JEE PYQ: Differentiation Question 6
Question 6 - 2020 (04 Sep Shift 1)
If $(a + \sqrt{2b}\cos x)(a - \sqrt{2b}\cos y) = a^2 - b^2$, where $a > b > 0$, then $\frac{dx}{dy}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is:
(1) $\frac{a - 2b}{a + 2b}$ (2) $\frac{a - b}{a + b}$ (3) $\frac{a + b}{a - b}$ (4) $\frac{2a + b}{2a - b}$
Show Answer
Answer: (3) $\frac{a+b}{a-b}$
Solution
Differentiating and evaluating at $(\frac{\pi}{4}, \frac{\pi}{4})$: $\frac{dy}{dx} = \frac{a-b}{a+b}$, so $\frac{dx}{dy} = \frac{a+b}{a-b}$.
BETA
