JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 10)
The domain of the definition of the function
$$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$ is
$$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$ is
(-1, 0) $$ \cup $$ (1, 2) $$ \cup $$ (2, $$\infty $$)
(-2, -1) $$ \cup $$ (-1,0) $$ \cup $$ (2, $$\infty $$)
(1, 2) $$ \cup $$ (2, $$\infty $$)
(-1, 0) $$ \cup $$ (1,2) $$ \cup $$ (3, $$\infty $$)
Explanation
Given $$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$
Let f1(x) = $${1 \over {4 - {x^2}}}$$ and f2(x) = $${\log _{10}}({x^3} - x)$$
Here in f1(x) denominator $$ \ne $$ 0
4 - x2 $$ \ne $$ 0
$$ \Rightarrow $$ x $$ \ne $$ $$ \pm $$ 2 ...........(1)
and x3 - x > 0
$$ \Rightarrow $$ x(x2 - 1) > 0
$$ \Rightarrow $$ x(x + 1)(x - 1) > 0
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x $$ \in $$ (-1, 0) $$ \cup $$ (1, $$\infty $$) ........(2)
Hence domain is intersection of (1) and (2)
x $$ \in $$ (-1, 0) $$ \cup $$ (1, 2) $$ \cup $$ (2, $$\infty $$)
Let f1(x) = $${1 \over {4 - {x^2}}}$$ and f2(x) = $${\log _{10}}({x^3} - x)$$
Here in f1(x) denominator $$ \ne $$ 0
4 - x2 $$ \ne $$ 0
$$ \Rightarrow $$ x $$ \ne $$ $$ \pm $$ 2 ...........(1)
and x3 - x > 0
$$ \Rightarrow $$ x(x2 - 1) > 0
$$ \Rightarrow $$ x(x + 1)(x - 1) > 0
x $$ \in $$ (-1, 0) $$ \cup $$ (1, $$\infty $$) ........(2)
Hence domain is intersection of (1) and (2)
x $$ \in $$ (-1, 0) $$ \cup $$ (1, 2) $$ \cup $$ (2, $$\infty $$)
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