JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 11)
If the functions are defined as $$f(x) = \sqrt x $$ and $$g(x) = \sqrt {1 - x} $$, then what is the common domain of the following functions :
f + g, f $$-$$ g, f/g, g/f, g $$-$$ f where $$(f \pm g)(x) = f(x) \pm g(x),(f/g)x = {{f(x)} \over {g(x)}}$$
f + g, f $$-$$ g, f/g, g/f, g $$-$$ f where $$(f \pm g)(x) = f(x) \pm g(x),(f/g)x = {{f(x)} \over {g(x)}}$$
$$0 \le x \le 1$$
$$0 \le x < 1$$
$$0 < x < 1$$
$$0 < x \le 1$$
Explanation
$$f + g = \sqrt x + \sqrt {1 - x} $$
$$ \Rightarrow x \ge 0$$ & $$1 - x \ge 0 \Rightarrow x \in [0,1]$$
$$f - g = \sqrt x - \sqrt {1 - x} $$
$$ \Rightarrow x \ge 0$$ & $$1 - x \ge 0 \Rightarrow x \in [0,1]$$
$$f/g = {{\sqrt x } \over {\sqrt {1 - x} }}$$
$$ \Rightarrow x \ge 0$$ & $$1 - x > 0 \Rightarrow x \in [0,1)$$
$$g/f = {{\sqrt {1 - x} } \over {\sqrt x }}$$
$$ \Rightarrow 1 - x \ge 0$$ & $$x > 0 \Rightarrow x \in (0,1]$$
$$g - f = \sqrt {1 - x} - \sqrt x $$
$$ \Rightarrow 1 - x \ge 0$$ & $$x \ge 0 \Rightarrow x \in [0,1]$$
$$ \Rightarrow $$ $$x \in (0,1)$$
$$ \Rightarrow x \ge 0$$ & $$1 - x \ge 0 \Rightarrow x \in [0,1]$$
$$f - g = \sqrt x - \sqrt {1 - x} $$
$$ \Rightarrow x \ge 0$$ & $$1 - x \ge 0 \Rightarrow x \in [0,1]$$
$$f/g = {{\sqrt x } \over {\sqrt {1 - x} }}$$
$$ \Rightarrow x \ge 0$$ & $$1 - x > 0 \Rightarrow x \in [0,1)$$
$$g/f = {{\sqrt {1 - x} } \over {\sqrt x }}$$
$$ \Rightarrow 1 - x \ge 0$$ & $$x > 0 \Rightarrow x \in (0,1]$$
$$g - f = \sqrt {1 - x} - \sqrt x $$
$$ \Rightarrow 1 - x \ge 0$$ & $$x \ge 0 \Rightarrow x \in [0,1]$$
$$ \Rightarrow $$ $$x \in (0,1)$$
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