Let f and g be real-valued functions. If lim | vedclass

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

Question

(D) Given that $\lim _{x \rightarrow 0} f(x)=1$, $\lim _{x \rightarrow 0} g(x)=\alpha$ and $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{2 / 3

Vedclass · Accepted Answer

continuous on $\left[0, \frac{\pi}{5}\right]$

Vedclass · Answer

(D) Given that $\lim _{x ightarrow 0} f(x)=1$, $\lim _{x ightarrow 0} g(x)=\alpha$ and $\lim _{x ightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{2 / 3}}=\frac{7}{4}$.Substituting the limits, we get $\frac{2(1)-\alpha}{(1+7)^{2 / 3}}=\frac{7}{4}$.$\Rightarrow \frac{2-\alpha}{8^{2 / 3}}=\frac{7}{4} \Rightarrow \frac{2-\alpha}{4}=\frac{7}{4} \Rightarrow 2-\alpha=7 \Rightarrow \alpha=-5$.Now, $h(x)= \begin{cases} \sin (-5x), & 0 \leq x \leq \frac{\pi}{10} \ \cos (-10x), & \frac{\pi}{10} Since $\sin (-5x)$ and $\cos (-10x)$ are continuous functions, we only need to check continuity at $x=\frac{\pi}{10}$.$LHL = \lim _{x ightarrow \frac{\pi}{10}^-} \sin (-5x) = \sin \left(-\frac{5\pi}{10}ight) = \sin \left(-\frac{\pi}{2}ight) = -1$.$h\left(\frac{\pi}{10}ight) = \sin \left(-\frac{5\pi}{10}ight) = -1$.$RHL = \lim _{x ightarrow \frac{\pi}{10}^+} \cos (-10x) = \cos \left(-\frac{10\pi}{10}ight) = \cos (-\pi) = -1$.Since $LHL = RHL = h\left(\frac{\pi}{10}ight)$, the function $h(x)$ is continuous at $x=\frac{\pi}{10}$.Thus, $h(x)$ is continuous on $\left[0, \frac{\pi}{5}ight]$.

Similar Questions

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

Examine the following function for continuity: $f(x) = \frac{1}{x-5}, x \neq 5$.

Let $f(x) = \begin{cases} 0, & x=0 \\ 2-x, & 0 < x < 1 \\ 2, & x=1 \\ \frac{1}{2}-x, & 1 < x < 2 \\ \frac{-3}{2}, & x \geq 2 \end{cases}$ then which of the following is true?

If $f(x) = \frac{\log_{\sin |x|} \cos^3 x}{\log_{\sin |3x|} \cos^3 (x/2)}$ for $|x| < \frac{\pi}{3}, x \neq 0$ and $f(0) = 4$,then the number of points of discontinuity of $f$ in $\left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module