Match List - I with List - II.
| List - I | List - II | ||
| Equation of curves | Slope of tangent at \(x=2\) | ||
| (A) | \(y=x^3-x\) | (I) | 8 |
| (B) | \(y=(x-2)^2\) | (II) | 2/3 |
| (C) | \(y=2x^2+3\) | (III) | 11 |
| (D) | \(y=\sqrt{4x+1}-7\) | (IV) | 0 |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The slope of the tangent to y = f(x) at x = 2 is the value of \(\frac{dy}{dx}\) at x = 2.
(A) \(y=x^3-x\)
This matches (III) 11.
\( \displaystyle \frac{dy}{dx} = 3x^2 - 1 \)
\(\text{at } x = 2:\ 3(4) - 1 = 11\)
(B) \(y=(x-2)^2\)
This matches (IV) 0.
\( \displaystyle \frac{dy}{dx} = 2(x-2) \)
\(\text{at } x = 2:\ 2(0) = 0\)
(C) \(y=2x^2+3\)
This matches (I) 8.
\( \displaystyle \frac{dy}{dx} = 4x \)
\(\text{at } x = 2:\ 4(2) = 8\)
(D) \(y=\sqrt{4x+1}-7\)
This matches (II) 2/3.
By the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{4}{2\sqrt{4x+1}} = \frac{2}{\sqrt{4x+1}} \)
\( \displaystyle \text{at } x = 2:\ \frac{2}{\sqrt{9}} = \frac{2}{3} \)