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

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

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Answer:

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
 

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