Class 11 Maths NCERT Solutions Chapter 2 Ex 2.3 – Functions

Every first element (2, 5, 8, 11, 14, 17) is distinct, each with exactly one image.

Every first element (2, 4, 6, 8, 10, 12, 14) is distinct, each with exactly one image.

The first element 1 appears twice, once with image 3 and once with image 5 — two different images for the same input.

|x| is defined for every real x, so the domain is all of R.

Since |x| ≥ 0 always, −|x| ≤ 0 always, and as x ranges over R, |x| takes every value in [0, ∞), so −|x| takes every value in (−∞, 0].

The expression under the square root must be non-negative: 9 − x² ≥ 0 ⟹ x² ≤ 9 ⟹ −3 ≤ x ≤ 3.

For x ∈ [−3, 3], x² ranges over [0, 9], so 9 − x² ranges over [0, 9], and √(9 − x²) ranges over [0, 3].

f(0) = 2(0) − 5 = −5

f(7) = 2(7) − 5 = 14 − 5 = 9

f(−3) = 2(−3) − 5 = −6 − 5 = −11

t(0) = (9×0)/5 + 32 = 0 + 32 = 32

t(28) = (9×28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4

t(−10) = (9×−10)/5 + 32 = −90/5 + 32 = −18 + 32 = 14

9C/5 + 32 = 212

9C/5 = 180

C = 180 × 5/9 = 100

As x ranges over (0, ∞), 3x ranges over (0, ∞), so 2 − 3x ranges over (−∞, 2).

x² ≥ 0 for every real x, so x² + 2 ≥ 2. As x ranges over R, x² takes every value in [0, ∞), so x² + 2 takes every value in [2, ∞).

This is the identity function — every real number is its own image.