Observe the following statements:
`"I. If "f(x)=ax^(41)+bx^(-40)," then "(f''(x))/(f(x))=1640x^(2)`
`"II. "(d)/(dx){tan^(-1)((2x)/(1-x^(2)))}=(1)/(1+x^(2))`
Which of the following is correct ?

Options

A. I is true, but II is false
B. Both I and II true
C. Neither I nor II is true
D. I is false, but II is true

Answer

To solve the problem, we need to analyze both statements one by one.

### Statement I:
If \( f(x) = ax^{41} + bx^{-40} \), then we need to check if

\[ \frac{f''(x)}{f(x)} = 1640x^2 \]

**Step 1: Find \( f'(x) \)**

Using the power rule of differentiation:

\[ f'(x) = \frac{d}{dx}(ax^{41}) + \frac{d}{dx}(bx^{-40}) \]
\[ = 41ax^{40} - 40bx^{-41} \]

**Hint for Step 1:** Remember that the derivative of \( x^n \) is \( nx^{n-1} \).

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**Step 2: Find \( f''(x) \)**

Now, differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}(41ax^{40}) + \frac{d}{dx}(-40bx^{-41}) \]
\[ = 41 \cdot 40ax^{39} + 40 \cdot 41bx^{-42} \]
\[ = 1640ax^{39} + 1640bx^{-42} \]

**Hint for Step 2:** Apply the power rule again for each term.

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**Step 3: Substitute \( f(x) \) and \( f''(x) \) into the equation**

Now, we need to find \( \frac{f''(x)}{f(x)} \):

\[ \frac{f''(x)}{f(x)} = \frac{1640ax^{39} + 1640bx^{-42}}{ax^{41} + bx^{-40}} \]

**Hint for Step 3:** Factor out common terms from the numerator and denominator.

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**Step 4: Simplify the expression**

Multiply numerator and denominator by \( x^{42} \) to eliminate negative powers:

\[ = \frac{1640a x^{81} + 1640b}{a x^{83} + b x^2} \]

Now, simplifying further leads to:

\[ = \frac{1640(a x^{81} + b)}{a x^{83} + b x^2} \]

This does not simplify to \( 1640x^2 \) for all values of \( a \) and \( b \). Thus, the first statement is **false**.

**Final Conclusion for Statement I:** False.

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### Statement II:
We need to check if

\[ \frac{d}{dx}\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) = \frac{1}{1+x^2} \]

**Step 1: Use the identity for \( \tan^{-1} \)**

Recall that

\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) \]

**Hint for Step 1:** Use trigonometric identities to simplify the expression.

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**Step 2: Differentiate \( 2\tan^{-1}(x) \)**

Now differentiate:

\[ \frac{d}{dx}(2\tan^{-1}(x)) = 2 \cdot \frac{1}{1+x^2} \]

**Hint for Step 2:** The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).

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**Step 3: Compare with the original statement**

Thus,

\[ \frac{d}{dx}\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) = \frac{2}{1+x^2} \]

This is not equal to \( \frac{1}{1+x^2} \), so the second statement is also **false**.

**Final Conclusion for Statement II:** False.

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### Final Answer:
Both statements are false, so the correct option is:

**Option C: Neither 1 nor 2 is true.**

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