Match List-I with List-II
|
List-I (Curve) |
List-II (Slope of tangent at $x = 4$) |
|
(A) $y = \sqrt{x^3}$ |
(I) -1 |
|
(B) $y = \sqrt{x}$ |
(II) 1 |
|
(C) $y=x^3-47x$ |
(III) 1/4 |
|
(D) $xy = 16$ |
(IV) 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
|
List-I (Curve) |
List-II (Slope of tangent at $x = 4$) |
|
(A) $y = \sqrt{x^3}$ |
(IV) 3 |
|
(B) $y = \sqrt{x}$ |
(III) 1/4 |
|
(C) $y=x^3-47x$ |
(II) 1 |
|
(D) $xy = 16$ |
(I) -1 |
Given curves and slopes to find at x = 4:
(A) $y = \sqrt{x^3} = x^{3/2}$
$\frac{dy}{dx} = \frac{3}{2} x^{1/2}$
At $x = 4$: $\frac{dy}{dx} = \frac{3}{2} \cdot 2 = 3$ → (A)-(IV)
(B) $y = \sqrt{x} = x^{1/2}$
$\frac{dy}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$
At $x = 4$: $\frac{dy}{dx} = \frac{1}{2*2} = 1/4$ → (B)-(III)
(C) $y = x^3 - 47x$
$\frac{dy}{dx} = 3x^2 - 47$
At $x = 4$: $\frac{dy}{dx} = 3*16 - 47 = 48 - 47 = 1$ → (C)-(II)
(D) $xy = 16 \Rightarrow y = 16/x$
$\frac{dy}{dx} = -16/x^2$
At $x = 4$: $\frac{dy}{dx} = -16/16 = -1$ → (D)-(I)