Match List-I with List-II
| List-I | List-II | ||
| (Curve) | (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)
The slope of the tangent at a point is the value of \(\frac{dy}{dx}\) at that point, here at \(x = 4\).
(A) \(y = \sqrt{x^3} = x^{3/2}\)
This matches (IV) 3:
\( \displaystyle \frac{dy}{dx} = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x} \)
\( \displaystyle \text{At } x = 4:\quad \frac{3}{2} \times 2 = 3 \)
(B) \(y = \sqrt{x} = x^{1/2}\)
This matches (III) 1/4:
\( \displaystyle \frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \)
\( \displaystyle \text{At } x = 4:\quad \frac{1}{2 \times 2} = \frac{1}{4} \)
(C) \(y = x^3 - 47x\)
This matches (II) 1:
\( \displaystyle \frac{dy}{dx} = 3x^2 - 47 \)
\(\text{At } x = 4:\quad 3(16) - 47 = 48 - 47 = 1\)
(D) \(xy = 16\), so \(y = 16x^{-1}\)
This matches (I) −1:
\( \displaystyle \frac{dy}{dx} = -\frac{16}{x^2} \)
\( \displaystyle \text{At } x = 4:\quad -\frac{16}{16} = -1 \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).