Relations & Functions
Domain and Range of Functions
Grade 12
Question:
<p>If <span style='display:inline-block'>\(f(x) = \frac{1}{2} - \tan\left(\frac{\pi x}{2}\right)\)</span>, where <span style='display:inline-block'>\(-1 < x < 1\)</span>, and <span style='display:inline-block'>\(g(x) = \sqrt{3 + 4x - 4x^2}\)</span>, then domain of <span style='display:inline-block'>\((f + g)\)</span> is given by</p>
<p>(a) <span style='display:inline-block'>\(\left[\frac{1}{2}, 1\right]\)</span></p>
<p>(b) <span style='display:inline-block'>\(\left[\frac{1}{2}, -1\right)\)</span></p>
<p>(c) <span style='display:inline-block'>\(\left[-\frac{1}{2}, 1\right)\)</span></p>
<p>(d) <span style='display:inline-block'>\(\left[-\frac{1}{2}, -1\right]\)</span></p>
Step-by-Step Solution
Key Concept: The domain of a sum of functions is the intersection of their individual domains. For functions involving square roots, ensure the expression under the radical is non-negative.
<p><strong>Step 1:</strong> Find domain of $f(x)$: Given that $-1 < x < 1$, so $d_1 = (-1, 1)$.</p><p><strong>Step 2:</strong> Find domain of $g(x)$: For $g(x) = \sqrt{3 + 4x - 4x^2}$ to be defined, we need $3 + 4x - 4x^2 \geq 0$.</p><p><strong>Step 3:</strong> Simplify the inequality: $3 + 4x - 4x^2 \geq 0 \Rightarrow -4x^2 + 4x + 3 \geq 0 \Rightarrow 4x^2 - 4x - 3 \leq 0$.</p><p><strong>Step 4:</strong> Factor: $(2x - 3)(2x + 1) \leq 0$, which gives $-\frac{1}{2} \leq x \leq \frac{3}{2}$. So $d_2 = \left[-\frac{1}{2}, \frac{3}{2}\right]$.</p><p><strong>Step 5:</strong> Domain of $(f + g)$ is $d_1 \cap d_2 = (-1, 1) \cap \left[-\frac{1}{2}, \frac{3}{2}\right] = \left[-\frac{1}{2}, 1\right)$.</p><p>∴ Answer is (c).</p>
Correct Answer: C