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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 11 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 10 Maths Textbook Solution.

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Answer:\text { The required points are }\left(\frac{5}{3}, \frac{4}{3}\right) \text { or }\left(\frac{4}{3}, \frac{4}{3}\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 { The curve is } y=3 x^{2}-9 x+8

Solution: y=3 x^{2}-9 x+8

\text { Differentiating the above with respect to } x

\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0

\frac{d y}{d x}=3(2) x^{2-1}-9

\frac{d y}{d x}=6 x-9 \quad \rightarrow(1)

\frac{d y}{d x}=\text { The slope of tangent }=\tan \theta

\text { The tangent is equally inclined with the axis } \theta=\frac{\pi}{4} \text { or } \frac{-\pi}{4}

\frac{d y}{d x}=\tan \left(\frac{\pi}{4}\right) \text { or } \tan \left(-\frac{\pi}{4}\right)

=1 \text { or }-1 \rightarrow(2)

From(1)&(2) we get

6 x-9=1 \text { or } 6 x-9=-1

6 x=1+9 \text { or } 6 x=-1+9

6 x=10 \quad \text { or } 6 x=8

x=\frac{10}{6} \quad \text { or } \quad x=\frac{8}{6}

x=\frac{5}{3} \quad \text { or } \quad x=\frac{4}{3}

Thus the requried points are\left(\frac{5}{3}, \frac{4}{3}\right) \text { or }\left(\frac{4}{3}, \frac{4}{3}\right)

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