<p>(a) If \(f(x) = \sin x\) and \(g(x) = \ln x\), then the range of \(g(f(x))\) is \([-1, 1]\)</p>
<p>(b) If \(x^2 + ax + 9 > x\) for all \(x \in \mathbb{R}\), then \(-5 < a < 7\)</p>
<p>(c) If \(f(x) = \left(2011 - x^{2012}\right)^{\frac{1}{2012}}\), then \(f(f(2)) = 2\)</p>
<p>(d) The function \(f: \mathbb{R} \to \mathbb{R}\) defined as \(f(x) = \frac{x^2 + 4x + 30}{x^2 - 8x + 18}\) is not surjective.</p>
Step-by-Step Solution
Key Concept: For each statement, determine whether it is correct or incorrect by carefully analyzing the domain, range, and function properties. The question asks which statements are INCORRECT, so we need to identify the false ones.
<p><strong>Step 1: Check Statement (a)</strong></p><p>f(x) = sin x, g(x) = ln x. For g(f(x)) = ln(sin x), we need sin x > 0.</p><p>Domain of g∘f: x where sin x ∈ (0, 1] (since ln requires positive argument).</p><p>When sin x ∈ (0, 1], we have ln(sin x) ∈ (-∞, 0] (as ln(u) → -∞ as u → 0⁺ and ln(1) = 0).</p><p>Range is (-∞, 0], NOT [-1, 1]. <strong>Statement (a) is INCORRECT.</strong></p><p><strong>Step 2: Check Statement (b)</strong></p><p>x² + ax + 9 > x for all x ∈ ℝ means x² + (a-1)x + 9 > 0 for all x ∈ ℝ.</p><p>For this quadratic to be always positive: Δ = (a-1)² - 36 < 0.</p><p>(a-1)² < 36 ⟹ |a-1| < 6 ⟹ -6 < a-1 < 6 ⟹ -5 < a < 7.</p><p><strong>Statement (b) is CORRECT.</strong></p><p><strong>Step 3: Check Statement (c)</strong></p><p>f(x) = (2011 - x^2012)^(1/2012). Find f(f(2)).</p><p>f(2) = (2011 - 2^2012)^(1/2012). Since 2^2012 >> 2011, we have 2011 - 2^2012 < 0.</p><p>The 2012-th root of a negative number is NOT real (2012 is even).</p><p>f(2) is not defined in ℝ, so f(f(2)) cannot equal 2. <strong>Statement (c) is INCORRECT.</strong></p><p><strong>Step 4: Check Statement (d)</strong></p><p>f(x) = (x² + 4x + 30)/(x² - 8x + 18). Let y = f(x), then x² + 4x + 30 = y(x² - 8x + 18).</p><p>x² + 4x + 30 = yx² - 8yx + 18y ⟹ (y-1)x² - (8y+4)x + (18y-30) = 0.</p><p>For real x: Δ = (8y+4)² - 4(y-1)(18y-30) ≥ 0.</p><p>Expanding: 64y² + 64y + 16 - 4(18y² - 30y - 18y + 30) ≥ 0.</p><p>64y² + 64y + 16 - 72y² + 192y - 120 ≥ 0 ⟹ -8y² + 256y - 104 ≥ 0 ⟹ y² - 32y + 13 ≤ 0.</p><p>Roots: y = (32 ± √(1024-52))/2 = (32 ± √972)/2 = 16 ± 3√27 ≈ 16 ± 15.59.</p><p>Range is a bounded interval, not all of ℝ. <strong>Statement (d) is CORRECT.</strong></p><p><strong>∴ Answer:</strong> a, c</p>
Correct Answer: a, c