Question

Let $f(x)=2+\cos x$ for all real $x$
STATEMENT-1: For each real $t$, there exists a point in $c$ in $\left[ t,t+\pi \right]$ such that ${{f}^{'}}(c)=0$
STATEMENT-2: $f(t)=f(t+2\pi )$ for each real $t$
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is the correct explanation of STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is not the correct explanation of STATEMENT-1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

Hint: We have to differentiate the given function $f(x)=2+\cos x$ and use the intermediate value theorem for checking STATEMENT-1. So, we have to find a value of c such that ${{f}^{'}}(c)$. For STATEMENT-2, we have to use periodic function concepts for $f(t)\And f(t+2\pi )$ and solve both of them.

Complete step-by-step answer:
Given $f(x)=2+\cos x$
If we differentiate$f(x)$ we get,
$\dfrac{d(f(x))}{dx}=\dfrac{d(2+\cos x)}{dx}$
${{f}^{'}}(x)=-\sin x$
We have to use IMVT i.e. intermediate value theorem.
Let$f$be a continuous function on the closed $\left[ a,b \right]$. Let $m$ be any number between $f(a)$ and $f(b)$. Then there is at least one number c in $\left[ a,b \right]$ which satisfies $f(c)=m$.
A function value $c$ exists between $\left[ a,b \right]$ where ${{f}^{'}}(c)=0$ and value of $a$ and $b$ should be $0$.
So in STATEMENT-1 we are given $\left[ t,t+\pi \right]$. So we know $-\sin x$ is a continuous function in all limits. So the value of $c$ will be there. Here ${{f}^{'}}(c)=0$, so we get
${{f}^{'}}(c)=-\sin c$
i.e. $-\sin c=0$
We know that the value of $\sin x$ is 0 when $x=n\pi ,n\in Z$ , therefore for value of $x=t$ , we have ${{f}^{'}}(t)=-\sin t\Rightarrow 0$ and also at $x=t+\pi $, we have ${{f}^{'}}\left( t+\pi \right)=-\sin \left( t+\pi \right)\Rightarrow \sin t\Rightarrow 0$.
So STATEMENT-1 is Correct.
Now in STATEMENT-2 $f(t)=f(t+2\pi )$
We have function as $f(x)=2+\cos x$. So we are going to use the periodic function, which is mentioned below.
A periodic function can be defined as a function that repeats its values in regular intervals or periods. So, if we consider a function $f$ , then it is said to be periodic if for some nonzero constant $P$, we have $f(x+P)=f(x)$; $P$ is called the period of the function.
So let $x=t$, then we get, $f(t)=2+\cos t$ and $\cos t$ is a periodic function, period = $2\pi $.
We know the property that $f(x)=f(a+x)$ where a is the period of $f(x)$.
So in our problem $2\pi $ is a period of $f(t)$.
So STATEMENT-2 is Correct.
But STATEMENT-2 is not the correct explanation of STATEMENT-1 because in STATEMENT-1 intermediate mean value theorem is used.
So option(B) is the correct answer.

Note: Don’t jumble the given options and choose the wrong answer, be thorough with the properties. So in STATEMENT-1, we had used property - Intermediate mean value theorem, you should be familiar with this property. For STATEMENT-2 $f(x)=f(a+x)$, we can use this one - periodic function and also the period. Suppose we know for $f(x)=\sin x$ we have the periodic function as $2\pi $ so $\sin 5x$ will have a period of $\dfrac{2\pi }{5}$.