Continuity and ...

$$f(x)= x\sin\dfrac{1}{x} , \  for x\neq 0$$
       $$= 0,\  for x=0$$

A function $$f$$ satisfies the relation $$f(x)=f'(x)+f"(x)+.....\infty$$ terms, where $$f(x)$$ is differentiable indefinitely, If $$f'(-1)=1$$ then $$f(-1)$$ is equal to

$$f(x) = \left\{\begin{matrix}(3/x^{2})\sin 2x^{2} & if x M 0 \\\dfrac {x^{2} + 2x + c}{1 - 3x^{2}}  & if\ x \geq 0, x \neq \dfrac {1}{\sqrt {3}}\\ 0 & x = 1/ \sqrt {3}\end{matrix}\right.$$ then in order that $$f$$ be continuous at $$x = 0$$, the value of $$c$$ is

Let $$g(x)$$ be a continuous function for all $$x$$, and $$f(x)=f(\alpha)+(x-\alpha).g(x)\ \forall \ x\ =\epsilon \ R$$. Then;

Let a function $$f: R\rightarrow R$$ be given by $$f(x+y)=f(x)f(y)$$ for all  $$x, y \in R$$ and $$f(x)\neq 0$$ for any $$x_{1}$$ function $$f(x)$$ is differentiable at $$x=0$$. Find $$f(x)    gi ven              f(0)=1$$.

$$\dfrac { d }{ dx } \left[ { cos }^{ -1 }\left( x\sqrt { x } -\sqrt { \left( 1-x \right) \left( 1-{ x }^{ 2 } \right)  }  \right)  \right] =$$

If f is twice differentianle such that f"$$\left( x \right) =  - f\left( x \right),f'\left( x \right) = g\left( x \right),h'\left( x \right) = {\left( {f\left( x \right)} \right)^2} + {\left( {g\left( x \right)} \right)^2}$$ and $$h\left( 0 \right) = 2,h\left( 1 \right) = 4$$,then equation of $$h\left( x \right)$$ represents ?

$$\frac{d}{{dx}}\left[ {a{{\tan }^{ - 1}}x + b\log \left( {\frac{{x - 1}}{{x + 1}}} \right)} \right] = \frac{1}{{{x^4} - 1}} \Rightarrow a - 2b = $$

Let $$f$$ be the function on $$[0,\,1]$$ given by $$f\left( x \right) = x\sin \frac{\pi }{x}\,for\,x \ne 0$$ and $$f(0)=0$$. Then 

If $$f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$$, then $$f(x)$$ is differentiable on