If c is a point at which Rolle's theorem hold | vedclass

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(B) For Rolle's theorem to hold,$f(3) = f(4)$.$\log_{e}\left(\frac{3^{2}+\alpha}{7(3)}\right) = \log_{e}\left(\frac{4^{2}+\alpha}{7(4)}\right)$$\frac{

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$\frac{1}{12}$

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(B) For Rolle's theorem to hold,$f(3) = f(4)$.$\log_{e}\left(\frac{3^{2}+\alpha}{7(3)}\right) = \log_{e}\left(\frac{4^{2}+\alpha}{7(4)}\right)$$\frac{9+\alpha}{21} = \frac{16+\alpha}{28} \Rightarrow 4(9+\alpha) = 3(16+\alpha) \Rightarrow 36+4\alpha = 48+3\alpha \Rightarrow \alpha = 12$.Now,$f(x) = \log_{e}(x^{2}+12) - \log_{e}(7x) = \log_{e}(x^{2}+12) - \log_{e}(7) - \log_{e}(x)$.$f'(x) = \frac{2x}{x^{2}+12} - \frac{1}{x} = \frac{2x^{2} - (x^{2}+12)}{x(x^{2}+12)} = \frac{x^{2}-12}{x(x^{2}+12)}$.For Rolle's theorem,$f'(c) = 0 \Rightarrow c^{2}-12 = 0 \Rightarrow c = \sqrt{12} = 2\sqrt{3}$ (since $c \in (3, 4)$ is not possible,let us re-evaluate).Wait,$c = 2\sqrt{3} \approx 3.464$,which is in $(3, 4)$.$f''(x) = \frac{d}{dx} \left( \frac{x^{2}-12}{x^{3}+12x} \right) = \frac{(2x)(x^{3}+12x) - (x^{2}-12)(3x^{2}+12)}{(x^{3}+12x)^{2}}$.At $c^{2}=12$,$f''(c) = \frac{(2c)(c^{3}+12c) - (0)}{(c^{3}+12c)^{2}} = \frac{2c}{c^{3}+12c} = \frac{2}{c^{2}+12} = \frac{2}{12+12} = \frac{2}{24} = \frac{1}{12}$.

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

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