f : R → R is a mapping where f(x) = x³ - 6, for all x ∈ R, R = set of real numbers. Prove that f is a bijective mapping.

f : R → R is a mapping where f(x) = x³ - 6, for al – Class 12 Mathematics

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Q. f : R \to R is a mapping where f(x) = x³ - 6, for all x \in R, R = set of real numbers. Prove that f is a bijective mapping.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2022Subjective· 4mImportance★★★★★

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Prove injectivity by showing equal outputs force equal inputs (cubing is one-one on $\mathbb R$ ), and prove surjectivity by exhibiting, for any target $y$ , an explicit real $x$ that maps to it.

Given $f:\mathbb R\to\mathbb R$ , $f(x)=x^3-6$ .

Injective (one-one): suppose $f(x_1)=f(x_2)$ for $x_1,x_2\in\mathbb R$ .

$x_1^3-6 = x_2^3-6 \Rightarrow x_1^3=x_2^3$

Since the cubing function $t\mapsto t^3$ is strictly increasing on $\mathbb R$ (its derivative $3t^2\ge0$ , and it's $0$ only at the single point $t=0$ , so it never flattens into a repeated value), $x_1^3=x_2^3$ forces $x_1=x_2$ . Hence $f$ is one-one.

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Q.f : R → R is a mapping where f(x) = x³ - 6, for all x ∈ R, R = set of real numbers. Prove that f is a bijective mapping.

Concept understanding — Onto (Surjective) Functions and Bijections

Q.f : R → R is a mapping where f(x) = x³ - 6, for all x ∈ R, R = set of real numbers. Prove that f is a bijective mapping.

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