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Name: Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 5 Maths Textbook Solution.
Uploaded: Thu, 2021-08-12 14:34
Description: Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 5 Maths Textbook Solution.

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

Answers (1)

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

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

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