Continuity and ...

If a curve is given by $$x=a\cos t+\displaystyle\frac{b}{2}\cos 2t$$ and $$y=\sin t +\displaystyle\frac{b}{2}\sin 2t$$, then the points for which $$\displaystyle\frac{d^2y}{dx^2}=0$$, are given by.

Let f be a function which is continuous and differentiable for all real x. If $$f(2)=-4$$ and $$f'(x)\geq 6$$ for all $$x\epsilon [2, 4]$$, then.

Let $$f(x)$$ be a differentiable function such that $$f(x) = x^{2} + \int_{0}^{x} e^{-t} f(x - t)dt$$, then $$\int_{0}^{1}f(x) dx=$$

If $$\sin ^{ -1 }{ \left( \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } }  \right)  } =\log { a } $$, then $$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } $$ equals

If $$y=\tan ^{ -1 }{ \left( \cfrac { 4x }{ 1+5{ x }^{ 2 } }  \right)  } +\tan ^{ -1 }{ \left( \cfrac { 2+3x }{ 2-3x }  \right)  } $$, then $$\cfrac { dy }{ dx } $$ is

Let $$f\left( x \right)$$ and $$g\left( x \right)$$ be defined and differentiable for $$x\ge { x }_{ 0 }$$ and $$f\left( { x }_{ 0 } \right) =g\left( { x }_{ 0 } \right) , f^{ ' }\left( x \right) >g^{ ' }\left( x \right) for\ x>{ x }_{ 0 }$$ then

Let g (x), $$x \geq 0$$, be a non-negative continuous function and let $$F(x) = \int_{0}^{x} f(t) dt$$, $$x \geq 0$$. If for some c > 0, f(x) $$\leq$$ c F(x) for all $$x \geq 0$$, then

If the function $$f:\left[ 0,8 \right] \rightarrow R$$ is differentiable, then for $$0<a,b<2,\int _{ 0 }^{ 8 }{ f(t) } dt$$ is equal to

A point where function $$f(x)$$ is not continuous where $$f(x)=\left[ \sin { \left[ x \right]  }  \right] $$ in $$\left( 0,2\pi  \right) $$; is ($$\left[ \ast  \right] $$ denotes greatest integer $$\le x$$)

Consider $$f(x)=\lim _{ n\rightarrow \infty  }{ \cfrac { { x }^{ n }-\sin { { x }^{ n } }  }{ { x }^{ n }+\sin { { x }^{ n } }  }  } $$ for $$x>0,x\neq 1,f(1)=0$$ then