Limits Continuity And Differentiability Question 69
Question: Let $ f(x)= \begin{cases} & \frac{1-{{\sin }^{3}}x}{3{{\cos }^{2}}x},x<\frac{\pi }{2} \\ & p,x=\frac{\pi }{2} \\ & \frac{q(1-\sin x)}{{{(\pi -2x)}^{2}}},x>\frac{\pi }{2} \\ \end{cases} . $ If f(x) is continuous at $ x=\frac{\pi }{2},(p,q)= $
Options:
A) $ (1,4) $
B) $ ( \frac{1}{2},2 ) $
C) $ ( \frac{1}{2},4 ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f[{{(\pi /2)}^{-}}]=\underset{h\to 0}{\mathop{\lim }}\frac{1-{{\sin }^{3}}[ (\pi /2)-h ]}{3{{\cos }^{2}}[ (\pi /2)-h ]} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{1-{{\cos }^{3}}h}{3{{\sin }^{2}}h}=\frac{1}{2} $
$ f[{{(\pi /2)}^{+}}]=\underset{h\to 0}{\mathop{\lim }}\frac{q[1-sin{(\pi /2)+h}]}{{{[\pi -2{(\pi /2)+h}]}^{2}}} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{q(1-cosh)}{4h^{2}}=\frac{q}{8} $
$ \therefore p=\frac{1}{2}=\frac{q}{8}\Rightarrow p=\frac{1}{2},q=4. $
BETA
