For the function $f(x)=(\cos x)-x+1, x \in \mathbb{R}$, between the following two statements (S1) $f(x)=0$…

For the function $f(x)=(\cos x)-x+1, x \in \mathbb{R}$, between the following two statements (S1) $f(x)=0$ for only one value of $x$ in $[0, \pi]$. (S2) $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right]$.
  1. Both (S1) and (S2) are correct.
  2. Both (S1) and (S2) are incorrect.
  3. Only (S2) is correct.
  4. Only (S1) is correct.

Solution

$\begin{aligned} & f^{\prime}(x)=\cos x-x+1 \\ & f^{\prime}(x)=-\sin x-1\end{aligned}$ $\mathrm{f}$ is decreasing $\forall \mathrm{x} \in \mathrm{R}$ $\begin{aligned} & \mathrm{f}(\mathrm{x})=0 \\ & \mathrm{f}(0)=2, \mathrm{f}(\pi)=-\pi\end{aligned}$ $\mathrm{f}$ is strictly decreasing in $[0, \pi]$ and $\mathrm{f}(0) . \mathrm{f}(\pi) < 0$ $\Rightarrow$ only one solution of $\mathrm{f}(\mathrm{x})=0$ $\mathrm{S} 1$ is correct and $\mathrm{S} 2$ is incorrect.

Asked in: JEE Main 2024 (08 Apr Shift 1)

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