If f(x) = begin{cases} frac{1-sin^3 x}{3 cos^ | vedclass

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

Question

(C) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $LHL = RHL = f(\frac{\pi}{2})$.First,calculate $LHL$:$LHL = \lim_{x 	o \frac{\pi^-}{2}}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $LHL = RHL = f(\frac{\pi}{2})$.First,calculate $LHL$:$LHL = \lim_{x 	o \frac{\pi^-}{2}} \frac{1-\sin^3 x}{3 \cos^2 x}$. This is a $\frac{0}{0}$ form.Using $L$'Hospital's Rule:$\lim_{x 	o \frac{\pi^-}{2}} \frac{-3 \sin^2 x \cos x}{3(2 \cos x)(-\sin x)} = \lim_{x 	o \frac{\pi^-}{2}} \frac{\sin x}{2} = \frac{1}{2}$.Thus,$\alpha = \frac{1}{2}$.Next,calculate $RHL$:$RHL = \lim_{x 	o \frac{\pi^+}{2}} \frac{\beta(1-\sin x)}{(\pi-2 x)^2} = \frac{1}{2}$.Let $x = \frac{\pi}{2} + h$,where $h 	o 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = -2h$.$\lim_{h 	o 0} \frac{\beta(1-\sin(\frac{\pi}{2} + h))}{(-2h)^2} = \lim_{h 	o 0} \frac{\beta(1-\cos h)}{4h^2} = \frac{1}{2}$.Using the limit formula $\lim_{h 	o 0} \frac{1-\cos h}{h^2} = \frac{1}{2}$:$\frac{\beta}{4} 	imes \frac{1}{2} = \frac{1}{2} \implies \frac{\beta}{8} = \frac{1}{2} \implies \beta = 4$.Therefore,$\alpha \beta = \frac{1}{2} 	imes 4 = 2$.

If $f(x) = \begin{cases} \frac{1-\sin^3 x}{3 \cos^2 x}, & x < \frac{\pi}{2} \\ \alpha, & x = \frac{\pi}{2} \\ \frac{\beta(1-\sin x)}{(\pi-2 x)^2}, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $\alpha \beta =$

Similar Questions

If $f(x) = \begin{cases} e^x; & x \le 0 \\ |1 - x|; & x > 0 \end{cases}$,then

Let $f(x) = x \left[ \frac{x}{2} \right]$,for $-10 < x < 10$,where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

If the function $f(x) = \begin{cases} a|\pi - x| + 1, & x \le 5 \\ b|\pi - x| + 3, & x > 5 \end{cases}$ is continuous at $x = 5$,then the value of $a - b$ is

Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the value of $\alpha$ for which the function $f(x)=\begin{cases} \frac{\sin [-x^2]}{[-x^2]}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$ is:

Let $f:[-1,2] \rightarrow \mathbb{R}$ be given by $f(x)=2x^2+x+[x^2]-[x]$,where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points where $f$ is not continuous is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module