Question

The slopes of the tangent and normal at $(0, 1)$ for the curve $y = \sin x + e^x$ are respectively

• A. $1$ and $-1$
• B. $- \frac{1}{2}$ and $2$
• C. $2$ and $- \frac{1}{2}$
• D. $-1$ and $1^-$

Answer 1

Answer: (C)

$2$ and $- \frac{1}{2}$

Answer 2

We have, $y = \sin x + e^x$
Differentiating w.r.t. x, we get
$\frac{dy}{dx} = \cos x +e^{x}$
$\left|\frac{dy}{dx}\right|_{\left(0,1\right)} = \cos\left(0\right)+e^{0} =1+1=2$
$- \frac{dx}{dy} = -\frac{1}{\cos x +e^{x}}$
$\left|-\frac{dx}{dy}\right|_{\left(0,1\right)} = -\frac{1}{\cos 0+e^{0} } = - \frac{1}{2}$
Hence, slope of tangent and normal at (0, 1) are 2 and $- \frac{1}{2}$ respectively.