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

We are given the function $$f(x)=\log_e\!\left(\dfrac{x^2+\alpha}{7x}\right)$$ on the closed interval $$[3,4]$$ and we are told that Rolle’s theorem holds for some real number $$\alpha$$. For Rolle’s theorem to be applicable, the function must be continuous on the closed interval, differentiable on the open interval and, most importantly, must satisfy $$f(3)=f(4)$$. Continuity and differentiability are already guaranteed by the form of the function, so we begin by imposing the equality of the end-point values.

First we evaluate the function at the two end points:

$$f(3)=\ln\!\left(\dfrac{3^2+\alpha}{7\cdot3}\right)=\ln\!\left(\dfrac{9+\alpha}{21}\right),$$ $$f(4)=\ln\!\left(\dfrac{4^2+\alpha}{7\cdot4}\right)=\ln\!\left(\dfrac{16+\alpha}{28}\right).$$

Rolle’s condition $$f(3)=f(4)$$ therefore becomes

$$\ln\!\left(\dfrac{9+\alpha}{21}\right)=\ln\!\left(\dfrac{16+\alpha}{28}\right).$$

Since the natural logarithm is one-to-one, the arguments must be equal:

$$\dfrac{9+\alpha}{21}=\dfrac{16+\alpha}{28}.$$

Cross-multiplying gives

$$28(9+\alpha)=21(16+\alpha).$$

Expanding both sides we get

$$252+28\alpha=336+21\alpha.$$

Subtracting $$21\alpha$$ and $$252$$ respectively, we obtain

$$7\alpha=84,$$ $$\alpha=12.$$

Now we know the exact function relevant to the problem:

$$f(x)=\ln\!\left(\dfrac{x^2+12}{7x}\right)=\ln(x^2+12)-\ln7-\ln x.$$

Next, Rolle’s theorem states that there exists some $$c\in(3,4)$$ with $$f'(c)=0$$. So we must first compute the derivative $$f'(x)$$.

Writing the logarithm as a difference, we differentiate term by term:

$$f(x)=\ln(x^2+12)-\ln7-\ln x,$$ $$f'(x)=\dfrac{2x}{x^2+12}-\dfrac{1}{x}.$$

We now set $$f'(c)=0$$:

$$\dfrac{2c}{c^2+12}-\dfrac{1}{c}=0.$$

Bringing terms to a common denominator and cross-multiplying yields

$$\dfrac{2c^2}{c^2+12}=1,$$ $$2c^2=c^2+12,$$ $$c^2=12,$$ $$c=2\sqrt3.$$

The positive root is taken because $$c$$ must lie in the interval $$[3,4]$$, and indeed $$2\sqrt3\approx3.464$$ satisfies $$3<2\sqrt3<4$$.

The problem asks for $$f''(c)$$, so we next compute the second derivative $$f''(x)$$. To do that we differentiate $$f'(x)$$ once more, taking each term separately.

We start with $$f'(x)=\dfrac{2x}{x^2+12}-\dfrac{1}{x}.$$ The derivative of the first fraction is found through the quotient rule. If we let $$g(x)=2x$$ and $$h(x)=x^2+12$$, then the quotient rule, $$\left(\dfrac{g}{h}\right)'=\dfrac{g'h-gh'}{h^2}$$, gives

$$\left(\dfrac{2x}{x^2+12}\right)'=\dfrac{2(x^2+12)-2x\cdot2x}{(x^2+12)^2}=\dfrac{2x^2+24-4x^2}{(x^2+12)^2}=\dfrac{-2x^2+24}{(x^2+12)^2}.$$

The derivative of $$-\dfrac1x$$ is $$+\dfrac1{x^2}$$, because $$\dfrac{d}{dx}(-x^{-1})=+x^{-2}$$.

Adding these two parts, we obtain the full second derivative:

$$f''(x)=\dfrac{-2x^2+24}{(x^2+12)^2}+\dfrac1{x^2}.$$

We now evaluate this at $$x=c=2\sqrt3$$. For this value we have

$$c^2=(2\sqrt3)^2=12,$$ $$c^2+12=12+12=24.$$

Substituting into the first fraction numerator we get

$$-2c^2+24=-2(12)+24=-24+24=0,$$ so the entire first fraction becomes

$$\dfrac{0}{(24)^2}=0.$$

The second term gives $$\dfrac1{c^2}=\dfrac1{12}.$$

Therefore

$$f''(c)=0+\dfrac1{12}=\dfrac1{12}.$$

Hence, the correct answer is Option B.