In $[0, 1]$ Lagrange's mean value theorem is $ NOT$ applicable to
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2} - x,\,\,\,\,\,\,\,\,\,x < \frac{1}{2}} \\
{{{\left( {\frac{1}{2} - x} \right)}^2},\,x \geqslant \frac{1}{2}}
\end{array}} \right.$
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin x}}{x}\,\,x \ne 0} \\
{1,\,\,\,\,\,\,\,\,x = \frac{1}{2}}
\end{array}} \right.$
$f(x) = x|x|$
$f(x) = |x|$
Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b],$ where $a=1$ and $b=4$
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 $t$.
Let $f(x) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} ;} $ $1 < x < 26$ be real valued function. Then $f\,'(x)$ for $1 < x < 26$ is
The function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfy the conditions of Rolle's theorem in $[1, 3]. $ The values of $a $ and $ b $ are
Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then, the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is