JEE MAIN - Mathematics (2003 - No. 5)
Domain of definition of the function f(x) = $${3 \over {4 - {x^2}}}$$ + $${\log _{10}}\left( {{x^3} - x} \right)$$, is
(-1, 0)$$ \cup $$(1, 2)$$ \cup $$(2, $$\infty $$)
(1, 2)
(-1, 0) $$ \cup $$ (1, 2)
(1, 2)$$ \cup $$(2, $$\infty $$)
Explanation
$$f\left( x \right) = {3 \over {4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)$$
$$4 - {x^2} \ne 0;\,\,\,{x^3} - x > 0;$$
$$x \ne \pm \sqrt 4 $$ and $$ - 1 < x < 0$$
or $$\,\,\,1 < x < \infty $$
$$\therefore$$ $$D = \left( { - 1,0} \right) \cup \left( {1,\infty } \right) - \left\{ {\sqrt 4 } \right\}$$
$$D = \left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right).$$
$$4 - {x^2} \ne 0;\,\,\,{x^3} - x > 0;$$
$$x \ne \pm \sqrt 4 $$ and $$ - 1 < x < 0$$
or $$\,\,\,1 < x < \infty $$
$$\therefore$$ $$D = \left( { - 1,0} \right) \cup \left( {1,\infty } \right) - \left\{ {\sqrt 4 } \right\}$$
$$D = \left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right).$$
Comments (0)
