If the value of the integral \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx is {k \over 6}, then k is equal to :

Suppose f(x) is a polynomial of degree four, having critical points at –1, 0, 1. If T = {x $$\in$$ R | f(x) = f(0)}, then the sum of squares of all the elements of T is :

Let $$f(x)=\left|x-2\right|$$ and g(x) = f(f(x)), $$x\in \left[0,4\right]$$. Then $$\underset{0}{\overset{3}{\int }}\left(g(x)-f(x)\right)dx$$ is equal to:

The integral \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} is equal to:

The value of \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} is:

If I1 = $$\underset{0}{\overset{1}{\int }}{\left(1-x^{50}\right)}^{100}dx$$ and I2 = $$\underset{0}{\overset{1}{\int }}{\left(1-x^{50}\right)}^{101}dx$$ such that I2 = $$\alpha$$I1 then $$\alpha$$ equals to :

The integral $$\underset{1}{\overset{2}{\int }}{e}^x.x^x\left(2+{\log }_{e}x\right)dx$$ equals :

If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then {1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx is equal to:

The value of $$\alpha$$ for which $$4\alpha \underset{-1}{\overset{2}{\int }}{e}^{-\alpha |x|}dx=5$$, is:

If $$\theta$$1 and $$\theta$$2 be respectively the smallest and the largest values of $$\theta$$ in (0, 2$$\pi$$) - {$$\pi$$} which satisfy the equation, 2cot2$$\theta$$ - {5 \over {\sin \theta }} + 4 = 0, then $$\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta$$ is equal to :

If I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}}, then :

The value of \int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx is equal to :

If for all real triplets (a, b, c), ƒ(x) = a + bx + cx2; then $$\underset{0}{\overset{1}{\int }}f(x)dx$$ is equal to :

Let a function ƒ : [0, 5] $$\to$$ R be continuous, ƒ(1) = 3 and F be defined as : $$F(x)=\underset{1}{\overset{x}{\int }}{t}^{2}g(t)dt$$ , where $$g(t)=\underset{1}{\overset{t}{\int }}f(u)du$$ Then for the function F, the point x = 1 is :

\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x} is equal to

Let z = x + iy be a non-zero complex number such that $$z^2=i{\left|z\right|}^2$$, where i = $$\sqrt{-1}$$ , then z lies on the :

The region represented by {z = x + iy $$\in$$ C : |z| – Re(z) $$\le$$ 1} is also given by the inequality : {z = x + iy $$\in$$ C : |z| – Re(z) $$\le$$ 1}

The value of {\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}} is :

If the four complex numbers $$z,\overline{z},\overline{z}-2\mathrm{Re}\left(\overline{z}\right)$$ and $$z-2Re(z)$$ represent the vertices of a square of side 4 units in the Argand plane, then $$|z|$$ is equal to :