Q68 / 125 · BITSAT 2012 · Mathematics

If the normal to the curve $y=f(x)$ at the point (3,4) makes an angle $3 \pi / 4$ with the positive $\mathrm{x}$ -axis, then $\mathrm{f}^{\prime}(3)=$

  • A
    -1
  • B
    $-3 / 4$
  • C
    $4 / 3$
  • Correct answer
    D
    1

Solution

Slope of normal to $y=f(x)$ at (3,4) is $\frac{-1}{f^{\prime}(3)}$. $\text { Thus, } \begin{array}{c} \frac{-1}{\mathrm{f}^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\ =-\cot \frac{\pi}{4}=-1 \Rightarrow \mathrm{f}^{\prime}(3)=1 \end{array}$

Hence, the correct answer is option: 'D'.

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