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Observe the following statements
I. $f(x)=a x^{41}+b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)}=1640 x^{-2}$
II. $\frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{1}{1+x^2}$
Which of the following is correct?
Options:
I. $f(x)=a x^{41}+b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)}=1640 x^{-2}$
II. $\frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{1}{1+x^2}$
Which of the following is correct?
Solution:
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Verified Answer
The correct answer is:
$\mathrm{I}$ is true, but $\mathrm{II}$ is false
I.
$\begin{aligned}
& f(x)=a x^{41}+b x^{-40} \\
& f^{\prime}(x)=41 a x^{40}-40 b x^{-41} \\
& f^{\prime \prime}(x)=1640 a x^{39}+1640 b x^{-42}
\end{aligned}$
Now, $\frac{f^{\prime \prime}(x)}{f(x)}=\frac{1640\left(a x^{39}+b x^{-42}\right)}{a x^{41}+b x^{-40}}=1640 x^{-2}$
II.
$\begin{aligned}
& \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \\
& =\frac{d}{d x} \tan ^{-1}(\tan 2 x) \\
& =\frac{d}{d x} 2 x \\
& =2
\end{aligned}$
$\therefore$ Statement I is true, but II is false.
$\begin{aligned}
& f(x)=a x^{41}+b x^{-40} \\
& f^{\prime}(x)=41 a x^{40}-40 b x^{-41} \\
& f^{\prime \prime}(x)=1640 a x^{39}+1640 b x^{-42}
\end{aligned}$
Now, $\frac{f^{\prime \prime}(x)}{f(x)}=\frac{1640\left(a x^{39}+b x^{-42}\right)}{a x^{41}+b x^{-40}}=1640 x^{-2}$
II.
$\begin{aligned}
& \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \\
& =\frac{d}{d x} \tan ^{-1}(\tan 2 x) \\
& =\frac{d}{d x} 2 x \\
& =2
\end{aligned}$
$\therefore$ Statement I is true, but II is false.
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