JEE Main 2023 — Application Of Derivatives Question with Solution

Given the function:

$$f(x)=x-2\sin x\cos x+\frac{1}{3}\sin 3x$$

We want to find the maximum value of this expression for $$0\le x\le \pi$$.

Step 1: Rewrite the expression

Notice that we can rewrite the expression as:

$$f(x)=x-\sin 2x+\frac{1}{3}\sin 3x$$

Step 2: Find the first derivative

Now, let's find the derivative of this expression with respect to $$x$$:

$$f^{\prime }(x)=1-2\cos 2x+\cos 3x$$

Step 3: Find the critical points

To find the critical points, we set $$f^{\prime }(x)=0$$:

$$1-2\cos 2x+\cos 3x=0$$

Step 4: Solve for x

This equation can be rewritten as:

$$(2\cos x+\sqrt{3})(2\cos x-\sqrt{3})(\cos x-1)=0$$

We get three possible solutions for $$x$$:

$$\cos x=\frac{-\sqrt{3}}{2},\frac{\sqrt{3}}{2},1$$

Which gives us: $$x=\frac{5\pi }{6},\frac{\pi }{6},0$$

Step 5: Find the second derivative

Now, let's find the second derivative of the expression:

$$f^{\prime \prime }(x)=4\sin 2x-3\sin 3x$$

Step 6: Analyze the critical points Evaluate the second derivative at the critical points:

1. $$f^{\prime \prime }\left(\frac{5\pi }{6}\right)=-2\sqrt{3}-\sqrt{3}<0$$, indicating a local maximum.
2. $$f^{\prime \prime }\left(\frac{\pi }{6}\right)=2\sqrt{3}-\sqrt{3}>0$$, indicating a local minimum.
3. $$f^{\prime \prime }(0)=0$$, inconclusive.

Step 7: Evaluate the function at the local maximum point and the boundary points Since we are looking for the maximum value of $$f(x)$$, we can now evaluate the function at the local maximum point and the boundary points:

1. $$f\left(\frac{5\pi }{6}\right)=\frac{5\pi }{6}+\frac{\sqrt{3}}{2}+\frac{1}{3}=\frac{5\pi +2+3\sqrt{3}}{6}$$
2. $$f(0)=0$$
3. $$f(\pi )=\pi$$

Step 8: Compare the values The maximum value of $$f(x)$$ is given by $$f\left(\frac{5\pi }{6}\right)=\frac{5\pi +2+3\sqrt{3}}{6}$$.