If the normal to the curve y=f(x) at (1, 2) make an angle 3π4 with positive x-axis, then f'(1)=

y=f(x) dydx=f1(x)dydx1−2=f1(x)Slope of normal =−1f(x)Normal of curve makes angle 3π4 with positive direction x- axis∴tan3π4=−1f1(1)⇒−1=−1f1(1) =f1(1)=1

If the normal to the curve y=f(x) at (1, 2) make an angle 3π4 with positive x-axis, then f'(1)=

y=f(x) dydx=f1(x)dydx1−2=f1(x)Slope of normal =−1f(x)Normal of curve makes angle 3π4 with positive direction x- axis∴tan3π4=−1f1(1)⇒−1=−1f1(1) =f1(1)=1

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If the normal to the curve $y = f (x)$ at (1, 2) make an angle $\frac{3 π}{4}$ with positive x-axis, then $f' (1) =$

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answer is B.

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Detailed Solution

$y = f (x) \begin{array}{l} \frac{d y}{d x} = f^{1} (x) \\ {(\frac{d y}{d x})}_{1 - 2} = f^{1} (x) \end{array}$
Slope of normal $= \frac{- 1}{f (x)}$
Normal of curve makes angle $\frac{3 π}{4}$ with positive direction x- axis
$\begin{array}{l} ∴ \tan \frac{3 π}{4} = \frac{- 1}{f^{1} (1)} \Rightarrow - 1 = \frac{- 1}{f^{1} (1)} \\ = f^{1} (1) = 1 \end{array}$