Application of Derivatives

Let $$f(x)$$ be a polynomial of degree four having extreme values at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :

A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:

The minimum distance of a point on the curve y = x²−4 from the origin is :

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 :

Let $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$, $$x \in R - \left\{ { - 1,0,1} \right\}$$. If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm²) of this cone is :

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x³ $$-$$ 9x² + 12x + 5 in the interval [0, 3]. Then M $$-$$m is equal to :

A helicopter is flying along the curve given by y – x^3/2 = 7, (x $$\ge$$ 0). A soldier positioned at the point $$\left( {{1 \over 2},7} \right)$$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -

If ƒ(x) is a non-zero polynomial of degree four, having local extreme points at x = –1, 0, 1; then the set S = {x $$\in$$ R : ƒ(x) = ƒ(0)} Contains exactly :

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is