If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is :

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point :

The integral $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ is equal to