Match List - I with List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | If \(y = x^2 - 8\) and \(\displaystyle\frac{dy}{dx} = 0\), then \(x = ?\) | (I) | 1 |
| (B) | If \(p(x) = 3x + 1\), then \(R(x)\) at \(x = 2\) | (II) | 0 |
| (C) | If \(y = x^3\), then \(\displaystyle\frac{dy}{dx}\) at \(x = -1\) | (III) | 14 |
| (D) | If \(C(x) = 100 + 5x\), \(R(x) = 102 + 3x\), then break-even point | (IV) | 3 |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A) If \(y = x^2 - 8\) and \(\frac{dy}{dx} = 0\), then x = ?
This matches (II) 0.
\( \displaystyle \frac{dy}{dx} = 2x = 0 \)
\(x = 0\)
(B) If p(x) = 3x + 1, then R(x) at x = 2
This matches (III) 14. Here p(x) is the price per unit when x units are sold, so the revenue is R(x) = x · p(x):
\(R(x) = x(3x + 1) = 3x^2 + x\)
\(R(2) = 12 + 2 = 14\)
(C) If \(y = x^3\), then \(\frac{dy}{dx}\) at x = −1
This matches (IV) 3.
\( \displaystyle \frac{dy}{dx} = 3x^2 \)
\(3(-1)^2 = 3\)
(D) If C(x) = 100 + 5x, R(x) = 102 + 3x, then break-even point
This matches (I) 1. At the break-even point R(x) = C(x):
\(102 + 3x = 100 + 5x\)
\(2x = 2\)
\(x = 1\)