$$25^{190}-19^{190}-8^{190}+2^{190}$$ is divisible by :

The absolute difference of the coefficients of $$x^{10}$$ and $$x^7$$ in the expansion of $${\left(2x^2+\frac{1}{2x}\right)}^{11}$$ is equal to :

If the coefficients of three consecutive terms in the expansion of $$(1+x{)}^n$$ are in the ratio $$1:5:20$$, then the coefficient of the fourth term is

If $${}^{2n}{C}_3:{}^n{C}_3=10:1$$, then the ratio $$\left(n^2+3n\right):\left(n^2-3n+4\right)$$ is :

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $${\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)}^{\mathrm{n}}$$ is $$\sqrt{6}:1$$, then the third term from the beginning is :

If the coefficient of $$x^7$$ in {\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}} and $$x^{-7}$$ in {\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}} are equal, then :

Among the statements : (S1) : $$2023^{2022}-1999^{2022}$$ is divisible by 8 (S2) : $$13(13{)}^n-12n-13$$ is divisible by 144 for infinitely many $$n\in \mathbb{N}$$

Let f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx}. If f(3) = {1 \over 2}({\log _e}5 - {\log _e}6), then $$f(4)$$ is equal to

For $$\alpha ,\beta ,\gamma ,\delta \in \mathbb{N}$$, if $$\int \left({\left(\frac{x}{e}\right)}^{2x}+{\left(\frac{e}{x}\right)}^{2x}\right){\log }_{e}xdx=\frac{1}{\alpha }{\left(\frac{x}{e}\right)}^{\beta x}-\frac{1}{\gamma }{\left(\frac{e}{x}\right)}^{\delta x}+C$$ , where e=\sum_\limits{n=0}^{\infty} \frac{1}{n !} and $$\mathrm{C}$$ is constant of integration, then $$\alpha +2\beta +3\gamma -4\delta$$ is equal to :

If $$I(x)=\int e^{{\sin }^{2}x}(\cos x\sin 2x-\sin x)dx$$ and $$I(0)=1$$, then I\left( {{\pi \over 3}} \right) is equal to :

The integral $$\int \left[{\left(\frac{x}{2}\right)}^x+{\left(\frac{2}{x}\right)}^x\right]\ln \left(\frac{ex}{2}\right)dx$$ is equal to :

Let $$I(x)=\int \frac{(x+1)}{x{\left(1+x{e}^x\right)}^2}dx,x>0$$. If \lim_\limits{x \rightarrow \infty} I(x)=0, then $$I(1)$$ is equal to :

Let $$I(x)=\int \frac{x^2\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx$$. If $$I(0)=0$$, then $$I\left(\frac{\pi }{4}\right)$$ is equal to :

Let $$f:\mathbb{R}-0,1\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to

Let $f:\mathbb{R}-\{2,6\}\to \mathbb{R}$ be real valued function defined as $f(x)=\frac{x^2+2x+1}{x^2-8x+12}$. Then range of $f$ is

The absolute minimum value, of the function $f(x)=\left|x^2-x+1\right|+\left[x^2-x+1\right]$, where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is :

Let f(x) = \left| {\matrix{ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr } } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

If the domain of the function $$f(x)=\frac{[x]}{1+x^2}$$, where $$[x]$$ is greatest integer $$\le x$$, is $$[2,6)$$, then its range is

The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is :

Consider a function $$f:\mathbb{N}\to \mathbb{R}$$, satisfying $$f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge 2$$ with $$f(1)=1$$. Then $$\frac{1}{f(2022)}+\frac{1}{f(2028)}$$ is equal to