Browse Questions

# Choose the correct answer in the slope of the normal to the curve $y = 2x^2 + 3 sin\: x\; at\;$ $x = 0$ is

$(A)\; 3 \quad (B)\;\frac{1}{3} \quad (C)\; -3 \quad (D)\; -\frac{1}{3}$

Can you answer this question?

Toolbox:
• If a tangent line to the curve $y=f(x)$ makes an angle $\theta$ with $x$-axis in the +ve direction,then $\large\frac{dy}{dx}$=slope of the tangent=$\tan \theta$
• Slope of the normal is $\large\frac{-1}{\Large\frac{dy}{dx}}$
Step 1:
Equation of the curve is $y=2x^2+3\sin x$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=4x+3\cos x Hence the slope of the tangent at x=0 \large\frac{dy}{dx}_{(x=0)}$$=4(0)+3\cos 0$
But $\cos 0=1$
$\qquad\quad\;\;=3\times 1=3$
Step 2:
Hence the slope of the tangent at $x=0$ is $3$
Therefore the slope of the normal is $\large\frac{-1}{3}$
The correct option is $D$
answered Jul 12, 2013