मराठी

The Equation of the Normal to the Curve Y = X + Sin X Cos X at X = π/2 is

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प्रश्न

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

पर्याय

  • = 2

  • x = π

  • x + π = 0

  • 2x = π

MCQ
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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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पाठ 15: Tangents and Normals - Exercise 16.5 [पृष्ठ ४२]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 15 Tangents and Normals
Exercise 16.5 | Q 2 | पृष्ठ ४२
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