In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

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

For a polynomial $g(x)$ with real coefficients,let $m_g$ denote the number of distinct real roots of $g(x)$. Suppose $S$ is the set of polynomials with real coefficients defined by $S = \{(x^2-1)^2(a_0+a_1x+a_2x^2+a_3x^3) : a_0, a_1, a_2, a_3 \in \mathbb{R}\}$. For a polynomial $f$,let $f'$ and $f''$ denote its first and second order derivatives,respectively. Then the minimum possible value of $(m_f + m_{f'})$,where $f \in S$,is

Let $f(x)=2+\cos x$ for all real $x$.
$STATEMENT-1$: For each real $t$,there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(c)=0$. because
$STATEMENT-2$: $f(t)=f(t+2\pi)$ for each real $t$.

$f:[2,10] \rightarrow R$ is defined as $f(x) = \begin{cases} \frac{1}{2}(x-6)^2-3, & x \leq 4 \\ x-5, & x > 4 \end{cases}$. Which of the following is true?

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