(i) [0, ∞ ) (ii)

(iii) [0, 4] (iv) {–1, 1}

Sol. (i) y = |x – 3|

Range y ∈ [0, ∞ )

(ii) y =

Domain x ∈ R

yx 2 – x + y = 0

quadratic in x has real roots as x ∈ R

∴ Discriminant D ≥ 0 ⇒ 1 – 4y 2 ≥ 0 ⇒ (2y – 1) (2y + 1) ≤ 0i ⇒ y ∈

Here at y = 0 quadratic vanishes. so we have to check this seperately

Put y = 0 ⇒ x = 0 (a point with in domain)

∴ y = 0 point is included in the range

Note : If there is no point of x in the domain for the value of y for which quadratic vanishes, we have to remove that point from range

Method 2

f(x) = = We know that ≥ 2 ⇒ 0 < ≤ ⇒ ∈ ∪

But divison by x is done by us, So at x = 0 , y = 0 ∴ Range y ∈

Method 3

f(x) = is an odd function ⇒ f ′ (x) = = 0 x = ± 1

> 0 x ∈ (–1, 1)

< 0 x ∈ (– ∞ , –1) ∪ (1, ∞ )

= 0 (0 + more accurately)

= 0 (0 – more accurately)

Range y ∈

(iii) f(x) = . Domain x ∈ [–4, 4] ⇒ f(x) > 0, y = ⇒ x 2 + y 2 = 16

Equation of semicircle

∴ Range y ∈ [0, 4]

(iv) f(x) = , x ≠ 4

f(x) =

∴ Range y ∈ {–1, 1}

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