The value of $c$ in the Lagrange's mean value theorem for the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-4 \mathrm{x}^{2}+8 \mathrm{x}+11$ when $\mathrm{x} \in[0,1]$ is
$\frac{2}{3}$
$\frac{\sqrt{7}-2}{3}$
$\frac{4-\sqrt{5}}{3}$
$\frac{4-\sqrt{7}}{3}$
If $f:R \to R$ and $f(x)$ is a polynomial function of degree ten with $f(x)=0$ has all real and distinct roots. Then the equation ${\left( {f'\left( x \right)} \right)^2} - f\left( x \right)f''\left( x \right) = 0$ has
If $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$
If $f(x)$ satisfies the conditions of Rolle’s theorem in $[1,\,2]$ and $f(x)$ is continuous in $[1,\,2]$ then $\int_1^2 {f'(x)dx} $ is equal to
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....................
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 $t$.