QJEEM23Ma75bd: JEE Main 2023 Mathematics - Class 12 JEEM23Ma75bd Solution: max 0 ≤ x ≤ π { x − 2 sin ⁡ x cos ⁡ x + 1 3 sin ⁡ 3 x } =

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 $0 \leq x \leq \pi$ .

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 $x$ :

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

Step 3: Find the critical points

To find the critical points, we set $f'(x) = 0$ :

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

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:

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

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

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