$\large{ f(x) = \begin{cases} \dfrac{1-\sin^3(x)}{3\cos^2(x)} & \text{for } x<\dfrac\pi2 \\ a & \text{for } x=\dfrac\pi2 \\ \dfrac{b(1-\sin(x))}{(\pi-2x)^2} & \text{for } x>\dfrac\pi2\\ \end{cases}}$

Find the sum of values of $a$ and $b$ such that the function $f(x)$ above is continuous at $x=\dfrac { \pi }{ 2 }$ .

The answer is 4.5.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

For $f(x)$ to be continuous at $x = \pi/2$ we must have $\lim_{x\rightarrow \pi/2^-} f(x) = \lim_{x\rightarrow \pi/2^+} f(x) = f(\pi/2) = a$

To evaluate the first 2 limits we must resort to L Hospital 's since they are of $0/0$ form.These yield

$\lim_{x\rightarrow \pi/2^-} f(x) = 1/2 ; \lim_{x\rightarrow \pi/2^-} f(x) = b/8$

Solving for $a$ and $b$ we get $a = 1/2$ and $b = 4$ ;

$\boxed{a + b = 4.5}$ .