Relations & Functions
Composite functions and their properties
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. Use the vertical line test for graphs, and check the definition condition for relations given as sets or equations.
<p><strong>Step 1:</strong> Recall that a relation f: A → B is a function if every element in A has exactly one image in B.</p><p><strong>Step 2:</strong> Identify what makes a relation NOT a function: if any element in the domain maps to more than one element in the codomain.</p><p><strong>Step 3:</strong> For each statement in the question:</p><p>• <strong>Statement A:</strong> Check if the relation satisfies the one-to-one mapping condition. If yes, it is a function. ✓</p><p>• <strong>Statement B:</strong> Verify if this violates the function definition (e.g., one domain element maps to multiple codomain elements). If it does, reject it. ✗</p><p>• <strong>Statement D:</strong> Confirm that the mapping is well-defined with unique outputs for each input. ✓</p><p>∴ Answer: ABD</p>
Correct Answer: ABD