Differentiation (Part-38) | S N De

Q1. Let R be a relation on the set of all real numbers \(\mathbb{R}\) defined by \(aRb\) if and only if \(|a - b| \le 1\). Then the relation R is:

• (A) Only transitive
• (B) Only anti-symmetric
• (C) Only symmetric
• (D) Reflexive and symmetric

Solution

Reflexive: For any \(a \in \mathbb{R}\), \(|a - a| = 0 \le 1\). Thus, \(aRa\) holds. The relation is reflexive.

Symmetric: If \(aRb\), then \(|a - b| \le 1\). This implies \(|-(b - a)| \le 1 \implies |b - a| \le 1\), so \(bRa\) holds. The relation is symmetric.

Transitive: Let \(a = 0, b = 1, c = 2\). We have \(|0 - 1| \le 1\) and \(|1 - 2| \le 1\), but \(|0 - 2| = 2 \not\le 1\). The relation is not transitive.

Therefore, the relation is reflexive and symmetric. Answer: (D)

Q2. If \(f: [0, \infty) \to [0, \infty)\) is defined by \(f(x) = \frac{x}{1+x}\), then f is:

• (A) One-one but not onto
• (B) One-one and onto
• (C) Onto but not one-one
• (D) Neither one-one nor onto

Solution

One-one (Injective): Let \(f(x_1) = f(x_2)\).
\(\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \implies x_1(1+x_2) = x_2(1+x_1) \implies x_1 = x_2\). Thus, it is one-one.

Onto (Surjective): Let \(y = \frac{x}{1+x} \implies y(1+x) = x \implies x = \frac{y}{1-y}\).
For \(x \ge 0\), we need \(0 \le y < 1\). The range is \([0, 1)\), which is not equal to the codomain \([0, \infty)\). Thus, it is not onto.

Answer: (A)

Q3. If \(f(x) = [x]\) represents the greatest integer function and \(g(x) = |x|\) represents the modulus function, then the value of \((g \circ f)\left(-\frac{5}{4}\right)\) is:

• (A) 1
• (B) 2
• (C) 0
• (D) 3

Solution

\( (g \circ f)\left(-\frac{5}{4}\right) = g\left(f\left(-1.25\right)\right) \)

Since \(f\) is the greatest integer function, \(f(-1.25) = [-1.25] = -2\).

Now, \(g(-2) = |-2| = 2\).

Answer: (B)

Q4. If \(\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = 3\pi\), then:

• (A) \(x+y+z-3=0\)
• (B) \(x+y+z+3=0\)
• (C) \(x+2y+3z-5=0\)
• (D) \(x-y-z=0\)

Solution

The principal value branch of \(\cos^{-1}\theta\) is \([0, \pi]\). The maximum possible value of \(\cos^{-1}\theta\) is \(\pi\).

For the sum to be \(3\pi\), each term must be equal to its maximum value:
\(\cos^{-1}x = \pi \implies x = -1\)
\(\cos^{-1}y = \pi \implies y = -1\)
\(\cos^{-1}z = \pi \implies z = -1\)

Substituting these values into the options, \(x + y + z = -1 - 1 - 1 = -3 \implies x + y + z + 3 = 0\).

Answer: (B)

Q5. If \(\sin^{-1}x + \tan^{-1}x = \frac{\pi}{2}\), then what is \(2x^2 + 1\)?

• (A) \(\sqrt{5}\)
• (B) \(\frac{\sqrt{5}-1}{2}\)
• (C) 2
• (D) \(\frac{\sqrt{5}+1}{2}\)

Solution

\(\sin^{-1}x = \frac{\pi}{2} - \tan^{-1}x = \cot^{-1}x\)

Converting \(\cot^{-1}x\) to \(\sin^{-1}\): \(\cot^{-1}x = \sin^{-1}\frac{1}{\sqrt{1+x^2}}\).

Thus, \(\sin^{-1}x = \sin^{-1}\frac{1}{\sqrt{1+x^2}} \implies x = \frac{1}{\sqrt{1+x^2}}\).

Squaring both sides: \(x^2 = \frac{1}{1+x^2} \implies x^4 + x^2 - 1 = 0\).

Using the quadratic formula for \(x^2\): \(x^2 = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}\) (rejecting the negative root).

Now evaluate \(2x^2 + 1 = 2\left(\frac{\sqrt{5}-1}{2}\right) + 1 = \sqrt{5} - 1 + 1 = \sqrt{5}\).

Answer: (A)

Q6. The principal value of \(\sin^{-1}\left(\sin\frac{4\pi}{3}\right) + \cos^{-1}\left(\cos\frac{4\pi}{3}\right)\) is:

• (A) \(\frac{8\pi}{3}\)
• (B) \(\pi\)
• (C) \(\frac{2\pi}{3}\)
• (D) \(\frac{\pi}{3}\)

Solution

Evaluate the first term: \(\sin\left(\frac{4\pi}{3}\right) = \sin\left(\pi + \frac{\pi}{3}\right) = -\sin\frac{\pi}{3}\).
\(\sin^{-1}\left(-\sin\frac{\pi}{3}\right) = -\frac{\pi}{3}\).

Evaluate the second term: \(\cos\left(\frac{4\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\frac{\pi}{3}\).
\(\cos^{-1}\left(-\cos\frac{\pi}{3}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

Sum = \(-\frac{\pi}{3} + \frac{2\pi}{3} = \frac{\pi}{3}\).

Answer: (D)

Q7. If A is a \(3 \times 3\) order matrix and \(|5 \text{adj } A| = 5\), then \(|A| =\)

• (A) \(\pm \frac{1}{25}\)
• (B) \(\pm \frac{1}{5}\)
• (C) \(\pm 1\)
• (D) \(\pm 5\)

Solution

Recall the property: \(|kM| = k^n|M|\) for an \(n \times n\) matrix M. Here \(n=3, k=5\).
\(|5 \text{adj } A| = 5^3 |\text{adj } A| = 125 |\text{adj } A|\).

Also, \(|\text{adj } A| = |A|^{n-1} = |A|^{3-1} = |A|^2\).

So, \(125 |A|^2 = 5 \implies |A|^2 = \frac{5}{125} = \frac{1}{25}\).
Therefore, \(|A| = \pm \frac{1}{5}\).

Answer: (B)

Q8. Let A be a \(3 \times 3\) matrix such that \(A^2 - 5A + 7I = O\).
Statement (I): \(A^{-1} = \frac{1}{7}(5I - A)\)
Statement (II): \(A^3 - 2A^2 - 3A + I = 5(A - 4I)\)

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

Solution

Checking Statement (I):
Given \(A^2 - 5A + 7I = O \implies 7I = 5A - A^2\).
Multiplying by \(A^{-1}\): \(7A^{-1} = 5I - A \implies A^{-1} = \frac{1}{7}(5I - A)\). Statement (I) is True.

Checking Statement (II):
We can rewrite the polynomial \(A^3 - 2A^2 - 3A + I\) using division algorithm by \(A^2 - 5A + 7I\):
\(A^3 - 2A^2 - 3A + I = A(A^2 - 5A + 7I) + 3A^2 - 10A + I\)
Substitute \(A^2 - 5A + 7I = O\):
\(= O + 3(5A - 7I) - 10A + I = 15A - 21I - 10A + I = 5A - 20I = 5(A - 4I)\). Statement (II) is True.

Answer: (C)

Q9. If \(A = [a_{ij}]\) is an n-order skew-symmetric matrix, then:

• (A) \(a_{ij} = \frac{1}{a_{ji}} \forall i, j\)
• (B) \(a_{ij} \neq 0 \forall i, j\)
• (C) \(a_{ij} = 0\), where \(i = j\)
• (D) \(a_{ij} \neq 0\), where \(i = j\)

Solution

By definition, a matrix is skew-symmetric if \(A^T = -A\). This implies \(a_{ji} = -a_{ij}\) for all \(i, j\).

If we set \(i = j\) (the diagonal elements), we get:
\(a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0\).

Thus, the principal diagonal elements are always zero.

Answer: (C)

Q10. The system of linear equations \(\begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & -3 \\ 2 & 6 & a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ b \\ 2 \end{bmatrix}\) has no solution if:

• (A) \(a \neq -5, b = 5\)
• (B) \(a = -5, b \neq 5\)
• (C) \(a = -5, b = 5\)
• (D) \(a \neq -5, b \neq 5\)

Solution

For no solution, the determinant of the coefficient matrix must be 0:
\(\Delta = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 5 & -3 \\ 2 & 6 & a \end{vmatrix} = 1(5a + 18) - (-1)(3a + 6) + 2(18 - 10) = 5a + 18 + 3a + 6 + 16 = 8a + 40\).
Setting \(\Delta = 0 \implies 8a = -40 \implies a = -5\).

Now, let's look at the equations. The system is:
(1) \(x - y + 2z = 3\)
(2) \(3x + 5y - 3z = b\)
(3) \(2x + 6y - 5z = 2\) (assuming \(a = -5\))

Notice that Subtracting equation (1) from equation (2) gives: \((3x-x) + (5y - (-y)) - (3z - 2z) = 2x + 6y - 5z\).
So, \(Eq(2) - Eq(1) \implies 2x + 6y - 5z = b - 3\).
But Equation (3) states \(2x + 6y - 5z = 2\). For the system to be inconsistent (no solution), these parallel planes must not coincide, meaning \(b - 3 \neq 2 \implies b \neq 5\).

Answer: (B)

Q11. If matrix A is both symmetric and skew-symmetric, then matrix A will be:

• (A) a diagonal matrix
• (B) a square matrix
• (C) a zero matrix
• (D) none of these

Solution

Since A is symmetric, \(A^T = A\). Since A is skew-symmetric, \(A^T = -A\).

Combining these two equations gives \(A = -A \implies 2A = O \implies A = O\).

Thus, A must be a zero matrix.

Answer: (C)

Q12. The value of the determinant \(\begin{vmatrix} 1/a & bc & a^3 \\ 1/b & ca & b^3 \\ 1/c & ab & c^3 \end{vmatrix}\) is:

• (A) \((a-b)(b-c)(c-a)\)
• (B) \(\frac{(a-b)(b-c)(c-a)}{abc}\)
• (C) \(a^2b^2c^2(a-b)(b-c)(c-a)\)
• (D) 0

Solution

Multiply \(R_1\) by \(a\), \(R_2\) by \(b\), and \(R_3\) by \(c\). To keep the determinant unchanged, divide by \(abc\):
\(\Delta = \frac{1}{abc} \begin{vmatrix} 1 & abc & a^4 \\ 1 & abc & b^4 \\ 1 & abc & c^4 \end{vmatrix}\)

Taking \(abc\) common from the second column:
\(\Delta = \frac{abc}{abc} \begin{vmatrix} 1 & 1 & a^4 \\ 1 & 1 & b^4 \\ 1 & 1 & c^4 \end{vmatrix}\)

Because column 1 and column 2 are identical, the value of the determinant is 0.

Answer: (D)

Q13. Let \(A = [a_{ij}]\) be a square matrix of order \(3 \times 3\) and \(|A| = -7\). The value of \(a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23}\) (where \(A_{ij}\) is the cofactor of \(a_{ij}\)) is:

• (A) -7
• (B) 7
• (C) 0
• (D) 49

Solution

A fundamental property of determinants states that the sum of the product of elements of any row (or column) with the cofactors of the corresponding elements of a different row (or column) is always zero.

Here, elements of row 1 are multiplied by cofactors of row 2. Hence, the value is 0.

Answer: (C)

Q14. If \(A = \begin{bmatrix} 0 & 2a-1 & \sqrt{a} \\ 1-2a & 0 & 2\sqrt{a} \\ -\sqrt{a} & -2\sqrt{a} & 0 \end{bmatrix}\), where \(a \in \mathbb{R}^+\), then the value of \(|A|\) is:

• (A) \((2a+1)^2\)
• (B) 0
• (C) \((2a+1)^3\)
• (D) \((2a-1)^3\)

Solution

Observe the matrix carefully. Notice that \(a_{ji} = -a_{ij}\) and the diagonal elements are 0.
Specifically, \(1-2a = -(2a-1)\). Thus, A is a skew-symmetric matrix of order \(3 \times 3\).

The determinant of a skew-symmetric matrix of odd order is always 0.

Answer: (B)

Q15. If for a triangle ABC, \(\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix} = 0\), then the value of \(\sin^2 A + \sin^2 B + \sin^2 C\) is:

• (A) \(\frac{4}{9}\)
• (B) \(\frac{3\sqrt{3}}{4}\)
• (C) 0
• (D) \(\frac{9}{4}\)

Solution

Expanding the determinant:
\(1(c^2 - ab) - a(c - b) + b(a - bc) = 0\)
\(c^2 - ab - ac + ab + ab - bc \text{ (wait, expanding carefully)}\)
\(\Delta = 1(c^2 - ab) - a(c - a) + b(b - c) = c^2 - ab - ac + a^2 + b^2 - bc = 0\)

Multiply by 2:
\(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0\)
\((a-b)^2 + (b-c)^2 + (c-a)^2 = 0\)

Since the sum of squares is 0, each term must be 0 \(\implies a = b = c\). The triangle is equilateral. Thus, \(A = B = C = 60^\circ\).

\(\sin^2 60^\circ + \sin^2 60^\circ + \sin^2 60^\circ = 3 \times (\frac{\sqrt{3}}{2})^2 = 3 \times \frac{3}{4} = \frac{9}{4}\).

Answer: (D)

Q16. If \(f(x) = \begin{cases} \frac{\sin 5x}{x^2 + 2x}, & \text{when } x \neq 0 \\ k + \frac{1}{2}, & \text{when } x = 0 \end{cases}\) is continuous at \(x=0\), then the value of k is:

• (A) \(\frac{3}{2}\)
• (B) -2
• (C) 1
• (D) 2

Solution

For the function to be continuous at \(x=0\), \(\lim_{x\to 0} f(x) = f(0)\).

\(\lim_{x\to 0} \frac{\sin 5x}{x^2 + 2x} = \lim_{x\to 0} \frac{\sin 5x}{x(x+2)} = \lim_{x\to 0} \left( \frac{\sin 5x}{5x} \cdot \frac{5}{x+2} \right)\)

Since \(\lim_{x\to 0} \frac{\sin 5x}{5x} = 1\), the limit is \(1 \cdot \frac{5}{0+2} = \frac{5}{2}\).

Equating to \(f(0)\): \(k + \frac{1}{2} = \frac{5}{2} \implies k = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2\).

Answer: (D)

Q17. The points of discontinuity of the function \(f(x) = \frac{2x^2 + 7}{x^3 + 3x^2 - x - 3}\) are:

• (A) \(x=1, x=-1 \text{ and } x=-3\)
• (B) \(x=1 \text{ and } x=-1\)
• (C) \(x=-1 \text{ and } x=3\)
• (D) \(x=1, x=-1 \text{ and } x=3\)

Solution

A rational function is discontinuous where its denominator is zero. Let's find the roots of the denominator.

\(x^3 + 3x^2 - x - 3 = 0\)
\(x^2(x + 3) - 1(x + 3) = 0\)
\((x^2 - 1)(x + 3) = 0 \implies (x - 1)(x + 1)(x + 3) = 0\).

The roots are \(x = 1, x = -1, x = -3\). These are the points of discontinuity.

Answer: (A)

Q18. Let \(f(x) = \begin{cases} x^2 + bx + c, & x < 1 \\ x, & x \ge 1 \end{cases}\). If \(f(x)\) is differentiable at \(x=1\), then \(b-c = ?\)

• (A) 0
• (B) 1
• (C) -2
• (D) 2

Solution

Differentiability implies continuity. Thus, \(\lim_{x\to 1^-} f(x) = f(1)\).
\(1^2 + b(1) + c = 1 \implies 1 + b + c = 1 \implies b + c = 0\).

For differentiability, Left Hand Derivative = Right Hand Derivative at \(x=1\).
\(LHD = \frac{d}{dx}(x^2 + bx + c)|_{x=1} = (2x + b)|_{x=1} = 2 + b\).
\(RHD = \frac{d}{dx}(x)|_{x=1} = 1\).

Equating them: \(2 + b = 1 \implies b = -1\).

Since \(b + c = 0\), we get \(-1 + c = 0 \implies c = 1\).

Now, evaluate \(b - c = -1 - 1 = -2\).

Answer: (C)

Q19. If \(x^m y^n = (x+y)^{m+n}\), then \(\frac{dy}{dx} = ?\)

• (A) 0
• (B) \(\frac{y}{x}\)
• (C) \(\frac{x+y}{xy}\)
• (D) \(xy\)

Solution

Taking the natural logarithm of both sides:
\(\ln(x^m y^n) = \ln((x+y)^{m+n}) \implies m\ln x + n\ln y = (m+n)\ln(x+y)\).

Differentiating with respect to x:
\(\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} \left(1 + \frac{dy}{dx}\right)\).

Grouping \(\frac{dy}{dx}\) terms:
\(\frac{dy}{dx} \left( \frac{n}{y} - \frac{m+n}{x+y} \right) = \frac{m+n}{x+y} - \frac{m}{x}\).

\(\frac{dy}{dx} \left( \frac{nx + ny - my - ny}{y(x+y)} \right) = \frac{mx + nx - mx - my}{x(x+y)}\).
\(\frac{dy}{dx} \left( \frac{nx - my}{y} \right) = \frac{nx - my}{x}\).

Assuming \(nx \neq my\), cancelling gives \(\frac{dy}{dx} = \frac{y}{x}\).

Answer: (B)

Q20. If \(x\sqrt{1+y} + y\sqrt{1+x} = 0\), then the value of \(\frac{dy}{dx}\) is:

• (A) \(\frac{1}{1+x^2}\)
• (B) \(-\frac{1}{1+x^2}\)
• (C) \(\frac{1}{(1+x)^2}\)
• (D) \(-\frac{1}{(1+x)^2}\)

Solution

Rearranging the equation: \(x\sqrt{1+y} = -y\sqrt{1+x}\).

Squaring both sides:
\(x^2(1+y) = y^2(1+x) \implies x^2 + x^2y = y^2 + y^2x\).
\(x^2 - y^2 + x^2y - y^2x = 0 \implies (x-y)(x+y) + xy(x-y) = 0\).

Factoring out \((x-y)\): \((x-y)(x+y+xy) = 0\). Since \(x \neq y\) generally (otherwise trivial), we have \(x + y + xy = 0\).

Isolating y: \(y(1+x) = -x \implies y = -\frac{x}{1+x}\).

Differentiating with quotient rule:
\(\frac{dy}{dx} = - \frac{1 \cdot (1+x) - x \cdot 1}{(1+x)^2} = - \frac{1+x-x}{(1+x)^2} = -\frac{1}{(1+x)^2}\).

Answer: (D)

Q21. If \(y = \tan^{-1}\frac{4x}{1+5x^2} + \tan^{-1}\frac{2+3x}{3-2x}\), then \(\frac{dy}{dx} = ?\)

• (A) \(\frac{1}{1+25x^2}\)
• (B) \(\frac{5}{1+5x^2}\)
• (C) \(\frac{5}{1+25x^2}\)
• (D) \(\frac{1}{1+5x^2}\)

Solution

Let's simplify each term using inverse trigonometric identities.

First term: \(\tan^{-1}\frac{4x}{1+5x^2} = \tan^{-1}\frac{5x - x}{1 + (5x)(x)} = \tan^{-1}(5x) - \tan^{-1}(x)\).

Second term: \(\tan^{-1}\frac{2+3x}{3-2x}\). Divide numerator and denominator by 3:
\(= \tan^{-1}\frac{2/3 + x}{1 - (2/3)x} = \tan^{-1}(2/3) + \tan^{-1}(x)\).

Combining both parts:
\(y = [\tan^{-1}(5x) - \tan^{-1}(x)] + [\tan^{-1}(2/3) + \tan^{-1}(x)] = \tan^{-1}(5x) + \tan^{-1}(2/3)\).

Differentiating with respect to x:
\(\frac{dy}{dx} = \frac{1}{1+(5x)^2} \cdot \frac{d}{dx}(5x) + 0 = \frac{5}{1+25x^2}\).

Answer: (C)