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

(A) Both (S1) and (S2) are correct

(C) Both (S1) and (S2) are incorrect

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Application of Derivatives

For the function

f (x) = (cos x) - x + 1, x \in R

, between the following two statements
(S1)

f (x) = 0

for only one value of

x

[0, π]

.
(S2)

f (x)

is decreasing in

[0, \frac{π}{2}]

and increasing in

[\frac{π}{2}, π]

Answer: (B)

f (x) = cos x - x + 1 \Rightarrow f^{'} (x) = - sin x - 1 \leq 0

, so

f

is decreasing on all of

R

(S2 is false). Since

f (0) = 2 > 0

and

f (π) = - π < 0

, there is exactly one root in

[0, π]

(S1 true). Only (S1) is correct — option (2).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option