For the function f(x) = e^{sin |x|} - |x|, x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The function is $f(x) = e^{\sin |x|} - |x|$.
First, check the differentiability at $x = 0$. The function $f(x)$ contains the term $|x|$, which is

Vedclass · Accepted Answer

Statement $I$ is false but Statement $II$ is true

Vedclass · Answer

(D) The function is $f(x) = e^{\sin |x|} - |x|$.
First, check the differentiability at $x = 0$. The function $f(x)$ contains the term $|x|$, which is not differentiable at $x = 0$. Therefore, $f(x)$ is not differentiable at $x = 0$. Thus, Statement $I$ is false.
Next, consider Statement $II$ in the interval $(-\pi, -\frac{\pi}{2})$. For $x < 0$, we have $|x| = -x$. Thus, $f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x$.
The derivative is $f'(x) = e^{-\sin x}(-\cos x) + 1 = 1 - e^{-\sin x}\cos x$.
For $x \in (-\pi, -\frac{\pi}{2})$, we have $\sin x \in (-1, 0)$ and $\cos x \in (-1, 0)$.
Since $\sin x$ is negative, $-\sin x$ is positive, so $e^{-\sin x} > e^0 = 1$. Also, $\cos x$ is negative.
Thus, $-e^{-\sin x}\cos x$ is positive. Therefore, $f'(x) = 1 + (	ext{positive value}) > 0$.
Since $f'(x) > 0$, the function $f(x)$ is strictly increasing in the interval $(-\pi, -\frac{\pi}{2})$. Thus, Statement $II$ is true.

Similar Questions

If $f(x) = \begin{cases} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then the correct statement is:

If $f: R \rightarrow R$ is defined as $f(x)=|x+1|+|x-1|$, then $f(x)$ is

If $f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \leq x \leq a \\ \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a < x \leq b \\ \frac{1}{3}\left(\frac{b^3 - a^3}{x}\right), & x > b \end{cases}$,then which of the following is true?

Assertion $(A)$: $f(x) = |x|$ is differentiable at $x = a \neq 0$ and continuous but not differentiable at $x = 0$.
Reason $(R)$: If a function is differentiable at a point,then it is continuous at the point. But the converse is not true.

Number of points where the function $f(x) = \text{maximum}(\sqrt{2x - x^2}, 2 - x)$ is non-differentiable is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module