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 $f (0)$

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is

The abscissa of the points of the curve $y = {x^3}$ in the interval $ [-2, 2]$, where the slope of the tangents can be obtained by mean value theorem for the interval  $[-2, 2], $ are

Let $f(x)$ satisfy all the conditions of mean value theorem in  $[0, 2]. $ If $ f (0) = 0 $ and $|f'(x)|\, \le {1 \over 2}$ for all  $x$ in  $[0, 2]$ then

Let $f (x)$ and $g (x)$ be two continuous functions defined from $R \rightarrow R$, such that $f (x_1) > f (x_2)$ and $g (x_1) < g (x_2), \forall x_1 > x_2$ , then solution set of $f\,\left( {\,g({\alpha ^2} - 2\alpha )\,} \right) >f\,\left( {\,g(3\alpha - 4)\,} \right)$ is