Let $f(x) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} ;} $ $1 < x < 26$ be real valued function. Then $f\,'(x)$ for $1 < x < 26$ is

If $(1 -x + 2x^2)^n$ = $a_0 + a_1x + a_2x^2+..... a_{2n}x^{2n}$ , $n \in N$ , $x \in R$ and $a_0$ , $a_2$ and $a_1$ are in $A$ . $P$ .,then there exists 

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)

$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$

$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$

$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$

$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

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

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 $t$.