Match List-I with List-II:
| List-I | List-II | ||
| (A) | The derivative of \(\log_e x\) with respect to \(\left(\displaystyle\frac{1}{x}\right)\) at \(x=5\) is | (I) | −5 |
| (B) | If \(x^3+x^2y+xy^2=21x\), then \(\displaystyle\frac{dy}{dx}\) at (1, 1) is | (II) | −6 |
| (C) | If \(f(x)=x^3\log_e\displaystyle\frac{1}{x}\), then \(f'(1)+f''(1)\) is | (III) | 5 |
| (D) | If \(y=f(x^2)\) and \(f'(x)=e^{\sqrt{x}}\), then \(\displaystyle\frac{dy}{dx}\) at \(x=0\) is | (IV) | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
(A) The derivative of \(\log_e x\) with respect to \(\frac{1}{x}\) at x = 5
This matches (I) −5. Let \(u = \log_e x\) and \(v = \frac{1}{x}\). The derivative of u with respect to v is the ratio of their derivatives with respect to x:
\( \displaystyle \frac{du}{dx} = \frac{1}{x},\quad \frac{dv}{dx} = -\frac{1}{x^2} \)
\( \displaystyle \frac{du}{dv} = \frac{1/x}{-1/x^2} = -x \)
At x = 5 this is −5.
(B) \(x^3+x^2y+xy^2=21x\), dy/dx at (1, 1)
This matches (III) 5. By implicit differentiation:
\( \displaystyle 3x^2 + 2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 21 \)
\( \displaystyle \left(x^2 + 2xy\right)\frac{dy}{dx} = 21 - 3x^2 - 2xy - y^2 \)
Putting x = 1, y = 1:
\( \displaystyle 3\,\frac{dy}{dx} = 21 - 3 - 2 - 1 = 15 \)
\( \displaystyle \frac{dy}{dx} = 5 \)
(C) \(f(x)=x^3\log_e\frac{1}{x}\), \(f'(1)+f''(1)\)
This matches (II) −6. Since \(\log_e\frac{1}{x} = -\log_e x\):
\(f(x) = -x^3\log_e x\)
\(f'(x) = -3x^2\log_e x - x^2\)
\(f''(x) = -6x\log_e x - 3x - 2x = -6x\log_e x - 5x\)
With \(\log_e 1 = 0\):
\(f'(1) = -1,\quad f''(1) = -5\)
\(f'(1) + f''(1) = -6\)
(D) \(y=f(x^2)\), \(f'(x)=e^{\sqrt{x}}\), dy/dx at x = 0
This matches (IV) 0. By the chain rule:
\( \displaystyle \frac{dy}{dx} = f'(x^2)\times 2x = 2x\,e^{\sqrt{x^2}} \)
At x = 0 this is \(0 \times e^0 = 0\).
So the matching is (A)-(I), (B)-(III), (C)-(II), (D)-(IV).
Note: It seems that there is a mistake in the original NTA paper: in List-I (B), the point (1, 1) does not lie on the curve \(x^3+x^2y+xy^2=21x\). Putting x = 1 and y = 1, the left side is 1 + 1 + 1 = 3, while the right side is 21. The question can still be solved, because the expression for dy/dx, worked out at x = 1, y = 1, gives 5, which is (III); also, (C) can only match (II), and Option (2) is the only option with (C)-(II).