Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function or relation | \(\displaystyle\frac{dy}{dx}\) | ||
| (A) | \(y = \log_e\sqrt{x},\ x \gt 0\) | (I) | \(-\displaystyle\frac{x}{y}\) |
| (B) | \(x^{2} + y^{2} = 4\) | (II) | \(\displaystyle\frac{y}{x}\) |
| (C) | \(x = 2t,\ y = t^{2}\) | (III) | \(\displaystyle\frac{x}{2}\) |
| (D) | \(\tan\left(\displaystyle\frac{x + y}{x - y}\right) = 5\) | (IV) | \(\displaystyle\frac{1}{2x}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) \(y = \log_e\sqrt{x},\ x \gt 0\)
This matches (IV) \(\frac{1}{2x}\). Since \(\log_e\sqrt{x} = \frac{1}{2}\log_e x\):
\( \displaystyle \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2x} \)
(B) \(x^{2} + y^{2} = 4\)
This matches (I) \(-\frac{x}{y}\). By implicit differentiation:
\( \displaystyle 2x + 2y\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{x}{y} \)
(C) \(x = 2t,\ y = t^{2}\)
This matches (III) \(\frac{x}{2}\). In parametric form:
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{2} = t \)
Since \(t = \frac{x}{2}\), this is \(\frac{x}{2}\).
(D) \(\tan\left(\frac{x + y}{x - y}\right) = 5\)
This matches (II) \(\frac{y}{x}\). Let \(u = \frac{x + y}{x - y}\). Differentiating \(\tan u = 5\):
\( \displaystyle \sec^{2} u \cdot \frac{du}{dx} = 0 \)
Since \(\sec^{2} u \ne 0\), \(\frac{du}{dx} = 0\). By the quotient rule, writing y′ for \(\frac{dy}{dx}\):
\( \displaystyle \frac{(1 + y')(x - y) - (x + y)(1 - y')}{(x - y)^{2}} = 0 \)
\((x - y) + y'(x - y) - (x + y) + y'(x + y) = 0\)
\(2xy' - 2y = 0\)
\( \displaystyle y' = \frac{y}{x} \)