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Answer:

$f^{\prime}(3)=1$

Hint:

First find the slope of given curve at point $\left ( 3,4 \right )$ , then compare with$\frac{3\pi}{4}$

Given:

Given curve,

$y=f(x)$

To find:

We have to find ${f}'\left ( 3 \right )$ , if the normal to the given curve at $\left ( 3,4 \right )$ makes an angle $\frac{3\pi}{4}$ with positive x-axis.

Solution:

Here we have,

$y=f(x)$

On differentiating with respect to $x$ , we get

$\Rightarrow \quad \frac{d y}{d x}=f^{\prime}(x)$

Slope of tangent at  $(3,4)=\frac{d y}{d x}=f^{\prime}(x)_{(3,4)}$

Therefore, slope of normal

\begin{aligned} &=\frac{-1}{f^{\prime}(x)_{(3.4)}} \\ &=\frac{-1}{f^{\prime}(3)} \end{aligned}

But  $\frac{-1}{f^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)$      [Given]

\begin{aligned} &\Rightarrow \quad \frac{-1}{f^{\prime}(3)}=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\\\ &\Rightarrow \quad \frac{-1}{f^{\prime}(3)}=-1 \end{aligned}

Hence   $f^{\prime}(3)=1$

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