#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 20 Maths Textbook Solution.

Hence Proved that the tangents to the curve $y=7 x^{3}+11 \text { at the points } x=2 \text { and } x=-2 \text { are parallel }$

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

Given: $\text { The equation of the curve is } y=7 x^{3}+11 \rightarrow(1)$

Solution$\text { Differentiating eqn(1) with respect to } x$

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0 \\ &\frac{d y}{d x}=7(3) x^{3-1}+0 \\ &\frac{d y}{d x}=21 x^{2} \end{aligned}

$\text { The slope of the tangent to a curve at }\left(x_{0}, y_{0}\right) \text { is } \frac{d y}{d x\left(x_{0}, y_{0}\right)}$

$\text { Therefore the slope of the tangent at the point where } x=2 \text { is given by }$

$\frac{d y}{d x_{x=2}}=21(2)^{2}$

\begin{aligned} &=21 \times 4 \\ &=84 \end{aligned}

$\text { and the slope of the tangent at the point where } x=-2 \text { is given by }$

$\frac{d y}{d x_{x}=-2}=21(-2)^{2}$

\begin{aligned} &=21 \times 4 \\ &=84 \end{aligned}

It is observed that the slopes of the tangent at the points where x=2 and x=-2 are equal.
Hence, the two tangents are parallel