For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\frac{a_{3}}{4}=0$,the equation $a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=0$ has a real root in the interval

Let $R$ be the set of all real numbers and $f:[-1,1] \rightarrow R$ be defined as $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$. Then:

Let $f(x)=(x-1)(x-2)(x-3)$,where $x \in [0,4]$. Find the values of $c$ if Lagrange's Mean Value Theorem $(LMVT)$ can be applied.

If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$,then $a+b=$

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

Let $f$ and $g$ be twice differentiable even functions on $(-2, 2)$ such that $f(\frac{1}{4}) = 0, f(\frac{1}{2}) = 0, f(1) = 1$ and $g(\frac{3}{4}) = 0, g(1) = 2$. Then,the minimum number of solutions of $f(x)g''(x) + f'(x)g'(x) = 0$ in $(-2, 2)$ is equal to