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Question

The equation of the normal to the curve y = x + sin x cos x at x = π/2 is

(a) x = 2
(b) x = π
(c) x + π = 0
(d) 2x = π

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Solution

(d) 2x = π

Given:y=x+sin x cosxOn differentiating both sides w.r.t. x, we getdydx=1+cos2 x-sin2 xSlope of the tangent=dydxx=π2=1+cos 2π2-sin2π2=1-1=0Slope of the normal, m=-10When x=π2, y=π2+cos π2 sin π2=π2Now, x1, y1=π2, π2Equation of the normal=y-y1=mx-x1y-π2=-10x-π2x=π22x=π

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