QJEEM24Ma21b0: JEE Main 2024 Mathematics - Class 12 JEEM24Ma21b0 Solution: For the function , between the following two statements (S1)…

Let's analyze the function $f(x) = (\cos x) - x + 1$ over the interval $[0, \pi]$ and the statements provided.

First, let's consider statement (S1):

(S1) $f(x)=0$ for only one value of $x$ in $[0, \pi]$ .

To examine this statement, we need to explore the zeros of the function $f(x)$ within the given interval. Let's define and analyze the function:

$f(x) = \cos x - x + 1$

We seek to determine if $f(x) = 0$ has only one solution in the interval $[0, \pi]$ . To do this, we can use the Intermediate Value Theorem and the behavior of the function's derivative. First, compute the derivative of $f(x)$ :

$f'(x) = \frac{d}{dx}(\cos x - x + 1) = -\sin x - 1$

The critical points occur when $f'(x) = 0$ :

$- \sin x - 1 = 0 \Rightarrow \sin x = -1$ .

The equation $\sin x = -1$ does not hold for any $x$ in $[0, \pi]$ . Note that:

• For $x \in [0, \frac{\pi}{2}]$ , $\sin x$ ranges from 0 to 1.


• For $x \in [\frac{\pi}{2}, \pi]$ , $\sin x$ ranges from 1 to 0.

Since $f'(x)$ is always negative (i.e., f'(x) < 0 ), $f(x)$ is a strictly decreasing function in $[0, \pi]$ . Moreover, we evaluate:

$f(0) = \cos 0 - 0 + 1 = 1 + 1 = 2$

$f(\pi) = \cos \pi - \pi + 1 = -1 - \pi + 1 = -\pi$

Given the continuous and strictly decreasing nature of $f(x)$ in $[0, \pi]$ , by the Intermediate Value Theorem, there is exactly one value of $x$ in the interval $[0, \pi]$ where $f(x) = 0$ , confirming (S1).

Now, let's consider statement (S2):

(S2) $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right]$ .

We already analyzed that $f'(x)$ shows that $f'(x) = -\sin x - 1$ is always less than zero in $[0, \pi]$ . No interval exists where the derivative is positive. This means that $f(x)$ is strictly decreasing throughout the entire interval of $[0, \pi]$ , invalidating (S2).

Therefore, the correct option is:

Option B: Only (S1) is correct.