Let $$f:\left[ {0,2} \right] \to R$$ be a function which is continuous on $$\left[ {0,2} \right]$$ and is differentiable on $$(0,2)$$ with $$f(0)=1$$. Let
$$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} $$ for $$x \in \left[ {0,2} \right]$$. If $$F'\left( x \right) = f'\left( x \right)$$ for all $$x \in \left[ {0,2} \right]$$, then $$F(2)$$ equals

Let $$f$$ and $$g$$ be real valued functions defined on interval $$(-1, 1)$$ such that $$g''(x)$$ is continuous, $$g\left( 0 \right) \ne 0.$$ $$g'\left( 0 \right) = 0$$, $$g''\left( 0 \right) \ne 0$$, and $$f\left( x \right) = g\left( x \right)\sin x$$

STATEMENT - 1: $$\mathop {\lim }\limits_{x \to 0} \,\,\left[ {g\left( x \right)\cot x - g\left( 0 \right)\cos ec\,x} \right] = f''\left( 0 \right)$$ and
STATEMENT - 2: $$f'\left( 0 \right) = g\left( 0 \right)$$

Let $$g\left( x \right) = \log f\left( x \right)$$ where $$f(x)$$ is twice differentible positive function on $$\left( {0,\infty } \right)$$ such that $$f(x+1)=x f(x)$$. Then, for $$N=1, 2, 3, ..........$$
$$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = $$

Let $$\,\,\,$$$$f\left( x \right) = 2 + \cos x$$ for all real $$X$$.

STATEMENT - 1: for eachreal $$t$$, there exists a point $$c$$ in $$\left[ {t,t + \pi } \right]$$ such that $$f'\left( c \right) = 0$$ because
STATEMENT - 2: $$f\left( t \right) = f\left( {t + 2\pi } \right)$$ for each real $$t$$.