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 ?
Rolles theorem is applicable to both $f, g$ and $b =\frac{3}{2}\,$
$LMVT$ is not applicable to $f$ and Rolles theorem if applicable to $g$ with $b =\frac{1}{2}\,$
$LMVT$ is applicable to $f$ and Rolles theorem is applicable to $g$ with $b = 1$
Rolles theorem is not applicable to both $f, g$ for any real $b.$
Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is....................
In $[0, 1]$ Lagrange's mean value theorem is $ NOT$ applicable to
Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.
$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and
$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.
Given $f (x) =4\,\, - \,\,{\left( {\frac{1}{2}\, - \,x} \right)^{2/3}}\,$ $g (x) = \left\{ \begin{array}{l}\frac{{\tan \,\,[x]}}{x}\,\,\,\,,\,\,x \ne \,0\\1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x\, = \,0\end{array} \right.$
$h (x) = \{x\}$ $k (x) = {5^{{{\log }_2}(x\, + \,3)}}$then in $[0, 1]$ Lagranges Mean Value Theorem is $NOT$ applicable 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 $c \in[0,1]$