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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the slope of the normal to the curve \( y = 2x^2 + 3 sin\: x\; at\;\) \(x = 0\) is

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

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1 Answer

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  • 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 by sreemathi.v

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