The slope of the normal to the curve y = 2 x 2 + 3 x at x = 0 is (A) 3 (B) 3 (C) - 3 (D) - 3

Option D is correct We have, y = 2 x 2 + 3 x …(i) = dx = 4x + 3 x Slope of the tangent to (i) at x = 0 is ( dx ) x = 0 = 4 0 + 3 0 = 3 So, slope of the normal to (i) at x = 0 is , , , , , , = - 3

Application of Derivatives — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 6.3, Q.26,Page 213

Question

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

(A) $3$

(B) $\cfrac{1}{3}$

(D) $- \cfrac{1}{3}$

Step-by-step Solution

Option D is correct

We have,
$y = 2{x^2} + 3\sin x$ …(i)
$= \cfrac{{dy}}{{dx}} = 4x + 3\cos x$

Slope of the tangent to (i) at $x = 0$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 0}} = 4 \cdot 0 + 3\cos 0 = 3$

So, slope of the normal to (i) at $x = 0$ is $\cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{tangent}}}} = - \cfrac{1}{3}$

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