Question

If $f\left(x\right)=\{\begin{array}{c}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x},\text{if}x<\frac{\pi }{2}\\ a,\text{if}x=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{(\pi -2x{)}^2},\text{if}x>\frac{\pi }{2}\end{array}$ so that $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$, then

• A. $a=\frac{1}{2}$
• B. $b=4$
• C. $a=1$
• D. $b=-4$

Answer 1

Answer: (A), (B)

$a=\frac{1}{2}$
$b=4$

Let $L={lim}_{x\to {\frac{\pi }{2}}^{-}}f\left(x\right)={lim}_{x\to \frac{\pi }{2}}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x}$

$L={lim}_{x\to \frac{\pi }{2}}\frac{(1-\sin x)(1+{\sin }^{3}x+\sin x)}{3(1-\sin x)(1+\sin x)}={lim}_{x\to \frac{\pi }{2}}\frac{(1+{\sin }^{2}x+\sin x)}{3(1+\sin x)}L=\frac{3}{3\left(2\right)}=\frac{1}{2}$

Let $M={lim}_{x\to {\frac{\pi }{2}}^{+}}f\left(x\right)={lim}_{x\to \frac{\pi }{2}}\frac{6(1-\sin x)}{{(\pi -2x)}^2}$

Let $\pi -2x=2z$

So, $x=\frac{\pi }{2}-z$

So, $M={lim}_{z\to 0}\frac{6(1-\cos z)}{4z^2}=\frac{6}{4}\times \frac{1}{2}=\frac{6}{8}$

and $f\left(\frac{\pi }{2}\right)=a$

Since for continunity,

${lim}_{x\to {\frac{\pi }{2}}^{-}}f\left(x\right)=f\left(\frac{\pi }{2}\right)={lim}_{x\to {\frac{\pi }{2}}^{+}}f\left(x\right)$

So, $\frac{1}{2}=a=\frac{b}{8}$

So, $b=4$

and $a=\frac{1}{2}$