The range of the function $f(x) = \frac{{\sqrt {1 - {x^2}} }}{{1 + \left| x \right|}}$ is
$\left[ {0,1} \right]$
$\left[ {0,\frac{1}{{\sqrt 2 }}} \right]$
$\left[ {0,\frac{1}{2}} \right]$
$\left[ {0,\frac{{\sqrt 3 }}{2}} \right]$
If domain of the function $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$ is $(\alpha, \beta) \cup(\gamma, \delta]$, then $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)$ is equal to $....$.
If $f(x) = \frac{1}{{\sqrt {x + 2\sqrt {2x - 4} } }} + \frac{1}{{\sqrt {x - 2\sqrt {2x - 4} } }}$ for $x > 2$, then $f(11) = $
If $f(x)$ and $g(x)$ are functions satisfying $f(g(x))$ = $x^3 + 3x^2 + 3x + 4$ $f(x)$ = $log^3x + 3$, then slope of the tangent to the curve $y = g(x)$ at $x = \ -1$ is
If $f(a) = a^2 + a+ 1$ , then number of solutions of equation $f(a^2) = 3f(a)$ is
If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x = 1}^n {f\left( x \right) = 120,} $ find the value of $n$