Questions

For $f (x) = \ln (\frac{x^{2} + α}{7 x})$ Roll's theorem is applicable on $[3, 4]$ with $c \in (3, 4)$ such that $f^{'} (c) = 0 .$ The value of $f^{''} (c)$ is

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\frac{1}{12}

- \frac{1}{12}

\frac{1}{6}

- \frac{1}{6}

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detailed solution

Correct option is A

Roll's theorem is applicable to $f (x) = \ln (\frac{x^{2} + α}{7 x})$ on $[3, 4] .$

$\begin{array}{ll} ∴ f (3) = f (4) \\ \Rightarrow \ln (\frac{9 + α}{21}) = \ln (\frac{16 + α}{28}) \end{array}$

$\Rightarrow \frac{9 + α}{21} = \frac{16 + α}{28} \Rightarrow 36 + 4 α = 48 + 3 α \Rightarrow α = 12$

Now, $f (x) = \ln (\frac{x^{2} + α}{7 x})$

$\begin{array}{l} \Rightarrow f (x) = \ln (x^{2} + 12) - \ln 7 x \\ \Rightarrow f^{'} (x) = \frac{2 x}{x^{2} + 12} - \frac{1}{x} = \frac{x^{2} - 12}{x (x^{2} + 12)} \\ ∴ f^{'} (x) = 0 \Rightarrow x^{2} - 12 = 0 \Rightarrow x = \pm \sqrt{12} \\ ∴ c = \sqrt{12} \end{array}$

Now, $f^{'} (x) = \frac{x^{2} - 12}{x (x^{2} + 12)}$

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