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Calculation:

Convert $(10110110)_2$ to decimal:

$(10110110)_2 = 1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0$

$= 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182$

Convert $(100011)_2$ to decimal:

$(100011)_2 = 1 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$

$= 32 + 0 + 0 + 0 + 2 + 1 = 35$

Add the two decimal numbers

$182 + 35 = 217$

Convert the sum back to binary

To convert $217$ into binary, divide by 2 and note down the remainders:

$217 \div 2 = 108 , \text{remainder} , 1$

$108 \div 2 = 54 , \text{remainder} , 0$

$54 \div 2 = 27 , \text{remainder} , 0$

$27 \div 2 = 13 , \text{remainder} , 1$

$13 \div 2 = 6 , \text{remainder} , 1$

$6 \div 2 = 3 , \text{remainder} , 0$

$3 \div 2 = 1 , \text{remainder} , 1$

$1 \div 2 = 0 , \text{remainder} , 1$

Reading the remainders from bottom to top, $217_{10} = (11011001)_2$

The sum of $(10110110)_2$ and $(100011)_2$ is $(11011001)_2$.

Therefore, the correct option is 1.

Calculation:

Given,

Cardinality of set X: |X| = 3n

Cardinality of set Y: |Y| = 2n

Cardinality of intersection: |X ∩ Y| = n

Calculate cardinality of set difference (X - Y):

|X - Y| = |X| - |X ∩ Y|

⇒ |X - Y| = 3n - n

⇒ |X - Y| = 2n

Calculate cardinality of set difference (Y - X):

|Y - X| = |Y| - |X ∩ Y|

⇒ |Y - X| = 2n - n

⇒ |Y - X| = n

The number of elements in Cartesian product (X - Y) × (Y - X) is:

|(X - Y) × (Y - X)| = |X - Y| · |Y - X|

⇒ |(X - Y) × (Y - X)| = (2n) · (n)

⇒ |(X - Y) × (Y - X)| = 2n²

∴ The number of elements is 2n², which is option (d).

Calculation:

Given sets:

$A = {-3, -2, -1, 0, 1, 2, 3}$

$B = {0, 1, 4, 9}$

We are asked to find the number of elements in the subset of $A \times B$ corresponding to the relation:

$R = {(x, y) : |x| < y}$

We evaluate the condition $|x| < y$ for each $y \in B$:

⇒ $y = 0 \Rightarrow |x| < 0$ No values of x 

⇒ $y = 1 \Rightarrow |x| < 1$ Only x = 0 

⇒ $y = 4 \Rightarrow |x| < 4$  All $x \in A$

⇒ $y = 9 \Rightarrow |x| < 9$ All $x \in A$

Count of valid pairs:

From y = 1: 1 pair → (0, 1)

From y = 4 : 7 pairs → all $x \in A$

From y = 9 : $7 pairs → all ( x \in A)$

Total $= 1 + 7 + 7 = 15$

Hence, the correct answer is Option 3.

Calculation:

We know that for any square matrix $A$ of order $n$:

$|kA| = k^{n}|A|$

Hence, for a matrix $X$ of order $n$:

$|mX| = m^{n}|X|$

So, Statement I is correct.

Let

$X= \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$

and

$Y= \begin{bmatrix} ma_{11} & ma_{12} & ma_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$

Then,

$|Y|= \begin{vmatrix} ma_{11} & ma_{12} & ma_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix}$

Taking $m$ common from the first row:

$|Y|=m \begin{vmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix}$

$|Y| = m|X|$

So, Statement II is also correct and Statement II is the correct explanation of Statement I.

Hence, the correct answer is option 1.

Calculation:

Given the matrix $M = \begin{pmatrix} 71 & 23 & 48 \ 57 & 28 & 29 \ 65 & 17 & 48 \end{pmatrix}$

Determinant of Matrix M

The inverse of a matrix exists if the determinant of the matrix is non-zero. So, we calculate the determinant of matrix M using the formula for the determinant of a 3x3 matrix:

$\text{det}(M) = 71 \times \begin{vmatrix} 28 & 29 \ 17 & 48 \end{vmatrix} - 23 \times \begin{vmatrix} 57 & 29 \ 65 & 48 \end{vmatrix} + 48 \times \begin{vmatrix} 57 & 28 \ 65 & 17 \end{vmatrix}$

After calculating the individual 2x2 determinants:

$\text{det}(M) = 71 \times 851 - 23 \times 851 + 48 \times (-851)$

$\text{det}(M) = 60421 - 19573 - 40848 = 0$

Since the determinant of the matrix $M$ is 0, this means the matrix $M$ is singular, and as a result, the inverse of $M$ does not exist.

Statement-I: "The inverse of M does not exist" is correct.

Statement-II: "M is non-singular" is incorrect because $M$ is singular (det(M) = 0).

Therefore, the correct answer is:

Statement-I is correct but Statement-II is not correct.

Hence, the correct answer is Option 3.

Calculation:

We are asked to evaluate the expression:

$\cot^{-1}(9) + \csc^{-1}\left(\frac{\sqrt{41}}{4}\right)$

Convert cotangent inverse to tangent inverse

⇒ $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$

⇒ $\cot^{-1}(9) = \tan^{-1}\left(\frac{1}{9}\right)$

Convert cosecant inverse to tangent inverse

Let $\theta = \csc^{-1}\left(\frac{\sqrt{41}}{4}\right)$, then:

$\csc(\theta) = \frac{\sqrt{41}}{4}$. We have Hypotenuse= $\sqrt{41}$ and Opposite= 4.

Adjacent side $a = \sqrt{(\sqrt{41})^2 - 4^2} = \sqrt{41 - 16} = \sqrt{25} = 5$.

⇒ $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{4}{5}$

⇒ $\csc^{-1}\left(\frac{\sqrt{41}}{4}\right) = \tan^{-1}\left(\frac{4}{5}\right)$

Add the two inverse tangents

Use the identity (since $\frac{1}{9} \cdot \frac{4}{5} < 1$):

⇒ $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$

Let $a = \frac{1}{9}, b = \frac{4}{5}$

⇒ $\frac{a + b}{1 - ab} = \frac{\frac{1}{9} + \frac{4}{5}}{1 - \frac{1}{9} \cdot \frac{4}{5}}$

$= \frac{\frac{5 + 36}{45}}{1 - \frac{4}{45}} = \frac{\frac{41}{45}}{\frac{45 - 4}{45}} = \frac{\frac{41}{45}}{\frac{41}{45}} = 1$

The expression equals: $\tan^{-1}(1)$

⇒ $\tan^{-1}(1) = \frac{\pi}{4}$

∴ The correct answer is $\frac{\pi}{4}$

Calculation:

We are given the conditions:

$\cot \theta = -\sqrt{3}$

$\csc \theta = -2$

Use identity for cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = -\sqrt{3} \Rightarrow \cos \theta = -\sqrt{3} \sin \theta$

Use identity for cosecant

$\csc \theta = \frac{1}{\sin \theta} = -2 \Rightarrow \sin \theta = -\frac{1}{2}$

Substitute into cotangent identity

$\cos \theta = -\sqrt{3} \cdot \left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$

So we now have:

$\sin \theta = -\frac{1}{2}$

$\cos \theta = \frac{\sqrt{3}}{2}$

Determine angle

These values correspond to the reference angle $\theta = \frac{\pi}{6}$, but we need the quadrant where:

sine is negative

cosine is positive

This occurs in Quadrant IV, so:

$\theta = -\frac{\pi}{6}$

The correct answer is 1 value of $\theta$ in the interval $-\pi < \theta < \pi$

Hence, the correct answer is option 3.

Calculation:

We are given that:

$x + \frac{1}{x} = 2 \cos\theta$

We need to find the value of:

$x^3 + \frac{1}{x^3}$

Cube both sides of the given equation:

⇒ $\left( x + \frac{1}{x} \right)^3 = \left( 2\cos\theta \right)^3$

Expanding the left-hand side:

⇒$\left( x + \frac{1}{x} \right)^3 = x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right)$

Thus:

⇒ $x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) = 8 \cos^3\theta$

Substitute  $x + \frac{1}{x} = 2 \cos\theta$ into the equation:

⇒ $x^3 + \frac{1}{x^3} + 3 \times 2 \cos\theta = 8 \cos^3\theta$

Simplifying:

⇒ $x^3 + \frac{1}{x^3} + 6 \cos\theta = 8 \cos^3\theta$

isolate $x^3 + \frac{1}{x^3}$

⇒ $x^3 + \frac{1}{x^3} = 8 \cos^3\theta - 6 \cos\theta$

Factor out$2 \cos\theta$

⇒ $x^3 + \frac{1}{x^3} = 2 \cos\theta (4 \cos^2\theta - 3)$

⇒ $x^3 + \frac{1}{x^3} = 2 \cos 3\theta$

Hence the correct answer is Option 3.

Calculation:

The problem is to find the number of values of $x$ in the interval $0 \le x \le \frac{\pi}{2}$ that satisfy the equation:

$\tan x + \sec x = 2\cos x$

Rewrite the equation in terms of $\sin x$ and $\cos x$

⇒ $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$

⇒ $\frac{\sin x + 1}{\cos x} = 2\cos x$

⇒ $\sin x + 1 = 2\cos^2 x$

Convert to a quadratic equation in $\sin x$

Using the identity $\cos^2 x = 1 - \sin^2 x$:

⇒ $\sin x + 1 = 2(1 - \sin^2 x)$

⇒ $\sin x + 1 = 2 - 2\sin^2 x$

⇒ $2\sin^2 x + \sin x - 1 = 0$

Solve the quadratic equation for $\sin x$

Factor the equation:

⇒ $(2\sin x - 1)(\sin x + 1) = 0$

Possible solutions for $\sin x$:

⇒ $\sin x = \frac{1}{2}$ or $\sin x = -1$

Find values of $x$ in the interval $[0, \frac{\pi}{2}]$

Case 1: $\sin x = \frac{1}{2}$

In $[0, \frac{\pi}{2}]$, the solution is $x = \frac{\pi}{6}$. (1 value)

Case 2: $\sin x = -1$

The solution is $x = \frac{3\pi}{2}$, which is outside $[0, \frac{\pi}{2}]$. (0 values)

The only valid value is $x = \frac{\pi}{6}$.

∴ The number of values of $x$ satisfying the equation is 1, which is option (b).

Calculation:

We are asked to evaluate the expression:

$\tan\left[\frac{1}{2} \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\right]$

Let $\theta = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Then $\sec \theta = \frac{2}{\sqrt{3}} \Rightarrow \cos \theta = \frac{\sqrt{3}}{2}$

This corresponds to $\theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$

Evaluate the half-angle

$\frac{1}{2} \theta = \frac{\pi}{12}$

So we need to compute $\tan\left(\frac{\pi}{12}\right)$

Use known value

$\tan\left(\frac{\pi}{12}\right) = \tan(15^\circ) = 2 - \sqrt{3}$

Hence, the correct answer is Option 1