JEE MAIN - Mathematics Bengali (2022 - 24th June Morning Shift - No. 13)
$$f\left( x \right) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}\left( {{x^2} - 3x + 2} \right)}}$$ অপেক্ষকটির সংজ্ঞার অঞ্চল হল :
$$\left( { - \,\infty ,\,1} \right) \cup \left( {2,\,\infty } \right)$$
$$\left( {2,\,\infty } \right)$$
$$\left[ { - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}},\,1} \right] \cup \left( {2,\,\infty } \right)$$
$$\left[ { - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}},\,1} \right] \cup \left( {2,\,\infty } \right) - \left\{ {{{3 + \sqrt 5 } \over 2},{{3 - \sqrt 5 } \over 2}} \right\}$$
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