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  • Need solution for RD Sharm Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 19

Need solution for RD Sharm Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 19

Answers (1)

Answer:

                \left ( 3,34 \right )

Hint:

When both curves touch then slope of both curves should be same.

Given:

Given curves,

                y=4 x^{2}+2 x-8  and  y=x^{3}-x+13  touch each other.

To find:

We have to find the point where the given curves touch each.

Solution:

The curves, y=4 x^{2}+2 x-8  and  y=x^{3}-x+13

For first curve say \left(\frac{d y}{d x}\right)_{1}

\therefore \left(\frac{d y}{d x}\right)_{1}=8x+2

For second curve  \left(\frac{d y}{d x}\right)_{2}=3 x^{2}-1

When both curves touch the slope both curves should be same

\begin{aligned} &\therefore \quad 8 x+2=3 x^{2}-1 \\ &\Rightarrow \quad 3 x^{2}-8 x-3=0 \end{aligned}

Solving the quadratic equation, we get

\begin{array}{ll} \Rightarrow \quad & (3 x+1)(x-3)=0 \\ \end{array}

\Rightarrow \quad x=\frac{-1}{3}, x=3

Now consider, x=3

For first curve,

                y\left ( 3 \right )=4\left ( 3 \right )^{2}+2\left ( 3 \right )-8=34

For second curve,

                y\left ( 3 \right )=\left ( 3 \right )^{3}-3+10=34   …(here there should be 10 inplace of 13)

Thus at  \left ( 3,34 \right )  both curves touch

Now consider,  x=\frac{-1}{3}

For first curve,

                y\left(\frac{-1}{3}\right)=4\left(\frac{-1}{3}\right)^{2}+2\left(\frac{-1}{3}\right)-5=\frac{-47}{9}

For second curve,

                y\left(\frac{-1}{3}\right)=\left(\frac{-1}{3}\right)^{3}-\left(\frac{-1}{3}\right)+13=\frac{8}{27}+13

Thus at  x=\frac{-1}{3}  both curves do not meet.

But their tangent are parallel

Hence the only point where both curves touch is  \left ( 3,34 \right )

Posted by

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