For the function f(x)=(cos x)-x+1, x in mathbb R , between the following two statements Statement I : f(x)=0 for only one value of x is [0, π] . Statement II : f(x) is decreasing in [0, π/2 ] and increasing in [π/2, π ] In the light of the above statements, choose the correct answer from the options given below :

Statement I is correct and statement Il is incorrect.

Both statement I and statement Il are correct.

Statement I is incorrect and statement Il is correct.

Both statements 1 and statements ll are incorrect.

Mathematics · Application of Derivatives

JEE Main 2024 — 8 April, Shift 1 — Question 17

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

Statement I : $f(x)=0$ for only one value of $x$ is $[0, \pi]$ .

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

In the light of the above statements, choose the correct answer from the options given below :

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

$f^{\prime}(\mathrm{x})=-\sin \mathrm{x}-1$

$f$ is decreasing $\forall x \in R$

$\mathrm{f}(\mathrm{x})=0$

$\mathrm{f}(0)=2, \mathrm{f}(\pi)=-\pi$

f is strictly decreasing in $[0, \pi]$ and $\mathrm{f}(0) . \mathrm{f}(\pi)<0$

$\Rightarrow$ only one solution of $\mathrm{f}(\mathrm{x})=0$

$S_1$ is correct and $S_2$ is incorrect.

Answer key and solution verified before publishing.

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