SATHEE: Application of Derivatives 1 Question 12

Application of Derivatives 1 Question 12

####12. If the normal to the curve $y = f (x)$ at the point $(3, 4)$ makes an angle $\frac{3 π}{4}$ with the positive $X$ -axis, then $f^{'} (3)$ is equal to

(2000, 1M)

(a) -1

(b) $- 3 / 4$

(d) 1

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Answer:

Correct Answer: 12. $y + x - 1 = 0$

Solution:

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

Therefore, slope of normal

$\begin{aligned} = - \frac{1}{f^{'} (x)_{(3, 4)}} & = - \frac{1}{f^{'} (3)} \\ But & - \frac{1}{f^{'} (3)} & = \tan \frac{3 π}{4} & [g i v e n] \\ \Rightarrow & \frac{- 1}{f^{'} (3)} & = \tan \frac{π}{2} + \frac{π}{4} = - 1 \\ ∴ & f^{'} (3) = 1 \end{aligned}$

Application of Derivatives 1 Question 12

Answer:

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Application of Derivatives 1 Question 12

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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