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 $$t$$.

If $$f(x)$$ is a twice differentiable function and given that $$f\left( 1 \right) = 1;f\left( 2 \right) = 4,f\left( 3 \right) = 9$$, then

If $$y$$ is a function of $$x$$ and log $$(x+y)-2xy=0$$, then the value of $$y'(0)$$ is equal to

Let $$f:\left( {0,\infty } \right) \to R$$ and $$F\left( x \right) = \int\limits_0^x {f\left( t \right)dt.} $$ If $$F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right)$$, then $$f(4)$$ equals