Which of the following is correct
$\sqrt {{x^2}} = \,\left| x \right|$
${x^{x + 1}}\, = \,x.{x^x}$
$\frac{{\left| x \right|}}{x} = \left\{ \begin{array}{l}
1:x > 0\\
- 1:x < 0
\end{array} \right.$
All
Let $a,b,c\; \in R.$ If $f\left( x \right) = a{x^2} + bx + c$ is such that $a + b + c = 3$ and $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy,$ $\forall x,y \in R,$ then $\mathop \sum \limits_{n = 1}^{10} f\left( n \right)$ is equal to :
Let $f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)$. If $f^{a+1}(x)=f\left(f^{n}(x)\right)$ for all $n \in N$, then $f^{\prime}(6)+f(7)$ is equal to
The range of the function,
$\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$
$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ is :
Which one of the following is not bounded on the intervals as indicated
If $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, for what value of $\alpha $ is $f(f(x)) = x$