For the function f(x)=( x)-x+1,x R, between the following two statements (S1) f(x)=0 for o…

Question

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 $x$ in $[0,\pi ]$.

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

Accepted Answer

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., ${f}^{'}(x)$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 $[0,\frac{\pi }{2}]$ and increasing in $[\frac{\pi }{2},\pi ]$.

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 $

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