Differentiation (Part-38) | S N De

Q1. Let \(X = \{1, 2, 3, 4, 5\}\) and \(Y = \{1, 3, 5, 7, 9\}\), \(R_1 = \{(x, y) : y = x + 2, x \in X, y \in Y\}\). Then relation \(R_1\) will be a relation from set X to set Y because —

• (A) \(R_1 \supseteq (X \times Y)\)
• (B) \(R_1 \subseteq (X \times Y)\)
• (C) \(R_1 \not\subseteq (X \times Y)\)
• (D) None

Solution

By definition, any relation \(R\) from a set \(A\) to a set \(B\) is a subset of the Cartesian product \(A \times B\).

Here, the relation \(R_1\) is from set \(X\) to set \(Y\). Therefore, it must satisfy the condition \(R_1 \subseteq (X \times Y)\).

Answer: (B)

Q2. Let \(\mathbb{N}\) be the set of natural numbers and a relation R on \(\mathbb{N}\) is defined as: \(nRm \iff n\) is a factor of \(m\).
Statement (A): R is not an equivalence relation.
Reason (R): R is not a symmetric relation.
Which of the following options correctly explains the given Statement (A) and Reason (R)?

• (A) Both A and R are true and R is the correct explanation of A
• (B) Both A and R are true but R is not the correct explanation of A
• (C) A is true but R is false
• (D) A is false but R is true

Solution

Reflexive: Every natural number is a factor of itself (\(n | n\)). So, it is reflexive.
Transitive: If \(n | m\) and \(m | p\), then \(n | p\). So, it is transitive.
Symmetric: If \(n\) is a factor of \(m\), \(m\) is NOT necessarily a factor of \(n\). (For example, 2 is a factor of 4, but 4 is not a factor of 2). So, it is NOT symmetric.

Since the relation fails the symmetry test, it cannot be an equivalence relation. Both the statement and the reason are true, and the lack of symmetry is the exact reason it is not an equivalence relation.

Answer: (A)

Q3. Let \(\mathbb{N}\) be the set of natural numbers and \(f : \mathbb{N} \to \mathbb{N}\) be a mapping defined by \(f(x) = 7x + 13\); then the mapping f is —

• (A) Injective (One-One)
• (B) Surjective (Onto)
• (C) Bijective
• (D) None

Solution

Injective check: Let \(f(x_1) = f(x_2)\).
\(7x_1 + 13 = 7x_2 + 13 \implies 7x_1 = 7x_2 \implies x_1 = x_2\). Thus, \(f\) is injective.

Surjective check: Let \(y \in \mathbb{N}\) (codomain). Then \(y = 7x + 13 \implies x = \frac{y - 13}{7}\).
For \(y = 1\), \(x = \frac{1 - 13}{7} = -\frac{12}{7} \notin \mathbb{N}\). Since not all elements in the codomain have pre-images in the domain, it is not surjective.

Therefore, the function is only injective.

Answer: (A)

Q4. Let \(f : \mathbb{Q} \to \mathbb{Q}\) be a mapping defined by \(f(x) = 6x^2 - 13x + 10\) where \(\mathbb{Q}\) is the set of Rational numbers. If \(A = \{x : f(x) = 4\}\), then the set A will be:

• (A) \(\{\frac{2}{3}\}\)
• (B) \(\{\frac{3}{2}\}\)
• (C) \(\{\frac{2}{3}, \frac{3}{2}\}\)
• (D) None

Solution

Given \(f(x) = 4\):
\(6x^2 - 13x + 10 = 4\)
\(6x^2 - 13x + 6 = 0\)

Now, factorize the quadratic equation:
\(6x^2 - 9x - 4x + 6 = 0\)
\(3x(2x - 3) - 2(2x - 3) = 0\)
\((3x - 2)(2x - 3) = 0\)

So, \(x = \frac{2}{3}\) or \(x = \frac{3}{2}\). Both values belong to the set of rational numbers \(\mathbb{Q}\).

Thus, \(A = \{\frac{2}{3}, \frac{3}{2}\}\).

Answer: (C)

Q5. Match Column-I with Column-II and choose the correct option:

Column-I	Column-II
(i) Value of \(\sin^{-1}(\sin\frac{5\pi}{6})\)	(a) \(\frac{17\pi}{20}\)
(ii) Value of \(\cos^{-1}(-\sin\frac{7\pi}{6})\)	(b) \(\frac{1}{\sqrt{2}}\)
(iii) Value of \(\cos\{\tan^{-1}(\tan\frac{15\pi}{4})\}\)	(c) \(\frac{\pi}{3}\)
(iv) Value of \(\cos^{-1}\{\frac{1}{\sqrt{2}}(\cos\frac{9\pi}{10} - \sin\frac{9\pi}{10})\}\)	(d) \(\frac{\pi}{6}\)

• (A) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)
• (B) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
• (C) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)
• (D) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a)

Solution

Evaluating (i): \(\sin^{-1}(\sin\frac{5\pi}{6}) = \sin^{-1}(\sin(\pi - \frac{\pi}{6})) = \sin^{-1}(\sin\frac{\pi}{6}) = \frac{\pi}{6}\). So, (i) \(\to\) (d).

Evaluating (ii): \(\cos^{-1}(-\sin\frac{7\pi}{6}) = \cos^{-1}(-\sin(\pi + \frac{\pi}{6})) = \cos^{-1}(-(-\sin\frac{\pi}{6})) = \cos^{-1}(\sin\frac{\pi}{6}) = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}\). So, (ii) \(\to\) (c).

Evaluating (iii): \(\cos\{\tan^{-1}(\tan\frac{15\pi}{4})\} = \cos\{\tan^{-1}(\tan(4\pi - \frac{\pi}{4}))\} = \cos\{\tan^{-1}(-\tan\frac{\pi}{4})\} = \cos(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\). So, (iii) \(\to\) (b).

Evaluating (iv): \(\cos^{-1}\{\cos\frac{\pi}{4}\cos\frac{9\pi}{10} - \sin\frac{\pi}{4}\sin\frac{9\pi}{10}\} = \cos^{-1}\{\cos(\frac{\pi}{4} + \frac{9\pi}{10})\} = \cos^{-1}\{\cos(\frac{23\pi}{20})\}\).
The principal value branch is \([0, \pi]\). \(\cos(\frac{23\pi}{20}) = \cos(2\pi - \frac{17\pi}{20}) = \cos(\frac{17\pi}{20})\).
So, \(\cos^{-1}(\cos\frac{17\pi}{20}) = \frac{17\pi}{20}\). Thus, (iv) \(\to\) (a).

Answer: (A)

Q6. If \(\sin^{-1}x \le \cos^{-1}x\), then —

• (A) \(x \in [0, -1]\)
• (B) \(x \in [-1, \frac{1}{\sqrt{2}}]\)
• (C) \(x \in [0, \frac{\sqrt{3}}{2}]\)
• (D) None

Solution

We know the identity: \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\).

Substitute \(\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x\) into the inequality:
\(\sin^{-1}x \le \frac{\pi}{2} - \sin^{-1}x\)
\(2\sin^{-1}x \le \frac{\pi}{2} \implies \sin^{-1}x \le \frac{\pi}{4}\).

Since the domain of \(\sin^{-1}x\) is \([-1, 1]\), the inequality becomes:
\(-1 \le x \le \sin(\frac{\pi}{4}) \implies -1 \le x \le \frac{1}{\sqrt{2}}\).

Thus, \(x \in [-1, \frac{1}{\sqrt{2}}]\).

Answer: (B)

Q7. If \(\cot^{-1}x + \cot^{-1}y + \cot^{-1}z = \frac{\pi}{2}\), \(x, y, z > 0\) and \(xy < 1\), then the value of \((x + y + z)\) is = ________.

• (A) \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\)
• (B) \(xyz\)
• (C) \(xy + yz + zx\)
• (D) \(2xyz\)

Solution

Convert \(\cot^{-1}\) to \(\tan^{-1}\) (since \(x, y, z > 0\)):
\(\tan^{-1}(\frac{1}{x}) + \tan^{-1}(\frac{1}{y}) + \tan^{-1}(\frac{1}{z}) = \frac{\pi}{2}\).

We know that if \(\tan^{-1}A + \tan^{-1}B + \tan^{-1}C = \frac{\pi}{2}\), then \(AB + BC + CA = 1\).

Substituting \(A = \frac{1}{x}\), \(B = \frac{1}{y}\), and \(C = \frac{1}{z}\):
\(\frac{1}{x} \cdot \frac{1}{y} + \frac{1}{y} \cdot \frac{1}{z} + \frac{1}{z} \cdot \frac{1}{x} = 1\)
\(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 1\).

Multiply the entire equation by \(xyz\):
\(z + x + y = xyz \implies x + y + z = xyz\).

Answer: (B)

Q8. If \(A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}\), then the value of \((A - 2I) \cdot (A - 3I)\) will be —

• (A) \(I\)
• (B) null matrix
• (C) \(A\)
• (D) \(6I\)

Solution

Expand the expression: \((A - 2I)(A - 3I) = A^2 - 3A - 2A + 6I = A^2 - 5A + 6I\).

By the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation, which is given by \(|A - \lambda I| = 0\).

\(\begin{vmatrix} 4 - \lambda & 2 \\ -1 & 1 - \lambda \end{vmatrix} = 0\)
\((4 - \lambda)(1 - \lambda) - (-2) = 0\)
\(\lambda^2 - 5\lambda + 4 + 2 = 0 \implies \lambda^2 - 5\lambda + 6 = 0\).

Substituting matrix \(A\) for \(\lambda\):
\(A^2 - 5A + 6I = O\) (where \(O\) is the null matrix).

Answer: (B)

Q9. If \(\begin{bmatrix} x-y & 2x+z \\ 2x-y & 3z+w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\), then the value of \((x, z)\) will be —

• (A) \((1, 3)\)
• (B) \((1, 2)\)
• (C) \((2, 4)\)
• (D) \((2, 3)\)

Solution

By equating the corresponding elements of the matrices:

1) \(x - y = -1\)
2) \(2x - y = 0 \implies y = 2x\)

Substitute \(y = 2x\) into equation 1:
\(x - 2x = -1 \implies -x = -1 \implies x = 1\).

Now, equate the top-right elements:
\(2x + z = 5\)
Substitute \(x = 1\): \(2(1) + z = 5 \implies z = 3\).

So, \((x, z) = (1, 3)\).

Answer: (A)

Q10. If \(A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix}\), then the value of \(A^{36}\) will be —

• (A) \(\begin{bmatrix} 1 & 0 \\ 0 & 36 \end{bmatrix}\)
• (B) \(\begin{bmatrix} 1 & 0 \\ 6 & 0 \end{bmatrix}\)
• (C) \(\begin{bmatrix} 1 & 0 \\ 18 & 1 \end{bmatrix}\)
• (D) \(\begin{bmatrix} 1 & 0 \\ 36 & 1 \end{bmatrix}\)

Solution

Let's calculate the first few powers of A to find a pattern:

\(A^2 = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} + \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\)

\(A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 + \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{bmatrix}\)

By observation, the \(n\)-th power follows the pattern: \(A^n = \begin{bmatrix} 1 & 0 \\ \frac{n}{2} & 1 \end{bmatrix}\).

For \(n = 36\): \(A^{36} = \begin{bmatrix} 1 & 0 \\ \frac{36}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 18 & 1 \end{bmatrix}\).

Answer: (C)

Q11. If \(A = \begin{bmatrix} 2 & 5 \\ 1 & 6 \end{bmatrix}\), then the value of \(A^{-1}\) will be —

• (A) \(\frac{1}{5}A + \frac{3}{5}I\)
• (B) \(-\frac{1}{6}A - \frac{3}{7}I\)
• (C) \(-\frac{1}{7}A + \frac{8}{7}I\)
• (D) \(\frac{1}{7}A - \frac{8}{7}I\)

Solution

Using the Cayley-Hamilton Theorem, the characteristic equation is:
\(\lambda^2 - \text{Trace}(A)\lambda + |A| = 0\)

\(\text{Trace}(A) = 2 + 6 = 8\)
\(|A| = (2)(6) - (5)(1) = 12 - 5 = 7\)

So, \(A^2 - 8A + 7I = 0\).

Multiply the entire equation by \(A^{-1}\):
\(A - 8I + 7A^{-1} = 0\)
\(7A^{-1} = -A + 8I \implies A^{-1} = -\frac{1}{7}A + \frac{8}{7}I\).

Answer: (C)

Q12. If the matrix \(\begin{bmatrix} -x & x & 2 \\ 2 & x & -x \\ x-2 & x & 0 \end{bmatrix}\) is non-singular, then the value of x is —

• (A) \(-2 \le x \le 2\)
• (B) \(x \neq \pm 2\)
• (C) \(x \ge 2\)
• (D) \(x \le -2\)

Solution

Note: Based on the mathematical context and options, the missing third element in the last row of the image is safely interpreted as 0.

A matrix is non-singular if its determinant is not equal to zero. Let's find the determinant \(\Delta\):

\(\Delta = -x \begin{vmatrix} x & -x \\ x & 0 \end{vmatrix} - x \begin{vmatrix} 2 & -x \\ x-2 & 0 \end{vmatrix} + 2 \begin{vmatrix} 2 & x \\ x-2 & x \end{vmatrix}\)

\(\Delta = -x(0 - (-x^2)) - x(0 - (-x(x-2))) + 2(2x - x(x-2))\)
\(\Delta = -x(x^2) - x(x^2 - 2x) + 2(2x - x^2 + 2x)\)
\(\Delta = -x^3 - x^3 + 2x^2 + 2(4x - x^2)\)
\(\Delta = -2x^3 + 2x^2 + 8x - 2x^2 = -2x^3 + 8x\)

Factoring out \(-2x\):
\(\Delta = -2x(x^2 - 4)\)

For the matrix to be non-singular, \(\Delta \neq 0\):
\(-2x(x^2 - 4) \neq 0 \implies x \neq 0\) and \(x^2 - 4 \neq 0 \implies x \neq \pm 2\).

Looking at the options, the required condition given is \(x \neq \pm 2\).

Answer: (B)

Q13. Statement (I): If A is a skew-symmetric matrix, then \(A^2\) is a symmetric matrix.
Statement (II): \(AA^T\) and \(A^TA\) matrices are both symmetric.

• (A) Statement (I) is true but Statement (II) is false
• (B) Statement (I) is false but Statement (II) is true
• (C) Both Statement (I) and Statement (II) are true
• (D) Both Statement (I) and Statement (II) are false

Solution

Checking Statement (I): If A is skew-symmetric, then \(A^T = -A\).
We need to check the transpose of \(A^2\):
\((A^2)^T = (A \cdot A)^T = A^T \cdot A^T = (-A) \cdot (-A) = A^2\).
Since \((A^2)^T = A^2\), \(A^2\) is symmetric. Statement (I) is True.

Checking Statement (II): We check the transpose of both matrices.
For \(AA^T\): \((AA^T)^T = (A^T)^T A^T = A A^T\). Thus, it is symmetric.
For \(A^TA\): \((A^TA)^T = A^T(A^T)^T = A^TA\). Thus, it is symmetric.
Statement (II) is True.

Answer: (C)