NDA-I 2024 (Mathematics) Official Paper (Held On: 21 Apr, 2024). (Previous Year Paper)

Explanation -

Given:

|A| = $\frac{1}{2\sqrt{2}}$  and |B| =$\frac{1}{729}$ 

1. Property of Adjugate:

$\text{adj}(A) = |A|^{n-1} \cdot A^{-1}$

For a 3 x 3 matrix A , $\text{adj}(A) = |A|^2 \cdot A^{-1}$

2. Determine $\text{adj}(3A):$

$\text{adj}(3A) = (3A)^{*} = 3^{n-1} \cdot \text{adj}(A)$

For a 3 x 3 matrix, $\text{adj}(3A) = 3^2 \cdot \text{adj}(A) = 9 \cdot \text{adj}(A)$

3. Determine |adj(A)|:

|adj(A)| = |A|^{n - 1}

For a 3 x 3 matrix, |adj(A)| = |A|²

Given $|A| = \frac{1}{2\sqrt{2}}$,

$|\text{adj}(A)| = \left( \frac{1}{2\sqrt{2}} \right)^2 = \frac{1}{8}$

4. Determine |adj(3A)|:

$|\text{adj}(3A)| = 9^3 \cdot |\text{adj}(A)| = 9^3 \cdot \frac{1}{8} = \frac{9^3}{8}$

5. Combine the Determinants: 

$|2B(\text{adj}(3A))| = 2^3 \cdot |B| \cdot |\text{adj}(3A)| = 8 \cdot \frac{1}{729} \cdot \frac{9^3}{8}$  

$|2B(\text{adj}(3A))| = \frac{8 \cdot 9^3}{729 \cdot 8} = \frac{729}{729} = 1$

So, the value of |2B(adj(3A))| is 1.

Explanation:

z is any complex number and iz3 + z2 - z + i = 0, where $i=\sqrt{-1}$

If z = i then 

iz3 + z2 - z + i = i(i)3 + (i)2 - i + i = 1 - 1 - i + i = 0

So, z = i satisfying the given equation.

Now, |z| = 1

So, (|z| + 1)2 = (1 + 1)² = 2² = 4.

Hence option (2) is true.

Concept -

If number of digit is n then the sum of all n digit numbers formed by using n digits without repetition of digits is given by 

(n-1)! × sum of digits × (11111111.... n times)

Explanation -

Now we have the digits 0, 1, 4, 5

So total digits = n = 4

Now Using the formula -

The sum of all four digit numbers formed by using all digits 0, 1, 4, 5 is

= (4-1)! × (0 + 1 + 4 + 5) × (1111)

= 3! × 10 × 1111

= 66660

But in this sum, there are the numbers which starts from 0 but those are not 4 digit number so we have to subtract the sum of those number from this sum.

Now for digits 1, 4, 5

We have n = 3

So total sum = 2! × 10 × 111 = 20 × 111 = 2220

Now the sum of all four digit numbers = 66660 - 2220 =64440

Hence Option (4) is correct.

Explanation -

The problem of finding the value of xy + yz + zx when x, y and z  are the cube roots of unity, we need to use some properties of the cube roots of unity.

Properties of Cube Roots of Unity:

The cube roots of unity are solutions to the equation x³ = 1

These roots are 1, ω and ω² where ω is a complex number satisfying:

ω = $e^{2\pi i / 3}$

ω³ = 1

1 + ω + ω² = 0

Let x = 1 , y = ω , and z = ω² .

xy + yz + zx = $(1 \cdot ω) + (ω \cdot ω^2) + (ω^2 \cdot 1)$

Simplify each term:

xy = 1 . ω = ω

yz = ω . ω² = ω³ = 1  (since ω³ = 1)

zx = ω² . 1 = ω²

4. Sum the Terms:

xy + yz + zx = ω + 1 + ω²

We know from the properties of the cube roots of unity that:

1 + ω + ω² = 0

Therefore:

xy + yz + zx = 0

So the correct answer is 0.

Explanation -

There are 4 possibilities

(i) 3 ladies from husband's side and 3 gentleman from wife's side .

No. of ways in this case = $4_{C_3} \times 4_{C_3} = 4 \times 4 =1 6$

ii) 3 gentleman from husband's side and 3 ladies from wife's side.

No. of ways in this case = $3_{C_3} \times 3_{C_3} = 1 \times 1 =1$

iii) Two ladies & one gentleman from husband's side and one lady and two gentleman from wife's side.

No. of ways in this case = $(4_{C_2} \times 3_{C_1}) ( 3_{C_1} \times 4_{C_2} ) = (3 \times 6)^2 = 324$

iv) One lady and two gentlemen from husband's side and 2 ladies and one gentleman from wife's side.

No. of ways in this case = $(4_{C_1} \times 3_{C_2}) ( 3_{C_2} \times 4_{C_1} ) = (4 \times 3)^2 =144$

Total no. of ways are = 16 + 1 + 324 + 144 = 485

Explanation -

The number of triangles that can be formed using points on the sides of a triangle PQR, we need to consider the given points and use combinatorial counting methods.

Given:

 3 points on side PQ

 4 points on side QR

 5 points on side RP

Total points on the sides: 3 + 4 + 5 = 12 points.

Combinatorial Approach:

To form a triangle, we need to select any 3 points out of these 12 points such that no three points lie on the same line. 

Total Ways to Select 3 Points Out of 12:

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

We need to subtract the number of ways we can select 3 collinear points (i.e., all points from the same side).

Invalid Selections for Each Side:

For PQ: Select 3 out of 3.

$\binom{3}{3} = 1$

For QR: Select 3 out of 4.

$\binom{4}{3} = 4$

For RP: Select 3 out of 5.

$\binom{5}{3} = 10$

Total Invalid Selections:

1 + 4 + 10 = 15

Total Valid Selections = $\binom{12}{3}$ - Total Invalid Selections

= 220 - 15 = 205

So, the number of triangles that can be formed using the points as vertices is 205.

Explanation -

Given:

$\log_b a = p$ , $\log_d c = 2p$ and  $\log_f e = 3p$  

Convert logarithmic forms to exponential forms:

$\log_b a = p \implies a = b^p$  

⇒$\log_d c = 2p \implies c = d^{2p}$

⇒ $\log_f e = 3p \implies e = f^{3p}$

Substitute the exponential forms into ace :

$ace = (b^p)(d^{2p})(f^{3p})$

$ace = b^p \cdot d^{2p} \cdot f^{3p}$

$(ace)^{\frac{1}{p}} = (b^p \cdot d^{2p} \cdot f^{3p})^{\frac{1}{p}}$

⇒ $(ace)^{\frac{1}{p}} = (b^p)^{\frac{1}{p}} \cdot (d^{2p})^{\frac{1}{p}} \cdot (f^{3p})^{\frac{1}{p}}$

= $b^{\frac{p}{p}} \cdot d^{\frac{2p}{p}} \cdot f^{\frac{3p}{p}}$

=$b^1 \cdot d^2 \cdot f^3$

= $bd^2 f^3$

So, the value of $(ace)^{\frac{1}{p}}$ is $bd^2 f^3$.

Concept -

The given roots are -√2 and √3, and we know that if a polynomial has real coefficients and a non-real root, its conjugate must also be a root.

Explanation -

Given Roots:

Roots: -√2, √3

Conjugate Roots (since coefficients are real):

Conjugate roots:  √2, -√3

The polynomial with these roots can be expressed as:

(x + √2)(x - √2)(x + √3)(x - √3)

⇒ (x² - 2)(x² - 3)

= $x^4 - 3x^2 - 2x^2 + 6 = x^4 - 5x^2 + 6$

The given polynomial is:

$a_0 + a_1 x + a_2 x^2 + a_3 x^3 + x^4 = 0$

By comparing coefficients:

The coefficient of x⁴ is 1.

The coefficient of x³ is 0.

The coefficient of x² is -5.

The coefficient of x is 0.

The constant term is 6.

Therefore:

$a_0 = 6, \quad a_1 = 0, \quad a_2 = -5, \quad a_3 = 0$

Hence Option(3) is correct.

Explanation:

Let z1 = 1 and z2 = i then

$\rm \left|\frac{z_1+z_2}{z_1-z_2}\right|$ = $\rm \left|\frac{1+i}{1-i}\right|$ = ${\sqrt 2\over \sqrt2}$ = 1

So, $\rm \left|\frac{z_1+z_2}{z_1-z_2}\right|=1$ satisfying.

Now, $\operatorname{Re}\left(\frac{z_1}{z_2}\right)+1$ = $\operatorname{Re}\left(\frac{1}{i}\right)+1$ = Re(-i) + 1 = 0 + 1 = 1

Hence option (3) is true.

Explanation:

$[{26\over2}]+[{26\over2^2}]+[{26\over2^3}]+[{26\over2^4}]$

= 13 + 6 + 3 + 1 ([.] is greatest integer function)

= 23

then

26! = a2²³

⇒ 26! = a(2³)⁷× 2²

⇒ 26! = 4a × 8⁷

Comparing with 26! = n8k we get maximum value of k is 7.

Hence option (2) is true.

Concept:

(i) Matrix multiplication is not communicative i.e., AB ≠ BA for two square non-singular matrix A and B.

(ii) For any non-singular matrix A of order n, A(adj A) = |A|× I_n.

(iii) adj(AB) = (adj B)(adj A)

Explanation:

we know that, 

adj(AB) = (adj B)(adj A)

and adj(BA) = (adj A)(adj B)

So, 1. and 2. are false.

A and B are non-singular matrices so, AB is non-singular.

So, using the formula A(adj A) = |A|× In we get

(AB) adj (AB) = |AB|In 

So, (AB) adj (AB) - |AB|In = 0 

i.e., (AB) adj (AB) - |AB|In is a null matrix of order n.

3. is true.

Hence option (2) is correct.

Concept:

Let A be a square matrix then (adj A)^T = adj(A^T)

Explanation:

1. A(adj AT) = A(adj A)T (as (adj A)T = adj(AT))

1. is true.

2. A2 = A

⇒ A2 - A = 0

⇒ A(A - I) = 0

⇒ A = 0 or A = I

Now, given A is non-singular matrix so A ≠ 0, A = I i.e., A is identity matrix of order n.

2. is true.

3. A3 = A

⇒ A3 - A = 0

⇒ A(A² - I) = 0

⇒ A(A + I)(A - I) = 0

⇒ A = 0, A = I, A = - I

Since A is non-singular matrix so A ≠ 0, 

So, its not necessary that A is identity matrix of order n.

3. is false.

Hence option (1) is true.

Concept -

1. The first digit (thousands place) cannot be 0 (since it wouldn't be a four-digit number).
2. The remaining three digits (hundreds, tens, and units place) can be any of the even digits.

Explanation -

1. First digit (thousands place):

   - Possible digits: 2, 4, 6, 8 (4 choices) 2. Remaining three digits (hundreds, tens, and units places):

   - Possible digits: 0, 2, 4, 6, 8 (5 choices for each)

Total number of four-digit natural numbers:

4 × 5 × 5 × 5 = 4 × 125 = 500

Thus, the total number of four-digit natural numbers such that all of the digits are even is 500.

Explanation -

The given equation $(z - 100)^3 + 1000 = 0$,

⇒ $(z - 100)^3 = -1000$

Next, we recognize that -1000 can be written as -10³.

Thus, we have:

$(z - 100)^3 = -(10)^3$

Taking the cube root of both sides, we get:

$z - 100 = ω \cdot (-10)$

where ω is a cube root of unity (ω³ = 1) and ω ≠ 1. The cube roots of unity are 1, ω, and ω², where ω and ω² are complex numbers and ω² = ω^-1.

So, the possible values for z - 100 are:

z - 100 = -10, -10ω, -10ω²

Solving for z, we get:

z = 100 - 10

z = 100 - 10ω

z = 100 - 10ω²

Therefore, the solutions are:

z = 90, 100 - 10ω, 100 - 10ω²

⇒ z = 90, 10(10 - ω), 10(10 - ω2)

Hence Option (2) is correct.

Concept -

(1 + x)2 = 1 + x² + 2x and i² = -1 

Explanation -

(1 + i)2 = 1 + i² + 2i = 1 - 1 + 2i = 2i

Now (1 + i)4 = 2i x 2i = 4 i² = - 4

and (1 - i)4 = -4

Put these values in given equation we get, 

(1 + i)4 + (1 - i)4  = - 4 - 4 = - 8

Hence Option (4) is correct.

Concept:

(i) A square matrix A is said to be skew-symmetric matrix if A^T = -A.

(ii) A matrix A is said to be orthogonal matrix if AA^T = I i.e., A^T = A^-1, where I is the identity matrix.

Explanation:

A is a skew-symmetric matrix A of order 3.

So, AT = - A

From the property of skew-symmetric matrix we know that,

All the elements of main diagonal elements of skew symmetric matrix are zero.

So, sum of the elements of all the diagonal elements of the matrix is zero. 

Therefore, 1. and 2. are true.

A can't be orthogonal matrix as it is possible only when AT = A-1.

So, 3. is false.

Hence option (1) is correct.

Concept Used:

A number is divisible by 4 if the last two digits form a number that is divisible by 4.

Solution:

The digits available are 1, 2, 3, and 5. Let's check the possible pairs:

• 12 - Divisible by 4
• 21 - Not divisible by 4
• 13 - Not divisible by 4
• 31 - Not divisible by 4
• 15 - Not divisible by 4
• 51 - Not divisible by 4
• 23 - Not divisible by 4
• 32 - Divisible by 4
• 25 - Not divisible by 4
• 52 - Divisible by 4
• 35 - Not divisible by 4
• 53 - Not divisible by 4

So, the valid pairs for the last two digits are 12, 32 and 52.

Now, let's calculate the number of four-digit numbers that can be formed with these pairs:

1. For the pair 32:

• We have two remaining digits: 1 and 5.

• The number of permutations of these two digits in the first two places is 2!=22!=22!=2

  .

• So, there are 2 numbers: 1532, 5132.

2. For the pair 52:

• We have two remaining digits: 1 and 3.

• The number of permutations of these two digits in the first two places is 2!=22!=22!=2

  .

• So, there are 2 numbers: 1352, 3152.

3. For the pair 12:

• We have two remaining digits: 3 and 5.
• The number of permutations of these two digits in the first two places is 2!=22!=22!=2

.

• So, there are 2 numbers: 3512, 5312.

Therefore, the total number of four-digit numbers that can be formed and are divisible by 4 is 2 + 2 + 2 = 6

Hence Option (4) is correct.

Concept -

Fermat's Little Theorem states that if p is a prime number and a is an integer such that a is not divisible by p, then:

$a^{p-1} \equiv 1 \pmod{p}$

Explanation -

For our problem: a = 2 and p = 7

According to Fermat's Little Theorem:

$2^{7-1} = 2^6 \equiv 1 \pmod{7}$

This implies that:

$2^6 \equiv 1 \pmod{7}$

We need to find the remainder when 2¹²⁰ is divided by 7.

Notice that 120 can be expressed as a multiple of 6:

120 = 20 ×  6

Using this, we can write:

$2^{120} = (2^6)^{20}$

Since $2^6 \equiv 1 \pmod{7}$:

$(2^6)^{20} \equiv 1^{20} \pmod{7}$

And since 1²⁰ = 1:

$2^{120} \equiv 1 \pmod{7}$

Thus, the remainder when 2¹²⁰ is divided by 7 is 1.

Concept:

Combination formula:

C(n, r) + C(n, r - 1) = C(n+1, r)

Explanation:

$\left|\begin{array}{ccc} C(9,4) & C(9,3) & C(10, n-2) \ C(11,6) & C(11,5) & C(12, n) \ C(m, 7) & C(m, 6) & C(m+1, n+1) \end{array}\right|$ = 0

Using C₁ = C₁ + C₂ and using the formula

$\left|\begin{array}{ccc} C(10,4) & C(9,3) & C(10, n-2) \ C(12,6) & C(11,5) & C(12, n) \ C(m+1, 7) & C(m, 6) & C(m+1, n+1) \end{array}\right|$ = 0

This determinant will be zero if 1^st and 3^rd column are same,

which is possible only when

n - 2 = 4 i.e., n = 6

Option (3) is correct

Explanation:

$\left|\begin{array}{ccc} \cos C & \sin B & 0 \ \tan A & 0 & \sin B \ 0 & \tan (B+C) & \cos C \end{array}\right|$

Let A = 30^∘, B = 90∘, C = 60∘ then the determinant is

$\left|\begin{array}{ccc} \frac12 &1 & 0 \ \frac1{\sqrt3} & 0 & 1 \ 0 & -\frac1{\sqrt3} &\frac12 \end{array}\right|$

= $\frac12(0+\frac1{\sqrt3})$ + 1$(0-\frac1{2\sqrt3})$

= 0 

Hence option (2) is true.

Explanation:

Types of matrices with 4 entries are 2 × 2, 4 × 1 and 1 × 4.

Number of 2 × 2 matrices using 1, 2, 3, 4 is 4⁴ = 256

Similarly, number of 4 × 1 matrices is 256 and number of 1 × 4 is 256.

So, total number of different matrices is 256 + 256 + 256 = 768.

Hence option (4) is true.

Concept:

One-one function: A function is said to be one-one function if it maps each element of the domain to a unique element of the range. 

On-to function: A function is said to be on-to function if every element in range has a pre-image in the domain.

Bijective function: A function is said to be bijective function if it is both one-one and on-to,

Explanation:

A = {x ∈ R: -1 < x < 1}.

1. f(x) = x|x|

From the graph we can see that each element from (-1,1) maps to a unique element. So it is one-one function.

Also, each element of range has a pre-image so it is on-to.

Hence f(x) = x|x| is bijective function.

1. is true.

2. g(x) = cos(πx)

$-\frac12, \frac12$ ∈ A are two different element 

But g(-$\frac12$) = cos(-π/2) = 0 and g($\frac12$) = cos(π/2) = 0

So, they maps to same element of the range.

Therefore, g(x) is not one-one and so not bijective.

2. is false.

Hence option (1) is correct.

Concept:

A relation R in a set A is said to be 

(i) reflexive if (x, x) ∈ R for every x ∈ A

(ii) Symmetric if (x, y) ∈ R implies (y, x) ∈ R for x, y ∈ A.

(iii) transitive if (x, y) ∈ R, (y, z) ∈ R implies (x, z) ∈ R for x, y, z ∈ A.

A relation R is said to be equivalence relation if it is reflexive, symmetric and transitive.

Explanation:

R is a relation on the open interval (-1, 1) and is given by

R = {(x, y) : |x + y| < 2}. 

(i) - 1 < x < 1

⇒ -1 - 1 < x + x < 1 + 1

⇒ -2 < x + x < 2

⇒ |x + x| < 2

So, (x, x) ∈ R.

R is reflexive.

(ii) Let (x, y) ∈ R then |x + y| < 2 ⇒ |y + x| < 2

So, (y, x) ∈ R

R is symmetric.

(iii) Let (x, y) ∈ R, (y, z) ∈ R

then |x + y| < 2 and |y + z| < 2 for x, y, z ∈ (-1, 1)

⇒ |x + z| < 2

So, (x, z) ∈ R.

R is transitive.

Therefore R is equivalence relation.

Hence option (4) is true.

Explanation:

Venn diagram of {(A - B) ∪ (B - A) ∪ (A ∩ B)} is

Hence we can see that {(A - B) ∪ (B - A) ∪ (A ∩ B)} = A ∪ B

So, (A ∪ B) - {(A - B) ∪ (B - A) ∪ (A ∩ B)} = (A ∪ B) - (A ∪ B) = ϕ 

Hence (A ∪ B) - {(A - B) ∪ (B - A) ∪ (A ∩ B)} is equal to null set.

Option (1) is true.

Concept:

Cosine rule: Let ABC be a triangle with angles A, B, C are corresponding sides a, b, c then 

Cos A = ${b^2+c^2-a^2 \over 2bc}$

Explanation:

$\left|\begin{array}{ccc} a^2 & b \sin A & c \sin A \ b \sin A & 1 & \cos A \ c \sin A & \cos A & 1 \end{array}\right|$

= a²(1 - cos²A) + b sin A(c sin A cos A - b sin A) + c sin A(b sin A cos A - c sin A)

= a2(1 - cos2A) + bc sin² A cos A - b² sin² A + bc sin² A cos A - c² sin² A

= a2sin2A + 2bc sin2 A cos A - (c² + b2)sin2 A

= - sin²A(b² + c² - a² - 2bc cos A)

= - sin2A(2bc cos A - 2bc cos A)

= 0 

Hence option (1) is true.

Concept:

Arithmetic progression (AP): Three number a, b, c are called in AP if 2b = a + c.

Geometric progression (GP): Three numbers a, b, c are called in GP if b² = ac.

Harmonic progression (HP): Three numbers a, b, c are called in HP if ${2\over b}=\frac1a+\frac1c$.

Explanation:

If we take a = 2, b = 4, c = 6, d = 9, e = 18 then 

a, b, c are in AP

b, c, d are in GP and

c, d, e are in HP

Now, c² = 6² = 36 = 2 × 18 = ae

So, a, c, e are in GP

1. is true.

${1\over c^2}=\frac1{36}$

and $\frac1a.\frac1e=\frac{1}{36}$ 

So, $\frac{1}{a}, \frac{1}{c}, \frac{1}{e}$ are in GP

2. is true.

Hence option (3) is the correct answer.

Concept:

(i) log_ax exits only if x > 0

(ii) $\log_{b^n}a=\frac1n\log_ba$

(iii) log_axⁿ = n log_ax 

Explanation:

log4(x - 1) = log2(x - 3)

⇒ log2²(x - 1) = log2(x - 3)

⇒ $\frac12$log₂(x - 1) = log₂(x - 3)

⇒ log2(x - 1)^1/2 = log2(x - 3)

⇒ (x - 1)1/2 = (x - 3)

Squaring both sides

⇒ x - 1 = x² - 6x + 9

⇒ x2 - 7x + 10 = 0

⇒ (x - 5)(x - 2) = 0

⇒ x = 2, 5

If we take x = 2 then the right hand side will be log₂(-1) which is not possible.

So x can't be 2

Hence x = 5 is the only solution i.e., number of solution is one.

Hence option (2) is true.

Concept:

(i) logax exits only if x > 0

(ii) $\log_{b^n}a=\frac1n\log_ba$

(iii) logaxn = n logax 

(iv) log_a x + log_a y = log_a xy

(v) loga x - loga y = loga (x/y)

(vi) log_ax = $\log x\over \log a$

Explanation:

$\log _x\left(\frac{x}{y}\right)+\log _y\left(\frac{y}{x}\right)$ = k

⇒ ${\log\left(\frac{x}{y}\right)\over \log x}+{\log\left(\frac{y}{x}\right)\over \log y}$ = k

⇒ ${\log x-\log y\over \log x}+{\log y-\log x\over \log y}$ = k

⇒ ${\log x.\log y-(\log y)^2+\log x\log y-(\log x)^2\over \log x\log y}$ = k

⇒ ${-{(\log x)^2-2\log x\log y+(\log y)^2}\over \log x\log y}$ = k

⇒ ${-(\log x-\log y)^2\over \log x\log y}$ = k

Now, since $x ≥slant y>1$ so

log x > 0, log y > 0

also, (log x - log y)² ≥ 0 for any x, y

Hence k can be zero or negative i.e., k can never be positive

Hence option (4) is true.

Concept:

(i) A square matrix A is called symmetric matrix if A = AT and is called skew symmetric if A = - A^T.

(ii) A square matrix A is called orthogonal if AAT = I.

Explanation:

A = $\left[\begin{array}{ccc}\sin 2 θ & 2 \sin ^2 θ-1 & 0 \ \cos 2 θ & 2 \sin θ \cos θ & 0 \ 0 & 0 & 1\end{array}\right]$

   = $\left[\begin{array}{ccc}\sin 2 θ & -\cos 2θ & 0 \ \cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]$ 

(as 2sin x cos x = sin2x and 2sin²x = 1- cos 2x)

A^T = $\left[\begin{array}{ccc}\sin 2 θ & \cos 2θ & 0 \ -\cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]$

i.e., A ≠  A^T

So, A is not symmetric matrix. 

2. is false.

Now, |A| = $\begin{vmatrix}\sin 2 θ & -\cos 2θ & 0 \ \cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{vmatrix}$ = 1(sin²θ + cos²θ) = 1

and adj(A) = $\left[\begin{array}{ccc}\sin 2 θ & \cos 2θ & 0 \ -\cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]$

So, A^-1 = ${1\over |A|}adj(A)$ = $\left[\begin{array}{ccc}\sin 2 θ & \cos 2θ & 0 \ -\cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]$

Hence, A-1 = AT and A-1 = adj(A) 

1. is true and 3. is true.

Hence option (3) is correct.

Explanation:

$\left(2-x^2-\frac{1}{x^2}\right)^{-5}$

= - $\left(x^2+\frac{1}{x^2}-2.x.\frac1x\right)^{-5}$

= - $\left[(x-\frac1x)^2\right]^{-5}$

= - $\left(x-\frac1x\right)^{-10}$

= - $\left(\frac{x^2-1}x\right)^{-10}$

= - $\left(\frac{x}{x^2-1}\right)^{10}$ = - $\frac{x^{10}}{(x^2-1)^{10}}$

So, $\left(1-x^2\right)^{20}\left(2-x^2-\frac{1}{x^2}\right)^{-5}$

= - (x² - 1)²⁰$\frac{x^{10}}{(x^2-1)^{10}}$

= - x¹⁰(x2 - 1)10

Hence the coefficient of x10 is

= - (constant term of the expression (x2 - 1)10)

= - 1

Option (1) is true.

Concept:

(i) The general term of the expression (a + b)ⁿ is

T_r+1 = $^nC_ra^{n-r}b^{r}$

(ii) $^nC_r={n!\over r!(n-r)!}$

Explanation:

4^th term in the expansion of $\left(m x+\frac{1}{x}\right)^n$ is 

T3+1 = $^nC_3(mx)^{n-3}.(\frac1x)^{3}$ = $^nC_3m^{n-3}.x^{n-6}$

By given condition, 4^th term is $\frac{5}{2}$ which is independent of x. 

So, n - 6 = 0 i.e., n = 6

Therefore

$^6C_3m^{3}$ = $\frac{5}{2}$

⇒ 20m³ = 5/2

⇒ m3 = 1/8

⇒ m = 1/2

Therefore mn = 6 × $\frac12$ = 3

Hence option (2) is true.

Explanation:

a, b, c are in GP so b² = ac...(i)

Also, a > 0 and c > 0 so, ac > 0

Now, roots of  ax2 + bx + c = 0 are

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$ = ${-b \pm \sqrt{ac-4ac} \over 2a}$ = ${-b \pm \sqrt{-3ac} \over 2a}$

Since ac > 0 so -3ac < 0

So roots are imaginary.

1. is true.

ratio of the roots

= ${-b + \sqrt{-3ac} \over 2a}\over {-b - \sqrt{-3ac} \over 2a}$ 

= ${-\sqrt{ac} + \sqrt{-3ac} \over -\sqrt{ac}-\sqrt{-3ac}}$ (using (i))

= ${-1 + \sqrt{3}i \over -1-\sqrt{3}i}$ 

= $2ω\over 2ω^2$ = $1\over \omega$

(as ω is a cube root of unity i.e., ω = $-1+\sqrt3i\over 2$ and ω² = $-1-\sqrt3i\over 2$)

2. is true.

The product of the roots

= ${-b + \sqrt{-3ac} \over 2a}$.${-b- \sqrt{-3ac} \over 2a}$ 

= ${b^2 -(-3ac) \over 4a^2}$ 

= ${b^2 -(-3ac) \over 4a^2}$ 

= ${b^2 +3b^2 \over 4a^2}$ (using (i))

= ${4b^2 \over 4a^2}$ = ${b^2 \over a^2}$

3. is true.

Hence option (4) is correct option

Explanation:

x2 + mx + n

= x2 + 2.$\frac m2$.x + $m^2\over4$+ n - $m^2\over4$

= (x + $\frac m2$)² + n - $m^2\over4$

So, x2 + mx + n will be an integer if

x + $\frac m2$ and n - $m^2\over4$ will be an integer for all integral value of x

Now, x + $\frac m2$ is integer implies $\frac m2$ is integer (say k)

Them m = 2k i.e., m is an even integer.

1. is true 

n - $m^2\over4$ is integer implies n - $4k^2\over 4$ is integer i.e., n - k² is integer so n is integer.

2. is true.

Hence option (3) is correct.

Concept: 

The rth term of the expansion (a + b)ⁿ is ⁿC_ra^n-rb^r.

Explanation:

 (x + y)2n + 1(x - y)2n + 1

=  {(x + y)(x - y)}2n + 1

= (x² - y²)2n + 1

if we take n = 1 then expression is (x2 - y2)3

Number of terms = 4 which is even

So, the expansion has 2 middle terms.

2^nd term and 3^rd term

2^nd term is - 3C1(x²)2y2 = - 3x⁴y²

3^rd term is 3C2x2y4 = 3x2y4

By give condition,

• 3x4y2 + 3x2y4 = 0

⇒ x4y2 = x2y4

⇒ x² = y² ⇒ $\left(\frac{x^2}{y^2}\right)$ = 1

Option (1) is true.

Concept:

Number of onto function from a set A (containing m elements) to a set B (containing n elements) (m ≥ n) is given by

n^m - ⁿC₁(n - 1)^m + nC2(n - 2)m - ...

Explanation:

A = {1, 2, 3, 4, 5}, B = {6, 7}

n(A) = m = 5 and n(B) = n = 2

So number of onto functions from A to B is

2⁵ - 2C1(2 - 1)5 

= 32 - 2 = 30

Option (3) is true.

Concept:

Formula: sin a cos b - cos a sin b = sin(a - b)

sin2x = 2 sin x cos x

Explanation:

$\frac{\sqrt{3} \cos 10^{\circ}-\sin 10^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}$

= $\frac{{\sqrt{3}\over2} \cos 10^{\circ}-\frac12\sin 10^{\circ}}{\frac12\sin 25^{\circ} \cos 25^{\circ}}$

= $\frac{\sin60^{\circ} \cos 10^{\circ}-\cos60^{\circ}\sin 10^{\circ}}{\frac12\sin 25^{\circ} \cos 25^{\circ}}$

= $\frac{2\sin(60^{\circ}-10^{\circ})}{\sin 25^{\circ} \cos 25^{\circ}}$ (using the formula)

= $\frac{2\sin50^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}$

= $\frac{2\sin(2\times25^{\circ})}{\sin 25^{\circ} \cos 25^{\circ}}$

= $\frac{2\times 2\sin 25^{\circ} \cos 25^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}$

= 4 

Option (4) is true.

Concept:

cos 36° = $\sqrt5+1\over 4$

Explanation:

(sin 9° - cos 9°)

= $\sqrt2$($\frac1{\sqrt2}$sin 9° - $\frac1{\sqrt2}$cos 9°)

= $\sqrt2$(sin 9° cos 45° - cos 9° sin 45°)

= $\sqrt2$sin(-36°) (sin a cos b - cos a sin b = sin(a - b))

= - $\sqrt2$ sin(36°)

= - $\sqrt2$ $\sqrt{1-\cos^236^{\circ}}$

= - $\sqrt2$(\sqrt{1-({\sqrt5+1\over 4})^2}\)

= - $\sqrt2$(\sqrt{{16-5-2\sqrt5+1\over 16}}\)

= - $\sqrt2$(\sqrt{{10-2\sqrt5\over 16}}\)

= - $\sqrt{{20-4\sqrt5\over 16}}$

= $-\frac{\sqrt{5-\sqrt{5}}}{2}$

Hence option (1) is true.

Concept:

(i) sine rule: Let ABC be a triangle with angles A, B, C and corresponding sides a, b, c then 

${\sin A\over a}={\sin B\over b}={\sin C\over c}$