A) \[(1,4)\]
B) \[\left( \frac{1}{2},2 \right)\]
C) \[\left( \frac{1}{2},4 \right)\]
D) None of these
Correct Answer: C
Solution :
[c] \[f[{{(\pi /2)}^{-}}]=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-{{\sin }^{3}}\left[ (\pi /2)-h \right]}{3{{\cos }^{2}}\left[ (\pi /2)-h \right]}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{3}}h}{3{{\sin }^{2}}h}=\frac{1}{2}\] \[f[{{(\pi /2)}^{+}}]=\underset{h\to 0}{\mathop{\lim }}\,\frac{q[1-sin\{(\pi /2)+h\}]}{{{[\pi -2\{(\pi /2)+h\}]}^{2}}}\] \[=\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{q(1-cos\,h)}{4{{h}^{2}}}=\frac{q}{8}\] \[\therefore p=\frac{1}{2}=\frac{q}{8}\Rightarrow p=\frac{1}{2},q=4.\]You need to login to perform this action.
You will be redirected in
3 sec