Limits and Deri...

Suppose the function $$f(x)-f(2x)$$ has the derivative $$5$$ at $$x=1$$ and derivative $$7$$ at $$x=2$$.The derivative  of the function $$f(x)-f(4x)$$ at $$x=1$$, has the value equal to 

Which one of the following statements is true?

If $$\displaystyle y=\frac { x }{ a+\displaystyle\frac { x }{ b+\displaystyle\frac { x }{ a+\displaystyle\frac { x }{ b+.....\infty  }  }  }  } $$, then $$\cfrac{dy}{dx} =$$

Let $$f(x)=\sin x$$, $$g(x)=\left [ x+1 \right ]$$ and $$g(f(x))=h(x)$$, where [.] is the greatest integer function. Then $$h^+\left ( \displaystyle \dfrac{\pi }{2} \right )$$ is

Find the solution of $$\displaystyle \frac{dy}{dx}= \frac{2x+2y-2}{3x+y-5}.$$

Given : $$f(x)=4x^3-6x^2\cos2a+3x \sin 2a.\sin 6a+\sqrt{\ln (2a-a^2)}$$ then 

$$\displaystyle \lim_{n\to\infty }\frac{n^{p}\sin ^{2}\left ( n! \right )}{n+1}$$, $$0<p<1$$, is equal to

$$f\left( x \right)=\begin{cases} \sin { x } \qquad ;\qquad x\neq n\pi ,n=0,\pm 1,\pm 2,\pm 3..... \\ 2\qquad \qquad ;\qquad otherwise \end{cases}$$ and $$g\left( x \right) =\begin{cases} { x }^{ 2 }+1\qquad ;\qquad x\neq 0 \\ 4\qquad \qquad ;\qquad x=0 \end{cases}.$$ 

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\left ( 1-\tan \dfrac{x}{2} \right )\left ( 1-\sin x \right )}{\left ( 1+\tan \dfrac{x}{2} \right )\left ( \pi -2x \right )^{3}}$$ is

Let $$f\left ( x \right )=\begin{cases}\sin x, x\neq n\pi 
                   \\ 2,  x=n\pi \end{cases}$$, where $$n\epsilon \mathbb{Z}$$ and
$$g\left ( x \right )=\begin{cases}x^{2}+1, x\neq 2 \\
              3, x=2 \end{cases}$$.
Then $$\displaystyle \lim_{x\to 0}g\left ( f\left ( x \right ) \right )$$ is