If the function f(x) = frac{2sqrt{2} - (cos x | vedclass

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

Question

(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Let $t = x - \frac{\pi}{4}$, so as $x 	o

Vedclass · Accepted Answer

$3\sqrt{2}$

Vedclass · Answer

(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Let $t = x - \frac{\pi}{4}$, so as $x 	o \frac{\pi}{4}$, $t 	o 0$. Then $x = t + \frac{\pi}{4}$.$\cos x + \sin x = \sqrt{2} \sin(x + \frac{\pi}{4}) = \sqrt{2} \sin(t + \frac{\pi}{2}) = \sqrt{2} \cos t$.$1 - \sin 2x = 1 - \sin(2t + \frac{\pi}{2}) = 1 - \cos 2t = 2 \sin^2 t$.$f(x) = \frac{2\sqrt{2} - (\sqrt{2} \cos t)^3}{2 \sin^2 t} = \frac{2\sqrt{2} - 2\sqrt{2} \cos^3 t}{2 \sin^2 t} = \sqrt{2} \frac{1 - \cos^3 t}{\sin^2 t}$.Using $1 - \cos^3 t = (1 - \cos t)(1 + \cos t + \cos^2 t)$ and $\sin^2 t = (1 - \cos t)(1 + \cos t)$:$f(x) = \sqrt{2} \frac{(1 - \cos t)(1 + \cos t + \cos^2 t)}{(1 - \cos t)(1 + \cos t)} = \sqrt{2} \frac{1 + \cos t + \cos^2 t}{1 + \cos t}$.As $t 	o 0$, $\cos t 	o 1$, so $f(\frac{\pi}{4}) = \sqrt{2} \frac{1 + 1 + 1}{1 + 1} = \sqrt{2} \cdot \frac{3}{2} = \frac{3\sqrt{2}}{2}$.Note: The provided options do not contain $\frac{3\sqrt{2}}{2}$. Re-evaluating the limit: $\lim_{t 	o 0} \sqrt{2} \frac{1 + \cos t + \cos^2 t}{1 + \cos t} = \frac{3\sqrt{2}}{2}$. Since this is not in the options, the question or options may be flawed.

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...

Similar Questions

The number of points at which the function $f(x) = \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}}{6-2\sqrt{2+|x|}}$ is discontinuous in $(-\infty, \infty)$ is

The function $f(x) = p[x + 1] + q[x - 1],$ where $[x]$ is the greatest integer function,is continuous at $x = 1$ if:

For the function $f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,which of the following is correct?

Let $f(x) = \frac{g(x)}{h(x)}$,where $g$ and $h$ are continuous functions on the open interval $(a, b)$. Which of the following statements is true for $a < x < b$?

If $f(x) = \frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}, x \neq 0$ is continuous at $x=0$, then the value of $f(0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module