Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{(1-{\sin }^{3}x)}{3{\cos }^{2}x},& x<\frac{\pi }{2}\\ a,& x=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{(\pi -2x{)}^2},& x>\frac{\pi }{2}\end{array}$ is continuous at $x=\frac{\pi }{2}$, then the value of ${\left(\frac{b}{a}\right)}^{5/3}$ is

• A. $1$
• B. $8$
• C. $32$
• D. $54$

Answer 1

The correct option is C $32$

$f\left(x\right)=\{\begin{array}{cc}\frac{(1-{\sin }^{3}x)}{3{\cos }^{2}x},& x<\frac{\pi }{2}\\ a,& x=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{(\pi -2x{)}^2},& x>\frac{\pi }{2}\end{array}$

If a function is continuous at x=c then its limit exist at a and is equal to $f\left(c\right)$

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

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

so Leftt side limit is $\frac{3}{2}$ So $a=f\left(\frac{\pi }{2}\right)=\frac{3}{2}$

Rightt side limit

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{b(1-\sin x)}{(\pi -2x{)}^2}=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{b(1-\sin x)(1+\sin x)}{(\pi -2x{)}^2(1+\sin x)}$

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{b{\cos }^{2}x}{(\pi -2x{)}^2(1+\sin x)}=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{b{\sin }^{(}\frac{\pi }{2}-x)}{4(\frac{\pi }{2}-x{)}^2(1+\sin x)}$

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{h\to 0^{+}}{lim}\frac{b{\sin }^{2}h}{4h^2(1+\sin (\frac{\pi }{2}+h\left)\right)}=\frac{b}{8}$

Hence $\frac{b}{8}=a⟹\frac{b}{a}=8$

$⟹(\frac{b}{a}{)}^{5/3}=8^{5/3}=2^5=32$

Hence answer is A