If the equation
${a_n}{x^{n - 1}} + \,{a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$
has a positive root $x= \alpha ,$ then the equation
$n{a_n}{x^{n - 1}} + \,(n - 1){a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1} = 0$
has a positive root which is
Equal to $\alpha$
Greater than or equal to $\alpha$
Smaller than $\alpha$
Greater than $\alpha$
Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$
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
Consider the function $f(x) = {e^{ - 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is
Examine the applicability of Mean Value Theorem:
$(i)$ $f(x)=[x]$ for $x \in[5,9]$
$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$
$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$
Let $\mathrm{f}$ be any continuous function on $[0,2]$ and twice differentiable on $(0,2)$. If $\mathrm{f}(0)=0, \mathrm{f}(1)=1$ and $f(2)=2$, then