#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 5 Maths Textbook Solution.

Answer:$\text { The points are }(2,-4) \&\left(\frac{-2}{3}, \frac{4}{27}\right)$

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

Given:$\text { Curve } y=x^{3}-2 x^{2}-2 x \text { and a line } y=2 x-3$

Solution:

First we will find the slope of tangent, we get

$y=x^{3}-2 x^{2}-2 x$

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}\right)-\frac{d}{d x}\left(2 x^{2}\right)-\frac{d}{d x}(2 x) \\ &\frac{d y}{d x}=3 x^{3-1}-2 \times 2\left(x^{2-1}\right)-2 \times x^{1-1} \\ &\frac{d y}{d x}=3 x^{2}-4 x-2 \quad \rightarrow(1) \end{aligned}

$y=2 x-3 \text { is the form of equation of a straight line } y=m x+c$

$\text { where } \mathrm{m} \text { is the slope of line. }$

$\text { So, the slope of the line is } y=2 \times x-3$

$\text { so, the slope is } 2 \rightarrow(2)$

$\text { from (1) \& (2) }$

\begin{aligned} &3 x^{2}-4 x-2=2 \\ &3 x^{2}-4 x=4 \\ &3 x^{2}-4 x-4=0 \end{aligned}

$\text { We will use factorization method to solve the above quadratic equation. }$

\begin{aligned} &3 x^{2}-6 x+2 x-4=0 \\ &3 x(x-2)+2(x-2)=0 \\ &(x-2)(3 x+2)=0 \end{aligned}

$x-2=0 \quad \& \quad 3 x+2=0$

$x=2 \quad \text { or } \quad x=\frac{-2}{3}$

$\text { Substitute } x=2 \& x=\frac{-2}{3} \text { in } y=x^{3}-2 x^{2}-2 x$

$\text { When } x=2$

\begin{aligned} &y=(2)^{3}-2 \times(2)^{2}-2 \times(2) \\ &y=8-2 \times 4-4 \\ &y=8-8-4 \\ &y=-4 \end{aligned}

$\text { When } x=\frac{-2}{3}$

$y=\left(\frac{-2}{3}\right)^{3}-2\left(\frac{-2}{3}\right)^{2}-2\left(\frac{-2}{3}\right)$

$y=\frac{-8}{27}-2 \times \frac{4}{9}+\frac{4}{3}$

$y=\frac{-8}{27}-2 \times \frac{4}{9}+\frac{4}{3}$

$y=\frac{-8}{27}-\frac{8}{9}+\frac{4}{3}$

$\text { L.C.M of } 27,3 \text { and } 9 \text { is } 27$

$y=\frac{(-8 \times 1)-(8 \times 3)+(4 \times 9)}{27}$

$y=\frac{-8-24+36}{27}=\frac{4}{27}$

$\text { Thus, the points are }(2,-4) \&\left(\frac{-2}{3}, \frac{4}{27}\right)$