JEE Main 2024 — Application Of Derivatives Question with Solution

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 $$f(x)$$:

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

The critical points occur when $$f^{\prime }(x)=0$$:

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

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 $$f^{\prime }(x)$$ is always negative (i.e., $$f^{\prime }(x)<0$$), $$f(x)$$ 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 $$f^{\prime }(x)$$ shows that $$f^{\prime }(x)=-\sin x-1$$ 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.