If the function f(x),defined below,is continu | vedclass

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

Question

(D) Since $f(x)$ is continuous in $[0, \pi]$,it must be continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.At $x = \frac{\pi}{4}$,$\lim_{x 	o

Vedclass · Accepted Answer

$a = \frac{\pi}{6}, b = \frac{-\pi}{12}$

Vedclass · Answer

(D) Since $f(x)$ is continuous in $[0, \pi]$,it must be continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.At $x = \frac{\pi}{4}$,$\lim_{x 	o \frac{\pi}{4}^-} f(x) = \lim_{x 	o \frac{\pi}{4}^+} f(x) = f(\frac{\pi}{4})$.$\frac{\pi}{4} + a\sqrt{2}(\sin \frac{\pi}{4}) = 2(\frac{\pi}{4})(\cot \frac{\pi}{4}) + b$$\frac{\pi}{4} + a = \frac{\pi}{2} + b \Rightarrow a - b = \frac{\pi}{4} \dots (I)$.At $x = \frac{\pi}{2}$,$\lim_{x 	o \frac{\pi}{2}^-} f(x) = \lim_{x 	o \frac{\pi}{2}^+} f(x) = f(\frac{\pi}{2})$.$2(\frac{\pi}{2})(\cot \frac{\pi}{2}) + b = a(\cos \pi) - b(\sin \frac{\pi}{2})$.Since $\cot \frac{\pi}{2} = 0$,we get $b = -a - b \Rightarrow a = -2b \dots (II)$.Substituting $(II)$ into $(I)$: $-2b - b = \frac{\pi}{4} \Rightarrow -3b = \frac{\pi}{4} \Rightarrow b = -\frac{\pi}{12}$.Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

Similar Questions

Given $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x > 0 \end{cases}$
If $f(x)$ is continuous at $x=0$,then the value of $a$ is:

If $f(x) = \frac{10^x + 7^x - 14^x - 5^x}{1 - \cos x}$ for $x \neq 0$ is continuous at $x = 0$,then the value of $f(0)$ is:

If the function $f(x) = \frac{2\sqrt{2} - (\cos x + \sin x)^3}{1 - \sin 2x}$ is continuous at $x = \frac{\pi}{4}$, then the value of $f(\frac{\pi}{4})$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module