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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 8 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 8 Maths Textbook Solution.

Answers (1)

Answer:

\text { The slope of the tangent is }-12

\text { The slope of the normal is } \frac{1}{12}

Hint:

\text { The slope of tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }

\text { The slope of normal is the normal to a curve at } P=(x, y) \text { is a line perpendicular to the tangent }

\text { at } P \text { and passing through } P \text { . }

Given:y=(\sin 2 x+\cot x+2)^{2} \text { at } x=\frac{\pi}{2}

Solution:

\text { First we have to find } \frac{d y}{d x} \text { of given function, }

f(x) \text { that is to find the derivative of } f(x)

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &y=(\sin 2 x+\cot x+2)^{2} \end{aligned}

\frac{d y}{d x}=2(\sin 2 x+\cot x+2)^{2-1}+\left\{\frac{d }{d x}(\sin 2 x)+\frac{d }{d x}(\cot x)+\frac{d }{d x}(2)\right\}

\frac{d y}{d x}=2(\sin 2 x+\cot x+2)\left\{(\cos 2 x) 2+\left(-\cos e c^{2} x\right)+(0)\right\}

\begin{aligned} &\frac{d}{d x}(\sin x)=\cos x \\ &\frac{d}{d x}(\cot x)=-\operatorname{cosec}^{2} x \\ &\frac{d y}{d x}=2(\sin 2 x+\cot x+2)\left(2 \cos 2 x-\cos e c^{2} x\right) \end{aligned}

\text { Since, } x=\frac{\pi}{2}

\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}=2\left(\sin 2\left(\frac{\pi}{2}\right)+\cot \left(\frac{\pi}{2}\right)+2\right)\left(2 \cos 2\left(\frac{\pi}{2}\right)-\operatorname{cosec}^{2}\left(\frac{\pi}{2}\right)\right)

\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}=2 \times(0+0+2)(2(-1)-1)

\text { We know that } \cot \left(\frac{\pi}{2}\right)=0 \text { and } \operatorname{cosec}\left(\frac{\pi}{2}\right)=1

\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}=2 \times(2)(-2-1)

\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}=4 \times(-3)=-12

\text { The slope of the tangent at } x=\frac{\pi}{2} \text { is }-12

\text { The slope of the normal }=\frac{-1}{\text { The slope of the tangent }}

\text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}}

\text { The slope of the normal }=\frac{1}{12}

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