If c is a point at which Rolle's theorem holds for the function,
f(x) = ${\log _e}\left( {{{{x^2} + \alpha } \over {7x}}} \right)$ in the interval [3, 4], where a $\in$ R, then ƒ''(c) is equal to

For Rolle’s theorem to be applicable in [3, 4] <br><br>ƒ(3) = ƒ(4) <br><br>$\Rightarrow$ $${\log _e}\left( {{{9 + \alpha } \over {21}}} \right) = {\log _e}\left( {{{16 + \alpha } \over {28}}} \right)$$ <br><br>$\Rightarrow$ $$\left( {{{9 + \alpha } \over {21}}} \right) = \left( {{{16 + \alpha } \over {28}}} \right)$$ <br><br>$\Rightarrow$ 36 + 4$\alpha$ = 48 + 3$\alpha$ <br><br>$\Rightarrow$ $\alpha$ = 12 <br><br>According to Rolle’s theorem, f'(c) = 0 <br>where c $\in$ (3, 4) <br><br>f'(x) = ${{7x} \over {{x^2} + 12}}$ $\times$ $$\left( {{{2x\left( {7x} \right) - \left( {{x^2} + 12} \right)7} \over {{7^2}{x^2}}}} \right)$$ <br><br>$\Rightarrow$ f'(x) = ${{{x^2} - 12} \over {\left( {{x^2} + 12} \right)x}}$ <br><br>$\therefore$ f'(c) = ${{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}$ <br><br>From Rolle's theorem <br><br>${{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}$ = 0 <br><br>$\Rightarrow$ c² = 12 <br><br>Now f''(c) = $${{2c\left( {{c^3} + 12c} \right) - \left( {{c^2} - 12} \right)\left( {3{c^2} + 12} \right)} \over {{{\left( {{c^2} + 12} \right)}^2}{c^2}}}$$ <br><br>= $${{2\left( {12} \right)\left( {24} \right) - 0} \over {{{\left( {24} \right)}^2} \times 12}}$$ <br><br>= ${1 \over {12}}$