SATHEE: Limit Continuity and Differentiability 7 Question 10

Limit Continuity and Differentiability 7 Question 10

10. Let $S = (t \in R : f (x) = | x - π | \cdot (e^{| x |} - 1) \sin | x |$ is not differentiable at $Extra close brace or missing open brace$ .Then, the set $S$ is equal to (2018 Main)

(a) $φ$ (an empty set)

(b) $0$

(d) $0, π$

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Answer:

Correct Answer: 10. (d)

Solution:

Let $y = \tan^{- 1} \frac{6 x \sqrt{x}}{1 - 9 x^{3}} = \tan^{- 1} \frac{2 \cdot (3 x^{3 / 2})}{1 - {(3 x^{3 / 2})}^{2}}$

$\begin{aligned} = 2 \tan^{- 1} (3 x^{3 / 2}) ∵ 2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}} \\ ∴ \frac{d y}{d x} & = 2 \cdot \frac{1}{1 + {(3 x^{3 / 2})}^{2}} \cdot 3 \times \frac{3}{2} (x)^{1 / 2} = \frac{9}{1 + 9 x^{3}} \cdot \sqrt{x} \\ ∴ g (x) & = \frac{9}{1 + 9 x^{3}} \end{aligned}$

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Limit Continuity and Differentiability 7 Question 10

10. Let S=(t∈R:f(x)=|x−π|⋅(e|x|−1)sin⁡|x| is not differentiable at Extra close brace or missing open braceExtra close brace or missing open brace.Then, the set S is equal to (2018 Main)

Answer:

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10. Let $S = (t \in R : f (x) = | x - π | \cdot (e^{| x |} - 1) \sin | x |$ is not differentiable at $Extra close brace or missing open brace$ .Then, the set $S$ is equal to (2018 Main)