JEE Main On 16 April 2018 Question 9

Question: If $ x=\sqrt{{2^{\cos e{c^{-1}}t}}} $ and $ y=\sqrt{{2^{{{\sec }^{-1t}}}}(|t|\ge 1)}, $ then $ \frac{dy}{dx} $ is equal to. [JEE Main 16-4-2018]

Options:

A) $ \frac{y}{x} $

B) $ -\frac{y}{x} $

C) $ -\frac{x}{y} $

D) $ \frac{x}{y} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ \frac{dx}{dt}=\frac{1}{2\sqrt{{2^{\cos e{c^{-1}}t}}}} $ $ {2^{^{\cos e{c^{-1t}}}}}\log 2.\frac{-1}{x\sqrt{x^{2}-1}} $ $ \frac{dy}{dt}=\frac{1}{2\sqrt{{2^{{{\sec }^{-1t}}}}}}{2^{{{\sec }^{-1t}}}}\log 2.\frac{1}{x\sqrt{x^{2}-1}} $ Thus $ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-\sqrt{{2^{{cosec^{-1}}t}}}}{\sqrt{{2^{{{\sec }^{-1t}}}}}}\frac{{2^{{{\sec }^{-1}}t}}}{{2^{\cos e{c^{-1}}t}}} $ $ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\sqrt{\frac{{2^{{{\sec }^{-1}}t}}}{{2^{\cos e{c^{-1}}t}}}}=\frac{-y}{x} $