SATHEE: Limit Continuity and Differentiability 7 Question 5

Limit Continuity and Differentiability 7 Question 5

5. Let $K$ be the set of all real values of $x$ , where the function $f (x) = \sin | x | - | x | + 2 (x - π) \cos | x |$ is not differentiable. Then, the set $K$ is equal to

(2019 Main, 11 Jan II)

(a) $0$

(b) $φ$ (an empty set)

(d) $0, π$

where, $a$ and $b$ are non-negative real numbers. Determine the compositie function gof. If $(g \circ f) (x)$ is continuous for all real $x$ determine the values of $a$ and $b$ . Further, for these values of $a$ and $b$ , is gof differentiable at $x = 0$ ? Justify your answer.

$(2002, 5$ M)

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Answer:

Correct Answer: 5. (a)

Solution:

Given expression is

$2 y = \cot^{- 1} \frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} = \cot^{- 1} \frac{\sqrt{3} \cot x + 1}{\cot x - \sqrt{3}}$

[dividing each term of numerator and denominator by $\sin x]$

$\begin{aligned} = \cot^{- 1} \frac{\cot \frac{π}{6} \cot x + 1}{\cot x - \cot \frac{π}{6}} \\ = \cot^{- 1} \cot \frac{π}{6} - x^{2} ∵ \cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A} \end{aligned}$

$\begin{aligned} = \frac{π}{6} - x^{2}, 0 < x < \frac{π}{6} \\ π + \frac{π}{6} - x, \frac{π}{6} < x < \frac{π}{2} \\ ∵ \cot^{- 1} (\cot θ) = \begin{array}{cc} π + θ, & - π < θ < 0 \\ θ, & 0 < θ < π \\ θ - π, & π < θ < 2 π \end{array} \\ \Rightarrow 2 y = \begin{matrix} \frac{π}{6} - x^{2}, 0 < x < \frac{π}{6} \\ \frac{7 π}{6} - x^{2}, \frac{π}{6} < x < \frac{π}{2} \end{matrix} \\ \Rightarrow 2 \frac{d y}{d x} = \begin{array}{ll} 2 \frac{π}{6} - x (- 1), & 0 < x < \frac{π}{6} \\ 2 \frac{7 π}{6} - x (- 1), & \frac{π}{6} < x < \frac{π}{2} \end{array} \\ \Rightarrow \frac{d y}{d x} = \begin{matrix} x - \frac{π}{6}, 0 < x < \frac{π}{6} \\ x - \frac{7 π}{6}, \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{aligned}$

Limit Continuity and Differentiability 7 Question 5

5. Let $K$ be the set of all real values of $x$ , where the function $f (x) = \sin | x | - | x | + 2 (x - π) \cos | x |$ is not differentiable. Then, the set $K$ is equal to

Answer:

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Limit Continuity and Differentiability 7 Question 5

5. Let K be the set of all real values of x, where the function f(x)=sin⁡|x|−|x|+2(x−π)cos⁡|x| is not differentiable. Then, the set K is equal to

Answer:

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5. Let $K$ be the set of all real values of $x$ , where the function $f (x) = \sin | x | - | x | + 2 (x - π) \cos | x |$ is not differentiable. Then, the set $K$ is equal to