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The slope of the normal to the curve $y=2 x^2+3 \sin x$ at $\mathbf{x}=\mathbf{0}$ is
(a) 3
(b) $\frac{1}{3}$
(c) $-3$
(d) $-\frac{1}{3}$
The slope of the normal to the curve $y=2 x^2+3 \sin x$ at $\mathbf{x}=\mathbf{0}$ is
(a) 3
(b) $\frac{1}{3}$
(c) $-3$
(d) $-\frac{1}{3}$
Solution:
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Verified Answer
(d) $\because y=2 x^2+3 \sin \mathrm{x} \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{x}+3 \cos \mathrm{x}$ at $\mathrm{x}=0$, $\frac{\mathrm{dy}}{\mathrm{dx}}=3, \therefore$ Slope $=3 \Rightarrow$ Slope of normal is $=-\frac{1}{3}$
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