The values of a and b,so that the function f( | vedclass

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

Question

(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,we have:At $x = \frac{\pi}{4}$:$\lim_{x 	o \frac{\pi}{4}^-} (x +

Vedclass · Accepted Answer

$\frac{\pi}{6}, -\frac{\pi}{12}$

Vedclass · Answer

(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,we have:At $x = \frac{\pi}{4}$:$\lim_{x 	o \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x 	o \frac{\pi}{4}^+} (2x \cot x + b)$$\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{\pi}{4} \cdot 1 + b$$\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4}$ . . . $(i)$At $x = \frac{\pi}{2}$:$\lim_{x 	o \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x 	o \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$$2 \cdot \frac{\pi}{2} \cdot 0 + b = a \cos \pi - b \sin \frac{\pi}{2}$$b = -a - b \implies a + 2b = 0$ . . . $(ii)$Solving $(i)$ and $(ii)$:From $(ii)$,$a = -2b$. Substituting in $(i)$:$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.Thus,$a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$.

Similar Questions

If $f(x) = \begin{cases} \frac{|x-2|}{x-2}, & x \neq 2 \\ 1, & x = 2 \end{cases}$,then which of the following statements is true?

The function $f(x) = \frac{x+1}{9x+x^3}$ is

If a function $f(x) = \begin{cases} \frac{\tan (\alpha + 1)x + \tan 2x}{x}, & \text{if } x > 0 \\ \beta, & \text{at } x = 0 \\ \frac{\sin 3x - \tan 3x}{x^{3}}, & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $|\alpha| + |\beta| =$

Let $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{cases}$. Then at $x=0$:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module