Find the area bounded by the curves $${x^2} = y,{x^2} = - y$$ and $${y^2} = 4x - 3.$$

If $$\left[ {\matrix{ {4{a^2}} & {4a} & 1 \cr {4{b^2}} & {4b} & 1 \cr {4{c^2}} & {4c} & 1 \cr } } \right]\left[ {\matrix{ {f\left( { - 1} \right)} \cr {f\left( 1 \right)} \cr {f\left( 2 \right)} \cr } } \right] = \left[ {\matrix{ {3{a^2} + 3a} \cr {3{b^2} + 3b} \cr {3{c^2} + 3c} \cr } } \right],\,\,f\left( x \right)$$ is a quadratic
function and its maximum value occurs at a point $$V$$. $$A$$ is a point of intersection of $$y=f(x)$$ with $$x$$-axis and point $$B$$ is such that chord $$AB$$ subtends a right angle at $$V$$. Find the area enclosed by $$f(x)$$ and chord $$AB$$.

$$f(x)$$ is a differentiable function and $$g(x)$$ is a double differentiable
function such that $$\left| {f\left( x \right)} \right| \le 1$$ and $$f'(x)=g(x).$$
If $${f^2}\left( 0 \right) + {g^2}\left( 0 \right) = 9.$$ Prove that there exists some $$c \in \left( { - 3,3} \right)$$
such that $$g(c).g''(c)<0.$$

If $$y\left( x \right) = \int\limits_{{x^2}/16}^{{x^2}} {{{\cos x\cos \sqrt \theta } \over {1 + {{\sin }^2}\sqrt \theta }}d\theta ,} $$ then find $${{dy} \over {dx}}$$ at $$x = \pi $$