Question

If \(a = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} \frac{2n}{{n^2 + k^2}}\)and
\(f(x) = \sqrt{\frac{1 - \cos x}{1 + \cos x}}, \quad x \in (0, 1)\) then

• A. \(2\sqrt2f(\frac{a}{2})=f^′(\frac{a}{2})\)
• B. \(f(\frac{a}{2})f^′(\frac{a}{2})=\sqrt2\)
• C. \(\sqrt2f(\frac{a}{2})=f^′(\frac{a}{2})\)
• D. \(f(\frac{a}{2})=\sqrt2f^′(\frac{a}{2})\)

Answer 1

The Correct Option is C

To find the value of \( a \) and evaluate the given options, let's solve the mathematical expressions step by step.

Step 1: Evaluate \( \lim_{{n \to \infty}} \sum_{{k=1}}^{n} \frac{2n}{n^2 + k^2} \)

The sum can be rewritten as:

\(\sum_{k=1}^{n} \frac{2n}{n^2 + k^2} = \sum_{k=1}^{n} \frac{2}{n} \cdot \frac{1}{1 + (\frac{k}{n})^2}\)

This can be viewed as a Riemann sum for the integral:

\(\int_{0}^{1} \frac{2}{1 + x^2} \, dx\)

where \( x = \frac{k}{n} \). This gives:

\(\int_{0}^{1} \! \frac{2}{1 + x^2} \, dx = 2 \cdot [\tan^{-1}(x)]_0^1 = 2 \cdot (\frac{\pi}{4} - 0) = \frac{\pi}{2}\)

Hence, the value of \( a \) is \( \frac{\pi}{2} \).

Step 2: Evaluate \( f(x) = \sqrt{\frac{1 - \cos x}{1 + \cos x}} \)

This function can be simplified using the identity:

\(\frac{1 - \cos x}{1 + \cos x} = \tan^2(\frac{x}{2})\)

Thus, \( f(x) = \sqrt{\tan^2(\frac{x}{2})} = \tan(\frac{x}{2}) \), assuming \( x \in (0, 1) \) where the square root yields a positive value.

Step 3: Compute \( f(\frac{a}{2}) \) and its derivative

Since \( a = \frac{\pi}{2} \), we evaluate \( f(\frac{\pi}{4}) \):

\(f(\frac{\pi}{4}) = \tan(\frac{\pi}{8})\)

Now, calculate the derivative of \( f(x) = \tan(\frac{x}{2}) \):

\(f'(x) = \frac{d}{dx} [\tan(\frac{x}{2})] = \frac{1}{2} \sec^2(\frac{x}{2})\)

Then, evaluate \( f'(\frac{\pi}{4}) \):

\(f'(\frac{\pi}{4}) = \frac{1}{2} \sec^2(\frac{\pi}{8})\)

Step 4: Verify the given options

We need to verify that:

\(\sqrt2 f(\frac{\pi}{4}) = f'(\frac{\pi}{4})\)

We know \( f(\frac{\pi}{4}) = \tan(\frac{\pi}{8}) \) and \( f'(\frac{\pi}{4}) = \frac{1}{2} \sec^2(\frac{\pi}{8}) \).

Using the identity, \( \sec^2(\theta) - 2\tan(\theta) = 0 \) gives us the relation:

\(\sqrt{2} \tan(\frac{\pi}{8}) = \frac{1}{2} \sec^2(\frac{\pi}{8})\)

Thus, the correct answer is:

\(\sqrt2 f(\frac{a}{2}) = f'(\frac{a}{2})\)