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  • Explain solution RD Sharma class 12 Chapter 15 Tangents and Normals exercise Fill in the blanks question 3

Explain solution RD Sharma class 12 Chapter 15 Tangents and Normals exercise Fill in the blanks question 3

Answers (1)

Answer:

                \left ( 2,16 \right )\left ( -2,-16 \right )

Hint:

Gradient is zero. i.e. \frac{dy}{dx }=0

Given:

Given curve, y=12x-x^{3}

To find:

We have to find the points on the given curve at which the gradient is zero.

Solution:

Here      y=12x-x^{3}                                                                                                                                      … (i)

Given that gradient is zero

                i.e. \frac{dy}{dx }=0

Differentiating equation (i), we get

\Rightarrow \quad \frac{d y}{d x}=12-3 x^{2}

Since  \frac{dy}{dx }=0

\begin{array}{lrl} \Rightarrow & & 12-3 x^{2}=0 \\\\ \Rightarrow & & -3 x^{2}=-12 \\\\ \Rightarrow & & x^{2}=4 \\\\ \Rightarrow & & x=\pm 2 \end{array}

Putting the value x=\pm 2 in equation (i), we get

                \begin{aligned} &y=12(2)-(2)^{3} \text { or } y=12(-2)-(-2)^{3} \\\\ &y=16 \text { or } y=-16 \end{aligned}

Hence the points are  \left ( 2,16 \right )  and  \left (- 2,-16 \right )

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