If f(x)= begin{cases}-2 sin x & , quad x leqs | 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 $x = -\frac{\pi}{2}$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o -\frac{\pi}{2}^-} (-2 \sin x

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) For $f(x)$ to be continuous at $x = -\frac{\pi}{2}$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o -\frac{\pi}{2}^-} (-2 \sin x) = \lim_{x 	o -\frac{\pi}{2}^+} (a \sin x + b)$ $-2 \sin(-\frac{\pi}{2}) = a \sin(-\frac{\pi}{2}) + b$ $-2(-1) = a(-1) + b \implies 2 = -a + b \quad (1)$ For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o \frac{\pi}{2}^-} (a \sin x + b) = \lim_{x 	o \frac{\pi}{2}^+} (\cos x)$ $a \sin(\frac{\pi}{2}) + b = \cos(\frac{\pi}{2})$ $a(1) + b = 0 \implies a + b = 0 \quad (2)$ Adding equations $(1)$ and $(2)$: $(-a + b) + (a + b) = 2 + 0 2b = 2 \implies b = 1$ Substituting $b = 1$ into equation $(2)$: $a + 1 = 0 \implies a = -1$ Now,calculate $2a + b$: $2(-1) + 1 = -2 + 1 = -1$ Thus,the value is $-1$.

If $f(x)= \begin{cases}-2 \sin x & , \quad x \leqslant-\frac{\pi}{2} \\ a \sin x+b & , \quad \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x & , \quad x \geqslant \frac{\pi}{2}\end{cases}$ is continuous at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$,then the value of $2a+b$ is

Similar Questions

If $f: R \rightarrow R$ defined as $f(x) = \frac{x^3+2x^2+x+2}{x^2+x-2}$ (when $x \neq -2$) is continuous at $x = -2$, then $f(-2)$ is equal to

If $f(x) = \begin{cases} x \sin x, & 0 < x \le \frac{\pi}{2} \\ \frac{\pi}{2} \sin(\pi + x), & \frac{\pi}{2} < x < \pi \end{cases}$,then

Find all points of discontinuity of the function $f$ defined by $f(x) = |x| - |x + 1|$.

Let $f:[-1,2] \rightarrow R$ be defined by $f(x)=[x^2-3]$ where $[\cdot]$ denotes the greatest integer function. The number of points of discontinuity for the function $f$ in the interval $(-1,2)$ is:

Let $[x]$ be the greatest integer less than or equal to $x$. At which of the following point$(s)$ is the function $f(x) = x \cos(\pi(x + [x]))$ discontinuous?
$[A]$ $x = -1$
$[B]$ $x = 0$
$[C]$ $x = 2$
$[D]$ $x = 1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module