SATHEE: Limit Continuity and Differentiability 7 Question 17

Limit Continuity and Differentiability 7 Question 17

17. The domain of the derivative of the functions $\tan^{- 1} x,$ if $| x | \leq 1$ $f (x) = \frac{1}{2} (| x | - 1),$ if $| x | > 1$ is

(2002, 2M)

(a) $R - 0$

(b) $R - 1$

(d) $R - - 1, 1$

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Answer:

Correct Answer: 17. (c)

Solution:

Given that, $\log (x + y) = 2 x y$

$∴$ At $x = 0, \Rightarrow \log (y) = 0 \Rightarrow y = 1$

$∴$ To find $\frac{d y}{d x}$ at $(0, 1)$

On differentiating Eq. (i) w.r.t. $x$ , we get

$\begin{aligned} \frac{1}{x + y} 1 + \frac{d y}{d y} = 2 x \frac{d y}{d x} + 2 y \cdot 1 \\ \Rightarrow \frac{d y}{d x} = \frac{2 y (x + y) - 1}{1 - 2 (x + y) x} \\ \Rightarrow {\frac{d y}{d x}}_{(0, 1)} = 1 \end{aligned}$

Limit Continuity and Differentiability 7 Question 17

17. The domain of the derivative of the functions $\tan^{- 1} x,$ if $| x | \leq 1$ $f (x) = \frac{1}{2} (| x | - 1),$ if $| x | > 1$ is

Answer:

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Limit Continuity and Differentiability 7 Question 17

17. The domain of the derivative of the functions tan−1⁡x, if |x|≤1 f(x)=12(|x|−1), if |x|>1 is

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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17. The domain of the derivative of the functions $\tan^{- 1} x,$ if $| x | \leq 1$ $f (x) = \frac{1}{2} (| x | - 1),$ if $| x | > 1$ is