Let $f(x)=2+\cos x$ for all real $x$.
$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.
Statement -$1$ is True, Statement -$2$ is True; Statement-$2$ is a correct explanation for Statement-$1$
Statement -$1$ is True, Statement - $2$ is True; Statement-$2$ is $NOT$ a correct explanation for Statement-$1$
Statement -$1$ is True, Statement -$2$ is False
Statement -$1$ is False, Statement -$2$ is True
If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............
Which of the following function can satisfy Rolle's theorem ?
Examine the applicability of Mean Value Theorem:
$(i)$ $f(x)=[x]$ for $x \in[5,9]$
$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$
$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$
If $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$
If the equation
${a_n}{x^{n - 1}} + \,{a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$
has a positive root $x= \alpha ,$ then the equation
$n{a_n}{x^{n - 1}} + \,(n - 1){a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1} = 0$
has a positive root which is