Question

The normal to the curve $y={\left(1+2x\right)}^x+{\sin }^{-1}\left({\sin }^3\left(x\right)\right)$ at $x=0$ is:

• A. $y=0$
• B. $x+y=1$
• C. $x=0$
• D. None of these

Answer 1

The correct option is C

$x=0$

Find the normal to the given curve

Given $y={\left(1+2x\right)}^x+{\sin }^{-1}\left({\sin }^3\left(x\right)\right)$

$$\begin{gathered} \frac{dy}{dx}=\frac{d{\left(1+2x\right)}^x}{dx}+\frac{d{\sin }^{-1}\left({\sin }^3\left(x\right)\right)}{dx} \\ ={\left(1+2x\right)}^x\log \left(1+2x\right)\frac{d\left(1+2x\right)}{dx}+\frac{1}{\sqrt{1-{\left({\sin }^3\left(x\right)\right)}^2}}\frac{d{\sin }^3\left(x\right)}{dx} \\ =2{\left(1+2x\right)}^x\log \left(1+2x\right)+\frac{3{\sin }^2\left(x\right)\cos \left(x\right)}{\sqrt{1-{\sin }^6\left(x\right)}} \end{gathered}$$ $\left[\because \frac{d{a}^x}{dx}=a^x\log \left(a\right)\mathrm{and}\frac{d{\sin }^{-1}\left(\mathrm{x}\right)}{d\mathrm{x}}=\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\right]$

At $x=0$

$\frac{dy}{dx}=0$

We know that equation of normal is

$\left(y-y_0\right)=-\frac{1}{{\displaystyle \frac{dy}{dx}}}\left(x-x_0\right)$

Line is parallel to the $x$ axis and normal is parallel to the $y$ axis.

But when $x=0$, $y=0$ the normal passes through $\left(0,0\right)$

It is the $y$ axis. itself, $x=0$

The normal is $x=0$.

Hence, the correct option is $\left(C\right)$