Animated Solution for Mathematics - Definite Integration: If $a={\lim }_{n\to \infty }{\sum }_{k=1}^n\frac{2n}{n^2+k^2}$ and $f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}},x\in (0,1)$, then :

From Discrete Sums to Elegant Derivatives

Welcome, future engineers! Today, we are going to dissect a problem that is a perfect microcosm of the JEE Advanced experience. It tests your ability to recognize patterns, your mastery of trigonometric identities, and your precision in calculus.

This problem is not just about finding an answer; it is about seeing the hidden connections between different branches of mathematics.

The Riemann Sum - Turning Discrete into Continuous

Let us look at the first part of our challenge: $a={\lim }_{n\to \infty }{\sum }_{k=1}^n\frac{2n}{n^2+k^2}$. When you see a limit of a summation as $n$ goes to infinity, your intuition should immediately scream 'Riemann Sum!'

We are essentially summing up the areas of infinitely many, infinitely thin rectangles to find the area under a curve. To unlock this, we need to force the expression into the form $\sum f(k/n)\cdot (1/n)$.

Let us manipulate the term $\frac{2n}{n^2+k^2}$. If we divide both the numerator and the denominator by $n^2$, we get:

Now, the structure becomes clear. We have a term $(1/n)$ which acts as our $dx$, and a function of $(k/n)$. The summation becomes an integral from $0$ to $1$ (since $k/n$ ranges from $1/n$ to $n/n$, which is $0$ to $1$ as $n\to \infty$).

Thus, our expression transforms into:

This is a standard integral! The antiderivative of $\frac{1}{1+x^2}$ is ${\tan }^{-1}(x)$. So, we have $a=2[{\tan }^{-1}(x){]}_0^1$.

Evaluating this, we get $2({\tan }^{-1}(1)-{\tan }^{-1}(0))=2(\pi /4-0)=\pi /2$. We have successfully tamed the limit!

The Trigonometric Simplification

Now, we turn our attention to the function $f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}}$. Many students would immediately reach for the quotient rule and the chain rule here, but that is a path to algebraic misery.

Remember, in trigonometry, identities are your best friends. Recall the half-angle identities: $1-\cos x=2{\sin }^2(x/2)$ and $1+\cos x=2{\cos }^2(x/2)$.

Substituting these into our function, we get:

Since $x\in (0,1)$, $x/2$ is in the first quadrant, so $\tan (x/2)$ is positive. Thus, $f(x)=\tan (x/2)$. Look at how much cleaner that is!

We have reduced a complex radical expression into a simple tangent function.

The Derivative and the Final Connection

Now, we need to find the relationship between $f(a/2)$ and $f^{\prime }(a/2)$. First, let us find the derivative $f^{\prime }(x)$.

Differentiating $\tan (x/2)$ with respect to $x$ gives us:

We know $a=\pi /2$, so $a/2=\pi /4$. Let us evaluate our function and its derivative at this point.

For the function: $f(\pi /4)=\tan (\pi /8)=\sqrt{2}-1$.

For the derivative: $f^{\prime }(\pi /4)=\frac{1}{2}{\sec }^2(\pi /8)=\frac{1}{2{\cos }^2(\pi /8)}$.

Here is a beautiful trick. Use the identity $2{\cos }^2(\theta )=1+\cos (2\theta )$.

So, $2{\cos }^2(\pi /8)=1+\cos (\pi /4)=1+1/\sqrt{2}=\frac{\sqrt{2}+1}{\sqrt{2}}$.

Therefore, $f^{\prime }(\pi /4)=\frac{1}{(\sqrt{2}+1)/\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}+1}$.

Rationalizing the denominator by multiplying by $(\sqrt{2}-1)$, we get:

Comparing this to our function value $f(\pi /4)=\sqrt{2}-1$, we see the elegant relationship: $f^{\prime }(\pi /4)=\sqrt{2}f(\pi /4)$.

This matches our third option perfectly! You see, the problem wasn't about brute force; it was about recognizing the structure of the limit, simplifying the trigonometry, and carefully applying the chain rule.