Relations & Functions
Composite functions, range, and types of functions
Grade 12
Question:
<p>Identify which of the following statement(s) is(are) <strong>correct</strong>?</p>
<p>If \(f(x) = \cos x\) and \(g(x) = \ln x\), then range of \(f(g(x))\) is \([-1, 1]\).</p>
<p>If \(f(x) = \dfrac{2}{\pi}(\sin^{-1} x + \cos^{-1} x)\) and \(g(x) = \text{sgn}(x^2 - x + 1)\), then \(f(g(x))\) and \(g(x)\) both are identical functions.</p>
<p>If \(f: R \to [-2, 2]\), \(f(x) = \dfrac{2x}{1+x^2}\), then \(f\) is a bijective function.</p>
<p>If \(f(x) = \sin^{-1} x\) and \(g(x) = \cos x\), then \(f(g(x))\) is odd and \(g(f(x))\) is even function.</p>
Step-by-Step Solution
Key Concept: A relation is a function if and only if each element in the domain maps to exactly one element in the codomain. Check vertical line test for graphical relations and one-to-one mapping for set relations.
<p><strong>Step 1:</strong> Recall that a relation f: A → B is a function if every element in domain A has exactly one image in B.</p><p><strong>Step 2:</strong> For each statement, verify:</p><ul><li>Does every element in domain map to exactly one element in codomain?</li><li>Is there any element mapping to multiple values? (If yes, NOT a function)</li><li>Can an element in codomain have multiple pre-images? (Yes, this is allowed in a function)</li></ul><p><strong>Step 3:</strong> Statement A satisfies the definition of a function - each domain element maps to exactly one codomain element. The function need not be bijective.</p><p>∴ Answer: A</p>
Correct Answer: A