JEE PYQ: Application Of Derivatives Question 29
Question 29 - 2020 (02 Sep Shift 2)
The equation of the normal to the curve $y = (1+x)^{2y} + \cos^2(\sin^{-1} x)$ at $x = 0$ is:
(1) $y + 4x = 2$
(2) $y = 4x + 2$
(3) $x + 4y = 8$
(4) $2y + x = 4$
Show Answer
Answer: (3)
Solution
At $x=0$: $y = 1 + 1 = 2$. $\frac{dy}{dx} = (1+x)^{2y}\left(2\ln(1+x)\frac{dy}{dx} + \frac{2y}{1+x}\right) - 2x$. At $x=0$: $\frac{dy}{dx} = 4$. Normal slope $= -\frac{1}{4}$. Equation: $y - 2 = -\frac{1}{4}(x) \Rightarrow x + 4y = 8$.
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