If f(x) = begin{cases} frac{1-sqrt{2} sin x}{ | vedclass

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

Question

(D) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x 	o \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}ight)$.Given $f\left(\fra

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(D) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x 	o \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}ight)$.Given $f\left(\frac{\pi}{4}ight) = a$.Now,$\lim_{x 	o \frac{\pi}{4}} f(x) = \lim_{x 	o \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4x}$.This is a $\frac{0}{0}$ form,so we apply $L$'Hospital's rule:$\lim_{x 	o \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4x)} = \lim_{x 	o \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$.Substituting $x = \frac{\pi}{4}$:$= \frac{-\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.Since the function is continuous,$a = \frac{1}{4}$.

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

Similar Questions

$19 \ g$ of a mixture containing $NaHCO_3$ and $Na_2CO_3$ on complete heating liberated $1.12 \ L$ of $CO_2$ at $STP$. The weight of the remaining solid was $15.9 \ g$. What is the weight (in $g$) of $Na_2CO_3$ in the mixture before heating?

$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow [0,1]$ be defined by $f(x) = \begin{cases} x & \text{for } x \in Q \\ 1-x & \text{for } x \notin Q \end{cases}$. Then,the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is equal to

The angle between the curves $y^2=4x+4$ and $y^2=36(9-x)$ is (in $^{\circ}$)

Define/Explain: $\alpha$-diversity.

Vedclass Test Series

Exam Paper Generator

Online Exam Module