eq(4)

= 1 + 1 + 1 

= 3

Given:
- The sum of 3 numbers is 285.
- The ratio between the 2nd and 3rd numbers is 6:5.
- The ratio between the 1st and 2nd numbers is 3:7.
To find:
The value of the 3rd number.

Let's assume the 1st number is 3x and the 2nd number is 7x (according to the given ratio).
The ratio between the 2nd and 3rd numbers is 6:5, so we can write it as:
(7x)/(5y) = 6/5
Cross-multiplying, we get:
35x = 6y
Similarly, the sum of the 3 numbers is given as 285, so we can write it as:
3x + 7x + 5y = 285
Combining the equations, we have:
3x + 7x + (35x/6) = 285
Simplifying the equation, we get:
(18x + 42x + 35x)/6 = 285
(95x)/6 = 285
95x = 1710
x = 1710/95
x = 18
Now, we can find the value of the 3rd number (5y) using the ratio between the 2nd and 3rd numbers:
(7x)/(5y) = 6/5
(7*18)/(5y) = 6/5
126/5y = 6/5
Cross-multiplying, we get:
30y = 630
y = 630/30
y = 21
Therefore, the value of the 3rd number is 5y = 5*21 = 105.
Answer:
The 3rd number is 105.

To find the roots of the equation x² - x + 1 = 0, we need to find the values of x that satisfy this equation.
Given that α and β are the roots of the equation x² + x + 1 = 0, we can use the fact that the sum of roots of a quadratic equation is equal to the negation of the coefficient of x divided by the coefficient of x².
The sum of the roots of x² + x + 1 = 0 is -1/1 = -1.
Therefore, the sum of the roots of x² - x + 1 = 0 is also -1.
Since the sum of the roots remains the same, it means that the roots of x² - x + 1 = 0 are not α and β.
Hence, the answer is D: None of these.

To find the values of p, we need to consider the equation (x – p)(x – 6) + 1 = 0.
Expanding the equation, we get x^2 - (6+p)x + 6p + 1 = 0.
Since the equation has integral coefficients, the discriminant of the quadratic equation must be a perfect square.
The discriminant is given by D = (6+p)^2 - 4(6p + 1).
Simplifying, we have D = 36 + 12p + p^2 - 24p - 4.
Combining like terms, we get D = p^2 - 12p + 32.
To find the values of p, we need to find the perfect squares that can be represented by p^2 - 12p + 32.
The perfect squares between 32 and 64 are 36 and 49.
Therefore, the possible values of p are:
- p = 6: (6)^2 - 12(6) + 32 = 0, which is not a perfect square.
- p = 8: (8)^2 - 12(8) + 32 = 64 - 96 + 32 = 0, which is a perfect square.
So the answer is p = 8.
Therefore, the correct option is A: 4 or 8.

To find the roots of the equation (x - p)(x - q) = r^2, we can expand the equation and set it equal to zero:
(x - p)(x - q) = r^2
x^2 - qx - px + pq = r^2
x^2 - (p + q)x + pq - r^2 = 0
We can now use the quadratic formula to find the roots of the equation:
x = (-(-p - q) ± √((-p - q)^2 - 4(pq - r^2))) / 2
x = (p + q ± √((p + q)^2 - 4(pq - r^2))) / 2
x = (p + q ± √(p^2 + 2pq + q^2 - 4pq + 4r^2)) / 2
x = (p + q ± √(p^2 - 2pq + q^2 + 4r^2)) / 2
x = (p + q ± √((p - q)^2 + 4r^2)) / 2
From the quadratic formula, we can see that the roots of the equation will be real if the discriminant ((p - q)^2 + 4r^2) is greater than or equal to zero. Let's analyze the different cases:
1. If r^2 > 0:
- The discriminant ((p - q)^2 + 4r^2) will always be greater than zero, as the sum of two positive numbers is always positive.
- Therefore, the roots will always be real.
2. If r^2 = 0:
- The discriminant ((p - q)^2 + 4r^2) simplifies to (p - q)^2.
- The roots will be real if (p - q)^2 is greater than or equal to zero, which is always true.
- Therefore, the roots will always be real.
3. If r^2 < 0:
- The discriminant ((p - q)^2 + 4r^2) will always be negative, as the sum of a positive number and a negative number is always negative.
- Therefore, the roots will always be complex.
Based on the analysis of the different cases, we can conclude that the roots of the equation (x - p)(x - q) = r^2 will always be real (option B).

A ² + B² = AA + BB
=A(BA) + B(AB)     (Given AB = B and BA = A)
=(AB)A + (BA)B
=BA + AB
=A + B

To determine whether the statement (AB)ⁿ = AⁿBⁿ is true, we need to consider different cases:
Case 1: AB = BA (i.e., A and B commute)
In this case, the statement (AB)ⁿ = AⁿBⁿ is true. This can be proved using mathematical induction.
Case 2: AB ≠ BA (i.e., A and B do not commute)
In this case, the statement (AB)ⁿ = AⁿBⁿ is not true. This can be shown by providing a counterexample.
Counterexample:
Let A = [[1, 0], [0, 2]] and B = [[2, 0], [0, 3]]. Here, AB ≠ BA.
If we calculate (AB)² and A²B², we get:
(AB)² = [[4, 0], [0, 18]]
A²B² = [[4, 0], [0, 36]]
Since (AB)² ≠ A²B², the statement (AB)ⁿ = AⁿBⁿ is false when AB ≠ BA.
Therefore, the correct answer is C: true only when AB = BA.

Given:
- A and B are symmetric matrices of the same order.
To find:
- The nature of the matrix (ABA)^T.
Explanation:
- A symmetric matrix is a matrix that is equal to its transpose. In other words, the elements of a symmetric matrix remain the same when they are reflected along the main diagonal.
- The transpose of a matrix is obtained by interchanging its rows with columns.
- A matrix multiplied by its transpose is always a symmetric matrix.
Now, let's analyze the given expression (ABA)^T:
1. First, perform the multiplication ABA:
- A * B = C, where C is a matrix of the same order as A and B.
- C * A = D, where D is also a matrix of the same order.
2. Take the transpose of D: D^T.
3. As mentioned earlier, the product of a matrix and its transpose is always a symmetric matrix.
- Therefore, (ABA)^T is a symmetric matrix.
Hence, the correct answer is Option A: symmetric matrix.

If each element in a row of a determinant is multiplied by the same factor r, then the value of the determinant will be multiplied by r.
Explanation:
To understand why the answer is option D, let's first review the concept of determinants:
- A determinant is a scalar value that can be calculated for a square matrix.
- The value of a determinant is used to determine various properties of the matrix, such as whether it is invertible or singular.
- The determinant of a matrix can be calculated by expanding along any row or column using cofactor expansion.
Now, let's consider the scenario where each element in a row of a determinant is multiplied by the same factor r. The key point to note here is that we are multiplying the elements of a single row, not the entire determinant.
When we multiply each element in a row by r, the determinant can be expanded along that row. The cofactors of the expanded determinant will have the same row, except for the elements that were multiplied by r. These elements will now be multiplied by r as well.
Here's a step-by-step explanation:
1. Let's assume the determinant is of size n x n, and we are multiplying the elements of the ith row by r.
2. The cofactor expansion of the determinant along the ith row will give us a sum of terms, where each term is a product of an element from the ith row, its cofactor, and the determinant of the remaining (n-1) x (n-1) matrix.
3. Since each element in the ith row has been multiplied by r, all the terms in the cofactor expansion will include a factor of r.
4. Therefore, when we simplify the expanded determinant, we can factor out r from each term.
5. The remaining cofactors and the determinant of the remaining (n-1) x (n-1) matrix will be unchanged.
6. Thus, the value of the determinant will be multiplied by r.
In conclusion, when each element in a row of a determinant is multiplied by the same factor r, the value of the determinant is multiplied by r. Therefore, the answer is option D.

(cotx+cosecx)-1/(cotx-cosecx+1)
=(cosecx+cotx)-(cosec²x-cot²x)/(cotx-cosecx+1)
=(cosecx+cotx)[1-(cosecx-cotx)]/(cotx-cosecx+1)
=(cosecx+cotx)(cotx-cosecx+1)/(cotx-cosecx+1)
=(cosecx+cotx)
=(1/sinx+cosx/sinx)
=1+cosx/sinx

Concept:
tan A = ^{sin A}⁄_{cos A} and cot A = ^{cos A}⁄_{sin A}
sin² A + cos² A = 1

Calculations:
Given, sin A cos A tan A + cos A sin A cot A
= sin A cos A (tan A + cot A)
= sin A cos A (^{sin A}⁄_{cos A} + ^{cos A}⁄_{sin A})
= sin A cos A (^{sin2 A + cos2 A}⁄_{sin A cos A})
= 1
Hence, sin A cos A tan A + cos A sin A cot A = 1

Since, A, B, C are in AP.

ie same cos ( π – 3α ) = - cos ( 3α )