Mathematics - Continuity - Differentiability Question with Solution | TestHub

$\mathrm{f}\left[(\pi /2{)}^{-}\right]={\lim }_{\mathrm{h}\to 0}\frac{1-{\sin }^3[(\pi /2)-\mathrm{h}]}{3{\cos }^2[(\pi /2)-\mathrm{h}]}$ $={\lim }_{\mathrm{h}\to 0}\frac{1-{\cos }^3\mathrm{h}}{3{\sin }^2\mathrm{h}}=\frac{1}{2}$ $\mathrm{f}\left[(\pi /2{)}^{+}\right]={\lim }_{\mathrm{h}\to 0}\frac{\mathrm{q}[1-\sin \{(\pi /2)+\mathrm{h}\}]}{[\pi -2\{(\pi /2)+\mathrm{h}\}{]}^2}$ $={\lim }_{\mathrm{h}\to 0}\frac{\mathrm{q}(1-\cosh )}{4{\mathrm{h}}^2}=\frac{\mathrm{q}}{8}$ $\therefore \mathrm{p}=\frac{1}{2}=\frac{\mathrm{q}}{8}\Rightarrow \mathrm{p}=\frac{1}{2},\mathrm{q}=4$