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

Answer 1

(d) 2x = π

$$\begin{gathered} \mathrm{Given}: \\ y=x+\sin x\cos x \\ \mathrm{On}\mathrm{d}\text{ifferentiating both sides w.r.t.}x,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=1+{\cos }^{2}x-{\sin }^{2}x \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{2}}{\text{=1+cos}}^2\frac{\mathrm{\pi }}{2}{\text{-sin}}^2\frac{\mathrm{\pi }}{2}\text{=1-1=0} \\ \text{Slope of the normal,}\mathit{\text{m}}\text{=}\frac{-1}{0} \\ \text{When}\mathit{\text{x}}\text{=}\frac{\mathrm{\pi }}{2}\text{,} \\ \text{y=}\frac{\mathrm{\pi }}{2}\text{+cos}\frac{\mathrm{\pi }}{2}\text{sin}\frac{\mathrm{\pi }}{2}\text{=}\frac{\mathrm{\pi }}{2} \\ \mathrm{Now}, \\ \left(x_1,y_1\right)=\left(\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \\ \therefore \text{Equation of the normal} \\ =y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{\mathrm{\pi }}{2}=\frac{-1}{0}\left(x-\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow 2x=\mathrm{\pi } \end{gathered}$$