Minimum integral value of $\alpha$ for which graph of $f(x) = ||x -2| -\alpha|-5$ has exactly four $x-$intercepts-
$6$
$4$
$7$
$5$
Let $[x]$ denote the greatest integer $\leq x$, where $x \in R$. If the domain of the real valued function $\mathrm{f}(\mathrm{x})=\sqrt{\frac{[\mathrm{x}] \mid-2}{\sqrt{[\mathrm{x}] \mid-3}}}$ is $(-\infty, \mathrm{a}) \cup[\mathrm{b}, \mathrm{c}) \cup[4, \infty), \mathrm{a}\,<\,\mathrm{b}\,<\,\mathrm{c}$, then the value of $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is:
If function $f(x) = \frac{1}{2} – \tan \left( {\frac{{\pi x}}{2}} \right)$; $( – 1 < x < 1)$ and $g(x) = \sqrt {3 + 4x – 4{x^2}} $, then the domain of $gof$ is
If the domain of the function $f(\mathrm{x})=\frac{\cos ^{-1} \sqrt{x^{2}-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}$ is the interval $(\alpha, \beta]$, then $\alpha+\beta$ is equal to:
Let $f\,:\,R \to R$ be a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$, $x \in R$. Then $f(2)$ equals
Which of the following function is surjective but not injective
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