1 Answer
Solution :-
(i) f (\( \dfrac{1}{2}\))
When, 0 ≤ x ≤ 1, f(x) = x
f (\( \dfrac{1}{2}\)) = \( \dfrac{1}{2}\)
(ii) f (-2)
When, x < 0, f(x) = x2
f (–2) = (–2)2
= 4
f (–2) = 4
(iii) f (1)
When, x ≥ 1, f (x) = \( \dfrac{1}{x}\)
f (1) = \( \dfrac{1}{1}\)
f(1) = 1
(iv) \( f(\sqrt{3})\)
We have \( \sqrt{3}\) = 1.732 > 1
When, x ≥ 1, f (x) = \( \dfrac{1}{x}\)
\(f( \sqrt{3})\)= \(\dfrac{1}{ \sqrt{3}}\)
(v) \(f(- \sqrt{3})\)
We know \(- \sqrt{3}\) is not a real number and the function f(x) is defined only when x ∈ R.
\(f(- \sqrt{3})\) does not exist.
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