Let f(x) = begin{cases} x+a sqrt{2} sin 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 $x = \frac{\pi}{4}$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{4}^-} (x +

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \frac{\pi}{4} + a \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} + a$$\lim_{x \rightarrow \frac{\pi}{4}^+} (2x \cot x + b) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b$Equating them: $\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4} \quad \dots(1)$For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{2}^-} (2x \cot x + b) = 2\left(\frac{\pi}{2}\right) \lim_{x \rightarrow \frac{\pi}{2}^-} \cot x + b = \pi(0) + b = b$$\lim_{x \rightarrow \frac{\pi}{2}^+} (a \cos 2x - b \sin x) = a \cos(\pi) - b \sin\left(\frac{\pi}{2}\right) = -a - b$Equating them: $b = -a - b \implies a + 2b = 0 \implies a = -2b \quad \dots(2)$Substitute $(2)$ into $(1)$: $-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.

Let $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x+b, & \frac{\pi}{4} \leq x < \frac{\pi}{2} \\ a \cos 2x-b \sin x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$. If $f(x)$ is continuous for $0 \leq x \leq \pi$,then:

Similar Questions

The function $f : R \rightarrow R$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$ is continuous for all $x$ in.

Let $f : [-1,3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + |x|, & 2 \leq x \leq 3 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is discontinuous at:

Find all the points of discontinuity of the greatest integer function defined by $f(x) = [x]$,where $[x]$ denotes the greatest integer less than or equal to $x$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is

In order that the function $f(x) = (x + 1)^{\cot x}$ is continuous at $x = 0$,$f(0)$ must be defined as

Vedclass Test Series

Exam Paper Generator

Online Exam Module