Match List-I with List-II
| List-I | List-II | ||
| (Parametric equations) | \(\left(\displaystyle\frac{dy}{dx}\right)\) | ||
| (A) | \(x = \displaystyle\frac{2}{t},\ y = 2t\) | (I) | \(4t^2\) |
| (B) | \(x = t^3,\ y = 3t + 2\) | (II) | \(2(t+1)\) |
| (C) | \(x = \log t,\ y = 2t^2\) | (III) | \(-t^2\) |
| (D) | \(x = e^t,\ y = 2te^t\) | (IV) | \(t^{-2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
For parametric equations,
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
(A) \(x = \frac{2}{t},\ y = 2t\)
This matches (III) \(-t^2\).
\( \displaystyle \frac{dx}{dt} = -\frac{2}{t^2},\quad \frac{dy}{dt} = 2 \)
\( \displaystyle \frac{dy}{dx} = \frac{2}{-2/t^2} = -t^2 \)
(B) \(x = t^3,\ y = 3t + 2\)
This matches (IV) \(t^{-2}\).
\( \displaystyle \frac{dy}{dx} = \frac{3}{3t^2} = t^{-2} \)
(C) \(x = \log t,\ y = 2t^2\)
This matches (I) \(4t^2\).
\( \displaystyle \frac{dy}{dx} = \frac{4t}{1/t} = 4t^2 \)
(D) \(x = e^t,\ y = 2te^t\)
This matches (II) \(2(t+1)\). By the product rule, \(\frac{dy}{dt} = 2e^t + 2te^t\):
\( \displaystyle \frac{dy}{dx} = \frac{2e^t(1 + t)}{e^t} = 2(t + 1) \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).