The Total solution for NCERT class 6-12
(i) f(1/2)
When, 0 ≤ x ≤ 1, f(x) = x
∴f (1/2) = ½
(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) = 1/x
f (1) = 1/1
∴ f(1)= 1
(iv) f(√3)
We have √3 = 1.732 > 1
∴f (√3) = 1/√3
(v) f(√-3)
We know √-3 is not a real number and the function f(x) isdefined only when x ∈ R.
∴ f(√-3) does not exist.