JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1
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 $\alpha\in\mathbf{R}$, then $f''(c)$ is equal to :
Differentiation and its Application
Solution
Concept. Rolle needs $f(3)=f(4)$, which fixes $\alpha$; then $c$ solves $f'(c)=0$. Formulas used. (1) $f=\ln(x^2+\alpha)-\ln7-\ln x$, $\ f'=\dfrac{2x}{x^2+\alpha}-\dfrac1x$Step 1 - Find $\alpha$. $\frac{9+\alpha}{21}=\frac{16+\alpha}{28}$ gives $4(9+\alpha)=3(16+\alpha)$, so $\alpha=12$. Step 2 - Find $c$. By (1), $f'=0$ gives $2x^2=x^2+12$, so $c=2\sqrt3\in[3,4]$. Step 3 - $f''(c)$. $$f''=\frac{24-2x^2}{(x^2+12)^2}+\frac{1}{x^2}.$$ At $x^2=12$: $0+\frac{1}{12}=\frac{1}{12}$. Answer: (B)