Consider f(x) = left[ frac{2(sin x - sin^3 x) | vedclass

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

Question

(A) Let $g(x) = \sin x - \sin^3 x = \sin x(1 - \sin^2 x) = \sin x \cos^2 x$.For $x \in (0, \pi)$,$\sin x > 0$ and $\cos^2 x \geq 0$. Thus,$g(x) \geq 0

Vedclass · Accepted Answer

$f$ is continuous and differentiable at $x = \frac{\pi}{2}$

Vedclass · Answer

(A) Let $g(x) = \sin x - \sin^3 x = \sin x(1 - \sin^2 x) = \sin x \cos^2 x$.For $x \in (0, \pi)$,$\sin x > 0$ and $\cos^2 x \geq 0$. Thus,$g(x) \geq 0$.Specifically,for $x 
eq \frac{\pi}{2}$,$g(x) > 0$.Since $g(x) > 0$ for $x \in (0, \pi) \setminus \{\frac{\pi}{2}\}$,we have $|g(x)| = g(x)$.Substituting this into the expression for $f(x)$:$f(x) = \left[ \frac{2g(x) + g(x)}{2g(x) - g(x)} ight] = \left[ \frac{3g(x)}{g(x)} ight] = [3] = 3$.Thus,$f(x) = 3$ for all $x \in (0, \pi) \setminus \{\frac{\pi}{2}\}$.Given $f(\frac{\pi}{2}) = 3$,the function is $f(x) = 3$ for all $x \in (0, \pi)$.$A$ constant function is continuous and differentiable everywhere in its domain.Therefore,$f$ is continuous and differentiable at $x = \frac{\pi}{2}$.

Consider $f(x) = \left[ \frac{2(\sin x - \sin^3 x) + |\sin x - \sin^3 x|}{2(\sin x - \sin^3 x) - |\sin x - \sin^3 x|} \right]$ for $x \in (0, \pi), x \neq \frac{\pi}{2}$,and $f(\frac{\pi}{2}) = 3$,where $[ \cdot ]$ denotes the greatest integer function. Then:

Similar Questions

If $f(x) = \begin{cases} \frac{81^{x}-9^{x}}{k^{x}-1} & x \neq 0 \\ 2 & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

Suppose $f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}$ and if $\lim_{x \to 1} f(x) = f(1)$,what are the possible values of $a$ and $b$?

If $f(x) = \begin{cases} \frac{1 - \cos x}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = $

If $f(x) = \begin{cases} \frac{\sqrt{1+mx} - \sqrt{1-mx}}{x}, & -1 \le x < 0 \\ \frac{2x+1}{x-2}, & 0 \le x \le 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $m$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module