Let $a > 0$ and $f$ be continuous in $[- a, a]$. Suppose that $f ' (x) $ exists and $f ' (x) \le 1$ for all $x \in (- a, a)$. If $f (a) = a$ and $f (- a) = - a$ then $f (0)$
equals $0$
equals $\frac{1}{2}$
equals $1$
is not possible to determine
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
If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is
If $g(x) = 2f (2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$, $\forall x \in R$ and $f"(x) > 0, \forall x \in R$ , then $g'(x) > 0$ for $x$ belonging to
If $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$
Consider $f (x) = | 1 - x | \,;\,1 \le x \le 2 $ and $g (x) = f (x) + b sin\,\frac{\pi }{2}\,x$, $1 \le x \le 2$ then which of the following is correct ?