For  f(x)=ln⁡x2+α7x  Roll’s theorem is applicable on  [3, 4] with  c∈(3,4)  such that  f′(c)=0.  The value of  f′′(c) is

# For  $f\left(x\right)=\mathrm{ln}\left(\frac{{x}^{2}+\alpha }{7x}\right)$  Roll's theorem is applicable on  with  such that  The value of  is

1. A

$\frac{1}{12}$

2. B

$-\frac{1}{12}$

3. C

$\frac{1}{6}$

4. D

$-\frac{1}{6}$

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### Solution:

Roll's theorem is applicable to  $f\left(x\right)=\mathrm{ln}\left(\frac{{x}^{2}+\alpha }{7x}\right)$ on

Now,  $f\left(x\right)=\mathrm{ln}\left(\frac{{x}^{2}+\alpha }{7x}\right)$

Now,  ${f}^{\mathrm{\prime }}\left(x\right)=\frac{{x}^{2}-12}{x\left({x}^{2}+12\right)}$

$\begin{array}{l}x\left({x}^{2}+12\right)\\ ⇒{f}^{\prime \prime }\left(x\right)=\frac{2{x}^{2}\left({x}^{2}+12\right)-\left({x}^{2}-12\right)\left({x}^{2}+12+2{x}^{2}\right)}{{x}^{2}{\left({x}^{2}+12\right)}^{2}}\\ ⇒{f}^{\prime \prime }\left(x\right)=\frac{-{x}^{4}+48{x}^{2}+144}{{x}^{2}{\left({x}^{2}+12\right)}^{2}}\\ ⇒{f}^{\prime \prime }\left(\sqrt{12}\right)=\frac{1}{12}\end{array}$

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