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

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                \left ( 3,34 \right )


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


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.


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,


For second curve,


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 )

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