Functions
General
Grade 12
Question:
If functions f(x) and g(x) are defined on \(\mathbb{R} \to \mathbb{R}\) such that f(x) = \(\begin{cases} 0, & x \in \text{rational} \\ x, & x \in \text{irrational} \end{cases}\) , g(x) = \(\begin{cases} 0, & x \in \text{irrational} \\ x, & x \in \text{rational} \end{cases}\) , then (f - g)(x) is -
one-one and onto
neither one-one nor onto
one-one but not onto
onto but not one-one
Step-by-Step Solution
Key Concept: General
<div class="solution"><p>Let y=sin(sin(x²)). Equation becomes y²=y → y=0 or y=1. But |y|≤sin1<1, so y=1 impossible. y=0 → sin(x²)=0 → x²=nπ → x=±√(nπ).</p><p><strong>Answer: <span class="math-inline">$x=\pm\sqrt{n\pi},\ n=0,1,2,\ldots$</span></strong></p><div class="trap-box"><strong>Trap:</strong> Don't jump from sin(sin(x²))=0 to sin(x²)=nπ. Inner value is in [-1,1] already.</div><div class="key-concept"><strong>Key Concept:</strong> Composition → new variable; y²=y forces y=0 or 1; range bound eliminates y=1</div></div>
Correct Answer: A