Let f:R→R be defined by f(x)=3x-5 and g:R→R by g(x)=(x+2)/3. i Prove that both f and g are one‑one and onto. ii Find the compositions (f° g)(x) and (g° f)(x). iii Determine the inverse function f^-1 and verify that f^-1(f(x))=x for all x∈R.

Let $f : R \to R$ be defined by $f (x) = 3 x - 5$ and $g : R \to R$ by $g (x) = \frac{x + 2}{3}$ . $i$ Prove that both $f$ and $g$ are one‑one and onto. $ii$ Find the compositions $(f \circ g) (x)$ and $(g \circ f) (x)$ . $iii$ Determine the inverse function $f^{- 1}$ and verify that $f^{- 1} (f (x)) = x$ for all $x \in R$ .

hard 4 marks Class 12 · Mathematics · Relations and Functions

Answer

(i) Both f and g are bijective. (ii) (f∘g)(x)=x and (g∘f)(x)=x. (iii) f^{-1}(x)=\dfrac{x+5}{3}, and f^{-1}(f(x))=x.

The correct answer is (i) Both f and g are bijective. (ii) (f∘g)(x)=x and (g∘f)(x)=x. (iii) f^{-1}(x)=\dfrac{x+5}{3}, and f^{-1}(f(x))=x. — Verify: f^{-1}(f(x)) = \dfrac{(3x-5)+5}{3}=\dfrac{3x}{3}=x, confirming the inverse property.

Step-by-step solution

For f(x)=3x-5, suppose f(a)=f(b). Then 3a-5=3b-5 ⇒ 3a=3b ⇒ a=b, so f is one‑one.
Since for any y∈ℝ we can solve y=3x-5 ⇒ x=\dfrac{y+5}{3}, every real y has a pre‑image; thus f is onto.
For g(x)=\dfrac{x+2}{3}, assume g(a)=g(b). Then (a+2)/3=(b+2)/3 ⇒ a+2=b+2 ⇒ a=b, so g is one‑one.
Given any y∈ℝ, solve y=\dfrac{x+2}{3} ⇒ x=3y-2; thus every y has a pre‑image, so g is onto.
Compute (f∘g)(x)=f(g(x))=f\bigl(\dfrac{x+2}{3}\bigr)=3\left(\dfrac{x+2}{3}\right)-5 = x+2-5 = x.
Compute (g∘f)(x)=g(f(x))=g(3x-5)=\dfrac{(3x-5)+2}{3}=\dfrac{3x-3}{3}=x.
To find f^{-1}, set y=3x-5 and solve for x: y+5=3x ⇒ x=\dfrac{y+5}{3}. Hence f^{-1}(y)=\dfrac{y+5}{3}. Replace y by x to write f^{-1}(x)=\dfrac{x+5}{3}.
Verify: f^{-1}(f(x)) = \dfrac{(3x-5)+5}{3}=\dfrac{3x}{3}=x, confirming the inverse property.

Common mistake: Mixing up the variable while solving for the inverse, leading to an expression like

(x - 5) /3

instead of

(x + 5) /3

Concept tested: Composition and inverse of bijective functions

Published 2 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

Let f:R→R be defined by f(x)=3x−5 and g:R→R by g(x)=3x+2​. i Prove that both f and g are one‑one and onto. ii Find the compositions (f∘g)(x) and (g∘f)(x). iii Determine the inverse function f−1 and verify that f−1(f(x))=x for all x∈R.

Step-by-step solution

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