Chose the correct answer.

The slope of the normal to the curve $y = 2 x^{2} + 3$ sinx at x=0 is

(a) 3 (b) $\frac{1}{3}$ (c)-3 (d)- $\frac{1}{3}$

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Solution

The equation of the given curve is $y = 2 x^{2} + 3$ sin x

On differentiating w.r.t.x, we get

$\frac{d y}{d x} = 4 x + 3$ cos x

Slope of the tangent to the given curve at x=0 is given by

$\frac{d y}{d x} = 4 \times 0 + 3$ cos 0=3

Slope of the normal to the given curve at x=0 is

$\frac{- 1}{S l o p e o f t a n g e n t} = - \frac{1}{3}$ .Hence, correct option is (d).

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Choose the correct answer in the following question:

The normal at the point (1, 1) on the curve $2 y + x^{2} = 3$ is

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Chose the correct answer. The slope of the normal to the curve y=2x2+3sinx at x=0 is (a) 3 (b) 13 (c)-3 (d)- 13

The equation of the given curve is y=2x2+3 sin x On differentiating w.r.t.x, we get dydx=4x+3 cos x Slope of the tangent to the given curve at x=0 is given by dydx=4×0+3 cos 0=3 Slope of the normal to the given curve at x=0 is −1Slope of tangent=−13.Hence, correct option is (d).

Chose the correct answer.

The slope of the normal to the curve $y = 2 x^{2} + 3$ sinx at x=0 is

(a) 3 (b) $\frac{1}{3}$ (c)-3 (d)- $\frac{1}{3}$