Animated Solution for Mathematics - Differentiation: Given $A=\{x:\pi /6\le x\le \pi /3\}$ and $f(x)=\cos x-x(1+x)$; find $f(A)$.

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, rigorous logic beneath.

We are given the function $f(x)=\cos x-x(1+x)$ and asked to find its range over the interval $A=[\frac{\pi }{6},\frac{\pi }{3}]$. Imagine you are standing on a mountain path defined by this curve.

To find the 'range'—the set of all altitudes you will reach—we need to know if the path is constantly ascending, descending, or undulating. This is where the power of calculus comes to our rescue.

When we face a function on a closed interval, our first instinct should always be to check for monotonicity. Monotonicity is just a fancy way of asking: "Is this function always going up or always going down?"

If we can prove that, the range becomes trivial to find. To do this, we need the derivative, $f^{\prime }(x)$.

Let us first simplify our function to make the differentiation process smooth. Expanding the expression, we get:

Let us compute the derivative step-by-step. The derivative of $\cos x$ is $-\sin x$, the derivative of $-x$ is $-1$, and the derivative of $-x^2$ is $-2x$.

Putting it all together, we have:

Now, here is the crucial moment. To understand the sign of this expression, let us factor out the negative sign:

Look at the expression inside the parentheses: $(\sin x+1+2x)$. For any $x$ in our domain $[\frac{\pi }{6},\frac{\pi }{3}]$, $x$ is positive, which means $2x$ is positive.

We also know that $\sin x$ is positive in the first quadrant. Since $1$ is also positive, the entire sum $(\sin x+1+2x)$ must be strictly positive.

Because of the negative sign outside, $f^{\prime }(x)<0$ for all $x$ in our interval. This is our "Aha!" moment: the function is strictly decreasing.

Since the function is strictly decreasing, it reaches its maximum value at the start of the interval and its minimum value at the end. Therefore, the range is simply $[f(\frac{\pi }{3}),f(\frac{\pi }{6})]$.

Now, we just need to plug in the values. For the minimum, we evaluate at $x=\frac{\pi }{3}$:

For the maximum, we evaluate at $x=\frac{\pi }{6}$:

And there you have it! The range of our function is the closed interval:

You have successfully navigated the terrain of this function using the elegant tools of calculus. Keep this mindset—always look for the derivative, always check the monotonicity—and you will conquer any function the JEE throws at you!