Let g(x) = cos² x, f(x) = √x, and α, β (α

Q: Why did we change the function from $\cos^2(x)$ to $\cos(x^2)$? Isn't that changing the question?

That is a brilliant observation! In the heat of a JEE exam, you might encounter typos. When you see options that are completely free of $\pi$ and involve simple trigonometric values, it is a huge hint that the function was intended to be $\cos(x^2)$. The composition $g(f(x))$ then becomes $\cos((\sqrt{x})^2) = \cos(x)$, which is mathematically elegant and consistent with the options provided.

Q: How do I know which root is $\alpha$ and which is $\beta$?

The problem explicitly defines $\alpha < \beta$. After you factor the quadratic into $(6x - \pi)(3x - \pi) = 0$, you get roots $\frac{\pi}{6}$ and $\frac{\pi}{3}$. Since $\frac{\pi}{6} \approx 0.52$ and $\frac{\pi}{3} \approx 1.04$, it is clear that $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$.

Q: Why can't we just use the area formula for a general curve?

You absolutely can! The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is defined by the definite integral $\int_{a}^{b} f(x) dx$. We are simply applying this fundamental theorem of calculus. There is no shortcut, but the beauty of this problem lies in how the complex-looking composition simplifies into a standard integral.

Question

Let $g(x) = \cos^2 x, f(x) = \sqrt{x}$, and $\alpha, \beta (\alpha < \beta)$ be the roots of the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Then the area (in sq. units) bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha, x = \beta$ and $y = 0$, is :

Options:

(A) $\frac{1}{2}(\sqrt{2}-1)$

(B) $\frac{1}{2}(\sqrt{3}-1)$

(C) $\frac{1}{2}(\sqrt{3}+1)$

(D) $\frac{1}{2}(\sqrt{3}-\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex-looking problem in the middle of a high-stakes exam. You see $g(x) = \cos^2 x$ and $f(x) = \sqrt{x}$, and your heart might skip a beat. But wait! Look at the options; they are clean, simple, and devoid of $\pi$.

This is a classic JEE moment where you must trust your intuition over the literal text. The function $g(x)$ was meant to be $\cos(x^2)$.

When we compose these functions, $y = (g \circ f)(x) = g(f(x))$, we get $y = \cos((\sqrt{x})^2)$. The square root and the square cancel out with satisfying precision, leaving us with the elegant curve $y = \cos(x)$.

### Hunting for the Boundaries

Now that we have our curve, we need to find the boundaries of our area. We are given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Do not let the $\pi$ scare you; it is just a constant.

We need to find the roots $\alpha$ and $\beta$. Let us split the middle term $-9\pi x$ into $-6\pi x$ and $-3\pi x$. This gives us:

$$18x^2 - 6\pi x - 3\pi x + \pi^2 = 0$$

By grouping the terms, we get $6x(3x - \pi) - \pi(3x - \pi) = 0$, which factors beautifully into $(6x - \pi)(3x - \pi) = 0$. This yields two roots: $x = \frac{\pi}{6}$ and $x = \frac{\pi}{3}$.

Since the problem states $\alpha < \beta$, we identify $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$. These are the vertical walls of our region.

### The Integral of Beauty

We are now ready to calculate the area. The area bounded by $y = \cos(x)$, $x = \alpha$, $x = \beta$, and $y = 0$ is simply the definite integral:

$$A = \int_{\alpha}^{\beta} \cos(x) dx$$

Substituting our limits, we have:

$$A = \int_{\pi/6}^{\pi/3} \cos(x) dx$$

The antiderivative of $\cos(x)$ is $\sin(x)$. So, we evaluate $A = [\sin(x)]_{\pi/6}^{\pi/3}$. This is the moment of truth; we plug in our upper limit $\frac{\pi}{3}$ and our lower limit $\frac{\pi}{6}$, giving us $A = \sin(\frac{\pi}{3}) - \sin(\frac{\pi}{6})$.

### The Final Triumph

Recall your trigonometric values: $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Substituting these, we get:

$$A = \frac{\sqrt{3}}{2} - \frac{1}{2}$$

Combining these, we arrive at **$A = \frac{1}{2}(\sqrt{3} - 1)$ square units**. This matches option two perfectly.

You have navigated the typo, solved the quadratic, and performed the integration. This is the essence of JEE Advanced: not just calculation, but the ability to see through the noise to the elegant mathematics waiting on the other side.

Select Answer:

Visualized Solution

Correcting the Composite Function

Setting up the Quadratic

Factorizing to find Roots

Identifying Limits $α$ and $β$

Formulating the Area Integral

Computing the Antiderivative

Evaluating the Definite Integral

The Final Area

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Hunting for the Boundaries

The Integral of Beauty

The Final Triumph

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