Differentiation (Part-38) | S N De

Q14. Match Column-I with Column-II and choose the correct option:

Column-I	Column-II
(i) \(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = \)	(a) \(xyz + xy + yz + zx\)
(ii) \(\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{vmatrix} = \)	(b) \(0\)
(iii) \(\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = \)	(c) \((x-y)(y-z)(z-x)(x+y+z)\)
(iv) \(\begin{vmatrix} 1 & x & x^2-yz \\ 1 & y & y^2-zx \\ 1 & z & z^2-xy \end{vmatrix} = \)	(d) \((x-y)(y-z)(z-x)\)

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

Solution

Evaluating (i): This is a standard Vandermonde determinant. Its value is \((x-y)(y-z)(z-x)\). So, (i) \(\to\) (d).

Evaluating (ii): This is another standard determinant identity. Expanding it yields \((x-y)(y-z)(z-x)(x+y+z)\). So, (ii) \(\to\) (c).

Evaluating (iii): Taking \((1+x), (1+y), (1+z)\) common or applying standard row/column operations simplifies it to \(xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = xyz + xy + yz + zx\). So, (iii) \(\to\) (a).

Evaluating (iv): We can split this determinant into two parts:
\(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} - \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}\).
The second determinant, after multiplying rows by \(x, y, z\) respectively and taking \(xyz\) common from the last column, becomes identical to the first. Thus, their difference is \(0\). So, (iv) \(\to\) (b).

Answer: (A)

Q15. Let \(A = \begin{vmatrix} x & x^2 & 1+px^3 \\ y & y^2 & 1+py^3 \\ z & z^2 & 1+pz^3 \end{vmatrix}\).
Statement (I): When \(P=0\), \(|A| = (x-y)(y-z)(z-x)\)
Statement (II): When \(P=1\), \(|A| = xyz(x-y)(y-z)(z-x)\)

• (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

By property of determinants, we can split \(|A|\) into two determinants:
\(|A| = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & px^3 \\ y & y^2 & py^3 \\ z & z^2 & pz^3 \end{vmatrix}\)

The first determinant requires two column swaps to become the standard Vandermonde determinant \(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}\), which equals \((x-y)(y-z)(z-x)\). Since there are two swaps, the sign remains positive.

For the second determinant, take \(p\), \(x\), \(y\), and \(z\) common from the respective columns/rows, which yields \(pxyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}\).
Thus, \(|A| = (1 + pxyz)(x-y)(y-z)(z-x)\).

Checking Statement (I): If \(p=0\), \(|A| = (1 + 0)(x-y)(y-z)(z-x) = (x-y)(y-z)(z-x)\). Statement (I) is True.

Checking Statement (II): If \(p=1\), \(|A| = (1 + xyz)(x-y)(y-z)(z-x)\). Statement (II) says it is \(xyz(x-y)\dots\), omitting the "+1". Thus, Statement (II) is False.

Answer: (A)

Q16. The value of the determinant \(\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{vmatrix}\) will be —

• (A) \(3\omega^2\)
• (B) \(3\omega(\omega - 1)\)
• (C) \(3\omega(1 - \omega)\)
• (D) \(3\omega\)

Solution

We know the properties of the cube roots of unity: \(1 + \omega + \omega^2 = 0\) and \(\omega^3 = 1\).

Substitute \(-1-\omega^2 = \omega\) and \(\omega^4 = \omega\cdot\omega^3 = \omega\) into the determinant:
\(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}\)

Expanding along the first row:
\(\Delta = 1(\omega^2 - \omega^4) - 1(\omega - \omega^2) + 1(\omega^2 - \omega)\)

Since \(\omega^4 = \omega\):
\(\Delta = (\omega^2 - \omega) - (\omega - \omega^2) + (\omega^2 - \omega)\)
\(\Delta = \omega^2 - \omega - \omega + \omega^2 + \omega^2 - \omega\)
\(\Delta = 3\omega^2 - 3\omega = 3\omega(\omega - 1)\).

Answer: (B)

Q17. Let \(f(x) = \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{vmatrix}\).
Statement (A): If \(\sin 2x = 1\), then \(f(x) = 0\)
Reason (R): If \(\sin x = \cos x\), then \(f(x) = 0\)
Which of the following correctly explains 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

First, evaluate the determinant \(f(x)\):
\(f(x) = 0(0 - \cos x\sin x) - \cos x(0 - \cos^2 x) + (-\sin x)(\sin^2 x - 0)\)
\(f(x) = \cos^3 x - \sin^3 x\).

Checking Reason (R): If \(\sin x = \cos x\), then \(\cos^3 x - \sin^3 x = \cos^3 x - \cos^3 x = 0\). Thus, \(f(x) = 0\). Reason (R) is True.

Checking Statement (A): If \(\sin 2x = 1\), then \(2\sin x\cos x = 1\).
This implies \(\sin^2 x + \cos^2 x - 2\sin x\cos x = 1 - 1 = 0\).
\((\cos x - \sin x)^2 = 0 \implies \cos x = \sin x\).
Since \(\cos x = \sin x\), we know from Reason (R) that \(f(x) = 0\). Statement (A) is True.

Because \(\sin 2x = 1\) directly leads to \(\sin x = \cos x\), Reason (R) correctly explains Statement (A).

Answer: (A)

Q18. If \(f(x) = \begin{cases} \frac{1 - \cos 2x}{2x^2}, & \text{when } x \neq 0 \\ \lambda, & \text{when } x = 0 \end{cases}\) is continuous at \(x = 0\), then the value of \(\lambda\) is:

• (A) \(-1\)
• (B) \(-2\)
• (C) \(+2\)
• (D) \(1\)

Solution

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

\(\lim_{x \to 0} \frac{1 - \cos 2x}{2x^2}\)
Use the trigonometric identity \(1 - \cos 2x = 2\sin^2 x\):
\(\lim_{x \to 0} \frac{2\sin^2 x}{2x^2} = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2\).

Since \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), the limit evaluates to \(1^2 = 1\).

Therefore, \(\lambda = 1\).

Answer: (D)

Q19. If \(f(x) = \begin{cases} x^2 + 3x + a, & \text{when } x \le 1 \\ bx + 2, & \text{when } x > 1 \end{cases}\) is differentiable everywhere, then the values of a and b are:

• (A) \(a = 5, b = 3\)
• (B) \(a = 3, b = 2\)
• (C) \(a = 3, b = 5\)
• (D) \(a = 5, b = 2\)

Solution

Since \(f(x)\) is differentiable everywhere, it must also be continuous everywhere, including at \(x = 1\).

Continuity at \(x = 1\):
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)\)
\((1)^2 + 3(1) + a = b(1) + 2\)
\(4 + a = b + 2 \implies a - b = -2\) (Equation 1)

Differentiability at \(x = 1\):
Left Hand Derivative (LHD) = \(\frac{d}{dx}(x^2 + 3x + a)\) at \(x=1\) \(\implies 2x + 3 = 2(1) + 3 = 5\).
Right Hand Derivative (RHD) = \(\frac{d}{dx}(bx + 2)\) at \(x=1\) \(\implies b\).

For differentiability, \(LHD = RHD \implies b = 5\).

Substitute \(b = 5\) into Equation 1:
\(a - 5 = -2 \implies a = 3\).

Answer: (C)

Q20. If \(f(x) = \begin{cases} \frac{x\log(\cos x)}{\log(1 + x^2)}, & \text{when } x \neq 0 \\ 0, & \text{when } x = 0 \end{cases}\), then the function \(f(x)\) is:

• (A) Continuous but not differentiable at \(x = 0\)
• (B) Differentiable but not continuous at \(x = 0\)
• (C) Continuous and differentiable at \(x = 0\)
• (D) Neither continuous nor differentiable at \(x = 0\)

Solution

Check Continuity at \(x = 0\):
\(\lim_{x \to 0} \frac{x\log(\cos x)}{\log(1 + x^2)} = \lim_{x \to 0} \left( \frac{x^2}{\log(1 + x^2)} \cdot \frac{\log(\cos x)}{x} \right)\).
We know \(\lim_{x \to 0} \frac{\log(1 + x^2)}{x^2} = 1\), so its reciprocal is also 1.
Now find \(\lim_{x \to 0} \frac{\log(\cos x)}{x}\). Using L'Hôpital's rule (0/0 form):
\(= \lim_{x \to 0} \frac{\frac{1}{\cos x}(-\sin x)}{1} = \lim_{x \to 0} -\tan x = 0\).
Since \(1 \cdot 0 = 0 = f(0)\), the function is continuous at \(x = 0\).

Check Differentiability at \(x = 0\):
\(f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h\log(\cos h)}{h\log(1 + h^2)} = \lim_{h \to 0} \frac{\log(\cos h)}{\log(1 + h^2)}\).
Using L'Hôpital's rule (0/0 form):
\(\lim_{h \to 0} \frac{-\tan h}{\frac{2h}{1+h^2}} = \lim_{h \to 0} \left(-\frac{\tan h}{h} \cdot \frac{1+h^2}{2} \right) = -1 \cdot \frac{1}{2} = -\frac{1}{2}\).
Since the limit exists and is finite, \(f(x)\) is differentiable at \(x = 0\).

Answer: (C)

Q21. In which interval is the function \(f(x) = \frac{x}{1+|x|}\) differentiable?

• (A) \((-\infty, \infty)\)
• (B) \((0, \infty)\)
• (C) \((-\infty, 0) \cup (0, \infty)\)
• (D) None

Solution

Rewrite \(f(x)\) without the absolute value:
For \(x < 0\): \(f(x) = \frac{x}{1-x}\) \(\implies f'(x) = \frac{1(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}\).
For \(x \ge 0\): \(f(x) = \frac{x}{1+x}\) \(\implies f'(x) = \frac{1(1+x) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}\).

Clearly, the function is differentiable for all \(x \neq 0\). Now check at \(x = 0\):
LHD at \(x = 0\): \(\lim_{x \to 0^-} \frac{1}{(1-x)^2} = 1\).
RHD at \(x = 0\): \(\lim_{x \to 0^+} \frac{1}{(1+x)^2} = 1\).

Since LHD = RHD, the function is also differentiable at \(x = 0\). Thus, it is differentiable everywhere in \(\mathbb{R}\).

Answer: (A)

Q22. Statement (A): For \(f(x) = \sin x\), \(f'(\pi) = f'(3\pi)\)
Reason (R): For \(f(x) = \sin x\), \(f(\pi) = f(3\pi)\)
Which of the following correctly explains 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

Evaluate Statement (A):
\(f'(x) = \cos x\).
\(f'(\pi) = \cos\pi = -1\) and \(f'(3\pi) = \cos 3\pi = -1\). Since \(-1 = -1\), Statement (A) is True.

Evaluate Reason (R):
\(f(\pi) = \sin\pi = 0\) and \(f(3\pi) = \sin 3\pi = 0\). Since \(0 = 0\), Reason (R) is True.

Does \(f(\pi) = f(3\pi)\) mathematically prove that their derivatives are equal? No. For example, if \(g(x) = x(x-\pi)(x-3\pi)\), \(g(\pi) = g(3\pi) = 0\), but their derivatives at those points are not equal. The true reason the derivatives are equal is because \(\cos x\) is periodic with period \(2\pi\).

Thus, R is not the correct explanation for A.

Answer: (B)

Q23. Statement (I): \(\frac{d}{dx} (x^y) = y x^{y-1}\)
Statement (II): \(\frac{d}{dx} (x^y) = y x^{y-1} + x^y \log x \cdot \frac{dy}{dx}\)

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

Solution

When differentiating \(x^y\) with respect to \(x\) (where \(y\) is an implicit function of \(x\)), we use logarithmic differentiation.

Let \(u = x^y \implies \log u = y \log x\).
Differentiating both sides with respect to \(x\):
\(\frac{1}{u} \frac{du}{dx} = y \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{dy}{dx}\)
\(\frac{1}{u} \frac{du}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}\)
\(\frac{du}{dx} = u \left( \frac{y}{x} + \log x \cdot \frac{dy}{dx} \right) = x^y \left( \frac{y}{x} + \log x \cdot \frac{dy}{dx} \right)\)
\(\frac{du}{dx} = y x^{y-1} + x^y \log x \cdot \frac{dy}{dx}\).

Therefore, Statement (II) is correctly derived, whereas Statement (I) treats \(y\) as a constant, which is incorrect in this context.

Answer: (B)

Q24. If \(f(x) = \sin 3x \cos 4x\), then the value of \(f''(\frac{\pi}{2})\) is:

• (A) \(24\)
• (B) \(25\)
• (C) \(-25\)
• (D) \(-24\)

Solution

Simplify \(f(x)\) using the trigonometric identity \(2\sin A\cos B = \sin(A+B) + \sin(A-B)\):
\(f(x) = \frac{1}{2} (2\sin 3x\cos 4x) = \frac{1}{2} [\sin(3x+4x) + \sin(3x-4x)] = \frac{1}{2} (\sin 7x - \sin x)\).

First derivative:
\(f'(x) = \frac{1}{2} (7\cos 7x - \cos x)\).

Second derivative:
\(f''(x) = \frac{1}{2} (-49\sin 7x + \sin x)\).

Substitute \(x = \frac{\pi}{2}\):
\(\sin(7\frac{\pi}{2}) = \sin(3\pi + \frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1\).
\(\sin(\frac{\pi}{2}) = 1\).

\(f''(\frac{\pi}{2}) = \frac{1}{2} [-49(-1) + 1] = \frac{1}{2} [49 + 1] = \frac{50}{2} = 25\).

Answer: (B)

Q25. If \(x = a\cos(nt) - b\sin(nt)\), then the value of \(\frac{d^2x}{dt^2}\) is:

• (A) \(n^2x\)
• (B) \(-nx\)
• (C) \(nx\)
• (D) \(-n^2x\)

Solution

Differentiate \(x\) with respect to \(t\):
\(\frac{dx}{dt} = \frac{d}{dt} [a\cos(nt) - b\sin(nt)] = -an\sin(nt) - bn\cos(nt)\).

Differentiate again with respect to \(t\):
\(\frac{d^2x}{dt^2} = \frac{d}{dt} [-an\sin(nt) - bn\cos(nt)]\)
\(\frac{d^2x}{dt^2} = -an(n\cos(nt)) - bn(-n\sin(nt))\)
\(\frac{d^2x}{dt^2} = -an^2\cos(nt) + bn^2\sin(nt)\)

Factor out \(-n^2\):
\(\frac{d^2x}{dt^2} = -n^2 [a\cos(nt) - b\sin(nt)]\)
Substitute \(x\) back in: \(\frac{d^2x}{dt^2} = -n^2x\).

Answer: (D)

Q26. In the adjacent figure, the height of the right circular cone (OC) = 20 cm. The semi-vertical angle \(\angle OCA = 30^\circ\). If the rate of increase of the semi-vertical angle is 2 rad/sec, then the rate of increase of its base radius is:

• (A) 30 cm/sec
• (B) \(\frac{160}{3}\) cm/sec
• (C) 10 cm/sec
• (D) 160 cm/sec

Solution

Let the height be \(h = 20\) cm (constant), base radius be \(r\), and semi-vertical angle be \(\alpha\).

From the right triangle OCA, \(\tan \alpha = \frac{r}{h} \implies r = h \tan \alpha = 20 \tan \alpha\).

Differentiate with respect to time \(t\):
\(\frac{dr}{dt} = 20 \sec^2 \alpha \cdot \frac{d\alpha}{dt}\).

We are given \(\alpha = 30^\circ\) and \(\frac{d\alpha}{dt} = 2\) rad/sec.
\(\sec 30^\circ = \frac{2}{\sqrt{3}} \implies \sec^2 30^\circ = \frac{4}{3}\).

Substitute the values:
\(\frac{dr}{dt} = 20 \left(\frac{4}{3}\right) (2) = \frac{160}{3}\) cm/sec.

Answer: (B)

Q27. The rate of change of the radius of a sphere is \(\frac{1}{2\pi}\). When the radius is 5 cm, the rate of change of its surface area is:

• (A) 20 sq cm
• (B) 10 sq cm
• (C) 15 sq cm
• (D) 25 sq cm

Solution

The surface area of a sphere is \(S = 4\pi r^2\).

Differentiate with respect to time \(t\):
\(\frac{dS}{dt} = 8\pi r \frac{dr}{dt}\).

Given \(r = 5\) cm and \(\frac{dr}{dt} = \frac{1}{2\pi}\) cm/s, substitute these values:
\(\frac{dS}{dt} = 8\pi (5) \left( \frac{1}{2\pi} \right) = 40\pi \cdot \frac{1}{2\pi} = 20\) sq cm/s.

Answer: (A)

Q28. If the straight line \(x + y - k = 0\) is a normal to the parabola \(y^2 = 12x\), then the value of k is:

• (A) 9
• (B) -9
• (C) 6
• (D) -6

Solution

The standard equation of a normal to the parabola \(y^2 = 4ax\) in terms of its slope \(m\) is:
\(y = mx - 2am - am^3\).

For \(y^2 = 12x\), \(4a = 12 \implies a = 3\).
The given line is \(x + y - k = 0 \implies y = -x + k\). The slope of this line is \(m = -1\).

Substitute \(a = 3\) and \(m = -1\) into the normal equation:
\(y = (-1)x - 2(3)(-1) - 3(-1)^3\)
\(y = -x + 6 + 3 = -x + 9\).

Rearranging this gives \(x + y - 9 = 0\). Comparing this with \(x + y - k = 0\), we get \(k = 9\).

Answer: (A)

Q29. The interval in which the function \(f(x) = 4x^3 + 6x^2 - 24x + 1\) is strictly decreasing is:

• (A) \((-1, 2)\)
• (B) \((2, 1)\)
• (C) \((-2, -1)\)
• (D) \((-2, 1)\)

Solution

First, find the derivative \(f'(x)\):
\(f'(x) = 12x^2 + 12x - 24 = 12(x^2 + x - 2)\).

Factorize the quadratic expression:
\(f'(x) = 12(x + 2)(x - 1)\).

For the function to be strictly decreasing, \(f'(x) < 0\):
\(12(x + 2)(x - 1) < 0\).

Using the wavy curve method, the critical points are \(x = -2\) and \(x = 1\). The expression is negative between the roots.
Thus, the interval is \((-2, 1)\).

Answer: (D)