Animated Solution for Mathematics - Functions: The larger of $\cos (\ln \theta )$ and $\ln (\cos \theta )$ if $-\pi /2<\theta <\pi /2$ is .........

Welcome, future engineers! Today, we are going to embark on a journey through the landscape of composite functions.

Often, in the heat of a JEE Advanced exam, we see a problem like this—comparing $\cos (\ln \theta )$ and $\ln (\cos \theta )$—and our instinct is to reach for the derivative, to sketch graphs, or to start solving complex inequalities. But pause. Take a breath.

The most beautiful solutions in mathematics are rarely the ones that require the most brute force. They are the ones that require the most insight. Let us dissect this problem layer by layer.

The Domain Trap

The problem presents us with the interval $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$. This is the first trap. It is a wide, inviting field, but it is filled with landmines.

Look at the first function: $f(\theta )=\cos (\ln \theta )$. The presence of the natural logarithm, $\ln \theta$, is a strict gatekeeper. We know from the fundamental laws of algebra that the logarithm of a non-positive number is undefined in the real number system.

Therefore, regardless of what the problem statement suggests, $\theta$ must be strictly greater than zero. When we intersect the given interval $(-\frac{\pi }{2},\frac{\pi }{2})$ with the domain constraint $\theta >0$, our effective domain shrinks dramatically to $(0,\frac{\pi }{2})$.

This is the first victory. We have successfully narrowed our focus to the first quadrant.

Analyzing $g(\theta )=\ln (\cos \theta )$

Now, let us examine our second function, $g(\theta )=\ln (\cos \theta )$. In our effective domain, $0<\theta <\frac{\pi }{2}$, what is the behavior of $\cos \theta$?

In the first quadrant, the cosine function starts at $1$ (when $\theta =0$) and decreases toward $0$ (as $\theta$ approaches $\frac{\pi }{2}$). Thus, for all $\theta$ in our domain, we have $0<\cos \theta <1$.

Now, apply the natural logarithm to this inequality. The logarithm of any number between $0$ and $1$ is strictly negative.

Think about it: $\ln (1)=0$, and as the input approaches $0$, the logarithm plunges toward $-\infty$. Therefore, $g(\theta )=\ln (\cos \theta )$ is always a negative value. It never crosses the x-axis; it lives entirely in the basement of the Cartesian plane.

Analyzing $f(\theta )=\cos (\ln \theta )$

Now, let us turn our attention to the first function, $f(\theta )=\cos (\ln \theta )$. This is where the magic happens. We need to understand the range of the inner function, $\ln \theta$.

As $\theta$ moves from $0$ to $\frac{\pi }{2}$, $\ln \theta$ moves from $-\infty$ to $\ln (\frac{\pi }{2})$. Since $\frac{\pi }{2}\approx 1.57$, $\ln (\frac{\pi }{2})$ is approximately $0.45$. So, the input to our cosine function is the interval $(-\infty ,0.45)$.

We know that the cosine function is positive whenever its argument lies between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$. Let us check if our argument, $\ln \theta$, spends time in this "positive zone."

Since $-\frac{\pi }{2}\approx -1.57$ and $\frac{\pi }{2}\approx 1.57$, and our upper bound is $0.45$, we can see that for a vast portion of our domain, the argument of the cosine function is indeed within the range where cosine is positive.

The Synthesis

We have two functions. One, $g(\theta )=\ln (\cos \theta )$, is trapped in the negative region, always less than zero. The other, $f(\theta )=\cos (\ln \theta )$, is positive for the vast majority of our domain.

In the world of real numbers, a positive quantity is always greater than a negative quantity. It is as simple as that. We do not need to calculate the exact intersection points or draw a perfect graph.

We have used the properties of the functions themselves to establish a hierarchy. The logic is ironclad:

Therefore, $\cos (\ln \theta )$ is the larger function.

This, my friends, is the essence of JEE Advanced problem-solving. It is not about who can calculate the fastest; it is about who can see the structure of the problem the clearest. You have successfully navigated the domain trap, analyzed the ranges, and synthesized the result. Keep this mindset—always look for the conceptual shortcut before you start the algebraic marathon.