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} $
