L 'Hopital 's rules says that the  
lim x →a f(x)/g(x)
⇒f '(a)/g '(a)
Using this, we get  
lim x →0 (1 −cosx)/x²
⇒−sin0/2(0)
Yet as the denominator is 0, this is impossible. So we do a second limit:
lim(x →0) sinx/2x
⇒cos0/2 = 1/2 = 0.5
So, in total lim x →0 (1 −cosx)/x²
⇒lim x →0 sinx/2x
⇒cosx/2
⇒cos0/2= 1/2

As all the three remaining statements are true for the given function.

f(g(x)) = x  
f '(g(x)) g '(x) = 1  
put x = a
f '(b) g '(a) = 1
2 f '(b) = 1
f '(b) = 1/2

The graph of f(x) = |x|
As observed from the graph, f(x) = |x| is continuous at x = 0.
As this curve is pointed at x = 0, f(x) is not derivable at x = 0.

Given x^p y^q  = (x+y)^p+q Taking log on both sides we get:
_p logx+_q logy = (p+q) log (x+y). Differentiating both sides w.r.t. x we get , 

y = ae^mx  + be^-mx  ⇒y1 = ame^mx  + (-m)be^-mx  ⇒y2

= am² e^mx _{+ (m}² )be^-mx  ⇒y2 = m²  (ae^mx  + be^-mx ) ⇒y2 = m² y