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प्रश्न
The equation of the normal to the curve y = x + sin x cos x at x = `π/2` is ___________ .
विकल्प
x = 2
x = π
x + π = 0
2x = π
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उत्तर
2x = π
\[\text { Given }: \]
\[y = x + \sin x \cos x\]
\[\text { On differentiating both sides w.r.t.x, we get }\]
\[\frac{dy}{dx} = 1 + \cos^2 x - \sin^2 x\]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{2}} {=1+cos}^2 \frac{\pi}{2} {-sin}^2 \frac{\pi}{2}=1-1=0\]
\[\text { Slope of the normal, } m=\frac{- 1}{0}\]
\[\text { When }x=\frac{\pi}{2},\]
\[y=\frac{\pi}{2}+cos\frac{\pi}{2}\sin\frac{\pi}{2}=\frac{\pi}{2}\]
\[\text { Now }, \]
\[\left( x_1 , y_1 \right) = \left( \frac{\pi}{2}, \frac{\pi}{2} \right)\]
\[ \therefore \text { Equation of the normal }\]
\[ = y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{\pi}{2} = \frac{- 1}{0}\left( x - \frac{\pi}{2} \right)\]
\[ \Rightarrow x = \frac{\pi}{2}\]
\[ \Rightarrow 2x = \pi\]
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