The equation of the normal to the curve y=(1+x)y+sin−1⁡sin2⁡x at x=0, is

A
$x + y = 2$
B
$x + y = 1$
C
$x - y = 1$
D
none of these

Solution:

We have,

$y = (1 + x)^{y} + \sin^{- 1} (\sin^{2} x)$ …(i)

When $x = 0$ , we have $y = 1$

Differentiating (i) w.r.t. x, we get

$\begin{array}{l} \frac{d y}{d x} = (1 + x)^{y} \{\frac{d y}{d x} \log (1 + x) + \frac{y}{1 + x}\} + \frac{\sin 2 x}{\sqrt{1 - \sin^{4} x}} \\ \Rightarrow {(\frac{d y}{d x})}_{(0, 1)} = 1 \end{array}$

$m_{1} m_{2} = - 1$

Here $m_{1} = 1$ then $m_{2} = - 1$

Equation of normal $y - y_{1} = m_{2} (x - x_{1})$

So, the equation of the normal at (0, 1) is

$y - 1 = - 1 (x - 0) \Rightarrow x + y = 1$

The equation of the normal to the curve $y = (1 + x)^{y} + \sin^{- 1} (\sin^{2} x) at x = 0$ , is

Solution:

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