Suppose that $f$ is differentiable for all $x$ and that $f '(x) \le 2$ for all x. If $f (1) = 2$ and $f (4) = 8$ then $f (2)$ has the value equal to

For every pair of continuous functions $f, g:[0,1] \rightarrow R$ such that $\max \{f(x): x \in[0,1]\}=\max \{g(x): x \in[0,1]\}$, the correct statement$(s)$ is (are) :

$(A)$ $(f(c))^2+3 f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$

$(B)$ $(f(c))^2+f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$

$(C)$ $(f(c))^2+3 f(c)=(g(c))^2+g(c)$ for some $c \in[0,1]$

$(D)$ $(f(c))^2=(g(c))^2$ for some $c \in[0,1]$

If $f$ is a differentiable function such that $f(2x + 1) = f(1 -2x)$ $\forall \,\,x \in R$ then minimum number of roots of the equation $f'(x) = 0$ in $x \in \left( { - 5,10} \right)$ ,given that $f(2) = f(5) = f(10)$ , is

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 $t$.

For a real number $x$ let $[x]$ denote the largest number less than or equal to $x$. For $x \in R$ let $f(x)=[x] \sin \pi x$. Then,

The function $f(x) = {(x - 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x - 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$  and $(4, 1)$  is