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  • Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 10 maths text book solution.

Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 10 maths text book solution.

Answers (1)

Answer : Equation of normal =2x = \pi

Hint : Use equation of normal,

          y-y_{1}=\frac{1}{\frac{d y}{d x}}\left(x-x_{1}\right)

Given :

Here given that the curve

    y=x+\sin x \cos x

We have to write the equation of normal to the given curve at x=\frac{\pi }{2}

Solution:

We have,

y=x+\sin x \cos x

On differentiating both sides with respect to  'x', we get

\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x

Now we know that,

\begin{aligned} &\text { slope of tangent }=\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}} \\ &=1+\cos ^{2}\left(\frac{\pi}{2}\right)-\sin ^{2}\left(\frac{\pi}{2}\right) \\ &=1+0-1 \\ &=0 \end{aligned}

Now,

When x=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}

\Rightarrow y=\frac{\pi }{2}

\therefore \quad\left(x_{1}, y_{1}\right)=\left(\frac{\pi}{2}, \frac{\pi}{2}\right)

So, equation of normal

          \begin{aligned} &y-y_{1}=\frac{-1}{\text { slope of tangent }}\left(x-x_{1}\right) \\ &y-\frac{\pi}{2}=\frac{-1}{0}\left(x-\frac{\pi}{2}\right) \end{aligned}

\begin{array}{ll} \Rightarrow \quad & x=\frac{\pi}{2} \\ \Rightarrow & 2 x=\pi \end{array}

Hence the equation of the given curve at x=\frac{\pi}{2} \text { is } 2 x=\pi.

 

Posted by

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