Let A = {1, 2} and B = {3, 6} and f and g be functions from A to B, defined by f(x) = 3x and g(x) = x² + 2. Show that f = g.
$[3]$

Open in App

Solution

$Since both f and g are defined from set A Therefore, domain (f) = domain (g) Now, for the co - domain of f (1) = 3 \times 1 = 3 f (2) = 3 \times 2 = 6 Co - domain of f = {3, 6}$
$[2]$

$And the co - domain of g : g (1) = 1^{2} + 2 = 3 g (2) = 2^{2} + 2 = 6 Co - domain of g = {3, 6} Co - domain of f = Co - domain of g for each x \in A Hence f (x) = g (x) for each x \in A$
$[1]$

Suggest Corrections

Similar questions

Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, g: A → B be functions defined by f(x) = x² − x, x ∈ A and. Are f and g equal?

Justify your answer. (Hint: One may note that two function f: A → B and g: A → B such that f(a) = g(a) &mnForE;a ∈A, are called equal functions).

f, g

R \to R

f (x) = x + 1, g (x) = 2 x - 3

f + g, f - g

\frac{f}{g}

Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and.

Join BYJU'S Learning Program