Using the relation $$2\left( {1 - \cos x} \right) < {x^2},\,x \ne 0$$ or otherwise,
prove that $$\sin \left( {\tan x} \right) \ge x,\,\forall x \in \left[ {0,{\pi \over 4}} \right]$$

If the function $$f:\left[ {0,4} \right] \to R$$ is differentiable then show that
(i)$$\,\,\,\,\,$$ For $$a, b$$$$\,\,$$$$ \in \left( {0,4} \right),{\left( {f\left( 4 \right)} \right)^2} - {\left( {f\left( 0 \right)} \right)^2} = gf'\left( a \right)f\left( b \right)$$
(ii)$$\,\,\,\,\,$$ $$\int\limits_0^4 {f\left( t \right)dt = 2\left[ {\alpha f\left( {{\alpha ^2}} \right) + \beta \left( {{\beta ^2}} \right)} \right]\forall 0 < \alpha ,\beta < 2} $$

If $$P(1)=0$$ and $${{dp\left( x \right)} \over {dx}} > P\left( x \right)$$ for all $$x \ge 1$$ then prove that
$$P(x)>0$$ for all $$x>1$$.

If $$f$$ is an even function then prove that
$$\int\limits_0^{\pi /2} {f\left( {\cos 2x} \right)\cos x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\sin 2x} \right)\cos x\,dx.} $$