Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p>The graph of the function \(f(x) = \cos x \cdot \cos(x+2) - \cos^2(x+1)\) is</p>
<p>(a) a straight line passing through \((0, -\sin^2 1)\) with slope 2</p>
<p>(b) a straight line passing through \((0, 0)\)</p>
<p>(c) a parabola with vertex \((1, -\sin^2 1)\)</p>
<p>(d) a straight line passing through the point \(\left(\frac{\pi}{2}, -\sin^2 1\right)\) and parallel to the \(x\)-axis</p>
Step-by-Step Solution
Key Concept: Expand the product using trigonometric identities and simplify to reveal the function's true form. The expression cos(x)·cos(x+2) - cos²(x+1) reduces to a constant after applying product-to-sum and double angle formulas.
<p><strong>Step 1:</strong> Use the product-to-sum formula: cos A · cos B = ½[cos(A+B) + cos(A-B)]</p><p>cos(x)·cos(x+2) = ½[cos(2x+2) + cos(-2)] = ½[cos(2x+2) + cos(2)]</p><p><strong>Step 2:</strong> Expand cos²(x+1) using the double angle formula: cos²θ = ½[1 + cos(2θ)]</p><p>cos²(x+1) = ½[1 + cos(2x+2)]</p><p><strong>Step 3:</strong> Substitute into f(x):</p><p>f(x) = ½[cos(2x+2) + cos(2)] - ½[1 + cos(2x+2)]</p><p>f(x) = ½cos(2x+2) + ½cos(2) - ½ - ½cos(2x+2)</p><p>f(x) = ½cos(2) - ½ = ½[cos(2) - 1]</p><p><strong>Step 4:</strong> Since cos(2) ≈ -0.416, we have f(x) ≈ ½[-0.416 - 1] = ½[-1.416] ≈ -0.708</p><p>The graph is a horizontal line at height ½[cos(2) - 1], a constant function.</p><p>∴ Answer: D</p>
Correct Answer: D