વિધેય fleft( x right) = frac{1}{{4 - {x^2}}} | vedclass

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Question

$f\left( x \right) = \frac{1}{{4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)$

Let ${f_1} = \frac{1}{{4 - {x^2}}}$   and

Vedclass · Accepted Answer

$\left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right)$

Vedclass · Answer

$f\left( x ight) = \frac{1}{{4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} ight)$

Let ${f_1} = \frac{1}{{4 - {x^2}}}$   and   ${f_2} = {\log _{10}}\left( {{x^3} - x} ight)$

$ \Rightarrow 4 - {x^2} 
e 0\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} - x > 0$

$ \Rightarrow x 
e  \pm 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x\left( {x + 1} ight)\left( {x + 1} ight) > 0$

$x \in \left( { - 1,0} ight) \cup \left( {1,\infty } ight) - \left\{ 2 ight\}$

$x \in \left( { - 1,0} ight) \cup \left( {1,2} ight) \cup \left( {2,\infty } ight)$

વિધેય $f\left( x \right) = \frac{1}{{4 - {x^2}}} + \log \,\left( {{x^3} - x} \right)$ નો પ્રદેશ મેળવો.

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