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Question 131

National Defence Academy (NDA) 2025• Mathematics

If A2+B2+C2=0, then what is the value of the following?

∣1cosCcosB cosC1cosA cosBcosA1∣\begin{vmatrix} 1& cosC& cosB\ cosC&1&cosA \ cosB&cosA&1 \end{vmatrix}

  1. AOption (a)

    -1

  2. BOption (b)

    0

  3. COption (c)

    1

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (b)B.

0

Why this answer is correct

Concept:

When A2 + B2 + C2 = 0, it implies A = B = C = 0 (since the squares of real numbers are non-negative).

Substitute the values of A, B, and C for determinant calculation into the matrix

Calculation:

∣1cos0cos0 cos01cos0 cos0cos01∣\begin{vmatrix} 1& cos0& cos0\ cos0&1&cos0 \ cos0&cos0&1 \end{vmatrix}

Since, Cos0 =1

Thus Matrix becomes 

∣111 111 111∣\begin{vmatrix} 1& 1& 1\ 1&1&1 \ 1&1&1 \end{vmatrix}

Now determinant = 1[(1×1 - 1×1)] - 1[(1×1 - 1×1)] + 1[(1×1 - 1×1)]

= = 1(0) - 1(0) + 1(0) = 0

∴ The value of the determinant is 0.

Hence, the correct answer is Option 2.

Question 132

National Defence Academy (NDA) 2025• Mathematics

If ω is a non-real cube root of unity, then what is a root of the following equation?∣x+1ωω2 ωx+ω21 ω21x+ω∣=0 \begin{vmatrix} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = 0

  1. AOption (a)

    x=0

  2. BOption (b)

    x=1

  3. COption (c)

    x=ω

  4. DOption (d)

    x=ω2

View answer & explanation

Correct answer

Answer: (a)A.

x=0

Why this answer is correct

Calculation:

Given,

Let ω be a non-real cube root of unity, so ω3=1 \omega^{3}=1 and 1+ω+ω2=0 1+\omega+\omega^{2}=0 .

Consider the determinant

Δ(x)=∣x+1ωω2 ωx+ω21 ω21x+ω∣=0. \Delta(x)= \begin{vmatrix} x+1 & \omega & \omega^{2}\ \omega & x+\omega^{2} & 1\ \omega^{2} & 1 & x+\omega \end{vmatrix}=0.

Step 1 — Column operation:  Replace the first column by C1−C2C_{1}-C_{2}:

Δ(x)=∣x2−1,2!k!+!1,1 k−1k+21 031∣. \Delta(x)= \begin{vmatrix} x^{2}-1 & ,2!k!+!1, & 1\ k-1 & k+2 & 1\ 0 & 3 & 1 \end{vmatrix}.

Step 2 — Expansion along the third row:

Δ(x)=−3!∣k2−11 k−11∣+∣k2−12k+1 k−1k+2∣, \Delta(x)= -3! \begin{vmatrix} k^{2}-1 & 1\ k-1 & 1 \end{vmatrix} + \begin{vmatrix} k^{2}-1 & 2k+1\ k-1 & k+2 \end{vmatrix},

which simplifies to

Δ(x)=x(x2−1)−x(ω+ω2)=x(x2−1)+x=x3. \Delta(x)=x(x^{2}-1)-x\bigl(\omega+\omega^{2}\bigr) =x(x^{2}-1)+x =x^{3}.

Step 3 — Equate to zero:

Δ(x)=0;⟹;x3=0;⟹;x=0. \Delta(x)=0 ;\Longrightarrow; x^{3}=0 ;\Longrightarrow; x=0.

∴ The root of the equation is  x=0 x = 0 .

Hence, the correct answer is Option 1.

Question 133

National Defence Academy (NDA) 2025• Mathematics

What is (3+i3−i)3 \left( \frac{\sqrt{3}+i}{\sqrt{3}-i} \right)^3  equal to?

  1. AOption (a)

    -1

  2. BOption (b)

    0

  3. COption (c)

    1

  4. DOption (d)

    3

View answer & explanation

Correct answer

Answer: (a)A.

-1

Why this answer is correct

Concept:

The key concept involves simplifying complex numbers in the polar form. We use the argument (angle) of the complex number and De Moivre's theorem, which states:

  (r(cosθ + i sinθ))n = rn(cos(nθ) + i sin(nθ))

Calculation:

⇒ (3+i3−i)3\left( \frac{\sqrt{3} + i}{\sqrt{3} - i} \right)^3

Multiply the numerator and the denominator by the conjugate of the denominator

⇒ 3+i3−i×3+i3+i\frac{\sqrt{3} + i}{\sqrt{3} - i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i}

⇒ (3+i)2(3−i)(3+i)\frac{(\sqrt{3} + i)^2}{(\sqrt{3} - i)(\sqrt{3} + i)}

 2+23i4=1+3i2\frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2}

Convert to polar form. The modulus r is 

r=(12)2+(32)2=14+34=1=1r = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1

The argument θ is 

θ=tan⁡−1(3212)=tan⁡−1(3)=π3\theta = \tan^{-1}\left( \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

So the polar form of the complex number is

1(cos⁡π3+isin⁡π3)1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)

To cube the complex number, we use De Moivre’s Theorem

In our case, r = 0 so,

(cos⁡π3+isin⁡π3)3=cos⁡π+isin⁡π=−1+0i=−1\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)^3 = \cos \pi + i \sin \pi = -1 + 0i = -1

Thus, the result of cubing the complex number is −1\boxed{-1}

Hence, the correct answer is Option 1.

Question 134

National Defence Academy (NDA) 2025• Mathematics

If x2− x +1 = 0, then what is (x−1x)2+(x−1x)4+(x−1x)8(x-\frac{1}{x})^2+(x-\frac{1}{x})^4+(x-\frac{1}{x})^8

  1. AOption (a)

    81

  2. BOption (b)

    85

  3. COption (c)

    87

  4. DOption (d)

    90

View answer & explanation

Correct answer

Answer: (c)C.

87

Why this answer is correct

Calculation:

Given,

The equation is x2−x+1=0 x^2 - x + 1 = 0

We need to find the value of the following expression:

(x−1x)2+(x−1x)4+(x−1x)8 \left( x - \frac{1}{x} \right)^2 + \left( x - \frac{1}{x} \right)^4 + \left( x - \frac{1}{x} \right)^8

The equation x2−x+1=0 x^2 - x + 1 = 0 is solved as follows:

x=1±−32=eiπ/3orx=e−iπ/3 x = \frac{1 \pm \sqrt{-3}}{2} = e^{i \pi / 3} \quad \text{or} \quad x = e^{-i \pi / 3}

Now, substitute the value of x x into the expression x−1x x - \frac{1}{x} :

x−1x=i3 x - \frac{1}{x} = i\sqrt{3}

Evaluate the powers of x−1x x - \frac{1}{x}

Now, let's evaluate each term in the expression:

(x−1x)2=(i3)2=−3 \left( x - \frac{1}{x} \right)^2 = (i \sqrt{3})^2 = -3

(x−1x)4=(−3)2=9 \left( x - \frac{1}{x} \right)^4 = (-3)^2 = 9

(x−1x)8=92=81 \left( x - \frac{1}{x} \right)^8 = 9^2 = 81

Now, sum the values:

−3+9+81=87 -3 + 9 + 81 = 87

∴ The value of the expression is 87.

Hence, the correct answer is Option 3.

Question 135

National Defence Academy (NDA) 2025• Mathematics

How many 7-letter words (with or without meaning) can be constructed using all the letters of the word CAPITAL so that all the consonants come together in each word?

  1. AOption (a)

    360

  2. BOption (b)

    300

  3. COption (c)

    288

  4. DOption (d)

    240

View answer & explanation

Correct answer

Answer: (c)C.

288

Why this answer is correct

Concept:

If all consonants are treated as a single unit, the problem reduces to arranging the unit (CPTL) and vowels (A, I, A).

Within the consonant unit, the consonants themselves can be permuted.

When letters repeat, the arrangements are divided by the factorial of the repeated letters.

Calculation:

⇒ Treat consonants CPTL as a single unit.

⇒ Total "super units" = {CPTL, A, I, A} = 4 units.

⇒ Arrangements of super units = 4! / 2! (since A repeats).

⇒ Arrangements of super units = 24 / 2 = 12. ...(1)

⇒ Internal arrangements of consonants CPTL = 4!.

⇒ Internal arrangements = 4 × 3 × 2 × 1 = 24. ...(2)

⇒ Total arrangements = (Arrangements of super units) × (Internal arrangements of consonants).

⇒ Total arrangements = 12 × 24.

⇒ Total arrangements = 288.

∴ The total number of 7-letter words = 288.

Hence, the correct answer is Option 3.

Question 136

National Defence Academy (NDA) 2025• Mathematics

If z≠0z\ne0 is a complex number, then what is amp(z)+amp(zˉ) equal to?

  1. AOption (a)

    0

  2. BOption (b)

    π/2

  3. COption (c)

    π

  4. DOption (d)

    2π

View answer & explanation

Correct answer

Answer: (a)A.

0

Why this answer is correct

Concept:

  1. The amplitude (or argument) of a complex number z=r(cos⁡θ+isin⁡θ) z = r(\cos\theta + i\sin\theta) is given by amp(z)=θ \text{amp}(z) = \theta .
  2. The conjugate of z z , denoted as z‾ \overline{z} , has an amplitude amp(z‾)=−θ \text{amp}(\overline{z}) = -\theta , because conjugating a complex number reflects it across the real axis in the Argand plane.

Formula Used:

amp(z)+amp(z‾)=θ+(−θ)=0 \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0 .

Calculation:

z=r(cos⁡θ+isin⁡θ) z = r(\cos\theta + i\sin\theta)

z‾=r(cos⁡θ−isin⁡θ) \overline{z} = r(\cos\theta - i\sin\theta)

amp(z)+amp(z‾)=θ+(−θ)=0 \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0

Conclusion:

∴amp(z)+amp(z‾)=0 \therefore \text{amp}(z) + \text{amp}(\overline{z}) = 0 .

Hence, the correct answer is Option 1.

Question 137

National Defence Academy (NDA) 2025• Mathematics

How many sides are there in a polygon that has 20 diagonals?

  1. AOption (a)

    6

  2. BOption (b)

    7

  3. COption (c)

    8

  4. DOption (d)

    10

View answer & explanation

Correct answer

Answer: (c)C.

8

Why this answer is correct

Calculation:

Given,

Number of diagonals, D = 20 

Formula for the number of diagonals in a polygon is:

D=n(n−3)2D = \frac{n(n - 3)}{2}

Substitute the value of D = 20  in the formula:

⇒ 20=n(n−3)220 = \frac{n(n - 3)}{2}

Multiply both sides by 2:

⇒ 40=n(n−3)40 = n(n - 3)

 

⇒ n2−3n−40=0n^2 - 3n - 40 = 0

We now solve this quadratic equation:

n=−b±b2−4ac2an = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

⇒ n=−(−3)±(−3)2−4(1)(−40)2(1)n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-40)}}{2(1)}

 

⇒ n=3±132n = \frac{3 \pm 13}{2}

The two solutions are:

⇒ n=3+132=8n = \frac{3 + 13}{2} = 8

or

⇒ n=3−132=−5n = \frac{3 - 13}{2} = -5

Since the number of sides cannot be negative, we choose n = 8.

∴ The polygon has 8 sides.

Hence, the correct answer is Option 3.

Question 138

National Defence Academy (NDA) 2025• Mathematics

In how many ways can the letters of the word DELHI be arranged, keeping the positions of vowels and consonants unchanged?

  1. AOption (a)

    6

  2. BOption (b)

    9

  3. COption (c)

    12

  4. DOption (d)

    14

View answer & explanation

Correct answer

Answer: (c)C.

12

Why this answer is correct

Calculation:

Number of vowels = 2 (E, I). Positions of vowels are fixed.

⇒ Arrangements of vowels = 2! = 2 × 1 = 2

Number of consonants = 3 (D, L, H). Positions of consonants are fixed.

⇒ Arrangements of consonants = 3! = 3 × 2 × 1 = 6

Total arrangements = Arrangements of vowels × Arrangements of consonants.

⇒ Total arrangements = 2 × 6 = 12

∴ The total number of arrangements is 12.

Hence, the correct answer is Option 3.

Question 139

National Defence Academy (NDA) 2025• Mathematics

What is the number of positive integer solutions of x+y+z=5?

  1. AOption (a)

    3

  2. BOption (b)

    5

  3. COption (c)

    6

  4. DOption (d)

    9

View answer & explanation

Correct answer

Answer: (c)C.

6

Why this answer is correct

Concept:

The number of positive integer solutions for the equation x + y + z = n is given by the combination formula:

Number of solutions = C(n - 1, k - 1), where k is the number of variables.

Formula Used:

C(n, r) = n! / [r! × (n - r)!]

Calculation:

Here, n = 5 and k = 3.

The number of solutions is:

⇒ C(5 - 1, 3 - 1) = C(4, 2)

⇒ C(4, 2) = 4! / [2! × (4 - 2)!]

⇒ C(4, 2) = 24 / [2 × 2]

⇒ C(4, 2) = 6

The correct answer is Option 3.

Question 140

National Defence Academy (NDA) 2025• Mathematics

What is the number of rational terms in the expansion of (3+514)12(\sqrt{3}+5^\frac{1}{4})^{12}

  1. AOption (a)

    2

  2. BOption (b)

    3

  3. COption (c)

    4

  4. DOption (d)

    6

View answer & explanation

Correct answer

Answer: (c)C.

4

Why this answer is correct

Concept:

A term in the binomial expansion of (a + b)n is given by Tk+1 = C(n, k) × an-k × bk.

For a term to be rational, the exponents of both √3 and 51/4 must be integers.

Formula Used:

In (√3)n-k, n-k must be even for it to be rational.

In (51/4)k, k must be a multiple of 4 for it to be rational.

Calculation:

Let n = 12:

⇒ For √3n-k to be rational, n-k must be even.

⇒ Since n = 12, k must also be even.

⇒ For (51/4)k to be rational, k must be a multiple of 4.

⇒ The values of k that satisfy both conditions (k is even and a multiple of 4) are:

⇒ k = 0, 4, 8, and 12.

⇒ These correspond to 4 rational terms in the expansion.

Hence, the Correct answer is Option 3.