If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following points lies on the curve ?

Let S be the set of all values of x for which the tangent to the curve y = ƒ(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment joining the points (1, ƒ(1)) and (–1, ƒ(–1)), then S is equal to :

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

Let ƒ : [0, 2] $$\to$$ R be a twice differentiable function such that ƒ''(x) > 0, for all x $$\in$$ (0, 2). If $$\varphi$$(x) = ƒ(x) + ƒ(2 – x), then $$\varphi$$ is :

Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is 2y \over x^2. If the curve passes through the centre of the circle x2 + y2 – 2x – 2y = 0, then its equation is :

The maximum volume (in cu.m) of the right circular cone having slant height 3 m is :

The shortest distance between the point \left( {{3 \over 2},0} \right) and the curve y = $$\sqrt{x}$$, (x > 0), is -

The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point

A helicopter is flying along the curve given by y – x3/2 = 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 -

The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = $$\left\{x\in R:x^2+30\le 11x\right\}$$ is :

Let f(x) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\, x $$\in$$ R, where a, b and d are non-zero real constants. Then :

If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a$$\in$$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, {{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right) is :

The tangent to the curve y = x2 – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point :

If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x $$\in$$ R, then :

The area (in sq. units) of the region A = { (x, y) $$\in$$ R × R| 0 $$\le$$ x $$\le$$ 3, 0 $$\le$$ y $$\le$$ 4, y $$\le$$ x2 + 3x} is :

If the area (in sq. units) bounded by the parabola y2 = 4$$\lambda$$x and the line y = $$\lambda$$x, $$\lambda$$ > 0, is {1 \over 9} , then $$\lambda$$ is equal to :

If the area (in sq. units) of the region {(x, y) : y2 $$\le$$ 4x, x + y $$\le$$ 1, x $$\ge$$ 0, y $$\ge$$ 0} is a $$\sqrt{2}$$ + b, then a – b is equal to :

The area (in sq.units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is :

The area (in sq. units) of the region A = {(x, y) : {{y{}^2} \over 2} $$\le$$ x $$\le$$ y + 4} is :-