First slide

Tangents and normals

Question

$T h e e q u a t i o n o f t h e n o r m a l t o t h e c u r v e$ $y = (1 + x)^{y} + \sin^{- 1} (\sin^{2} x) at x=0 is$

Moderate

A

$x + y = 1$
B

$x + y + 1 = 0$
C

$2 x - y + 1 = 0$
D

$x + 2 y + 2 = 0$

Solution

$\begin{array}{l} Given y = {(1 + x)}^{y} + \sin^{- 1} (\sin^{2} x) \\ When x = 0, y = 1 \\ To get the slope of the tangent, differentiate both sides \\ \frac{d y}{d x} = {(1 + x)}^{y} \{\frac{d y}{d x} \log (1 + x) + \frac{y}{1 + x}\} + \frac{\sin 2 x}{\sqrt{1 - \sin^{4} x}} \\ \frac{d y}{d x} |_{(0, 1)} = 1 \\ The slope of the normal is - 1 \\ Equation of the normal is x + y - 1 = 0 \end{array}$

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