Trigonometry & Inverse Trigonometry
Range of trigonometric functions
Grade 11
Question:
<p>The minimum and maximum values of <span>\(a\sin x + b\sqrt{1 - a^2}\cos x + c\)</span> (where <span>\(|a| < 1, b > 0\)</span>) respectively are</p>
<p>(a) <span>\(\{b - c, b + c\}\)</span></p>
<p>(b) <span>\(\{b + c, b - c\}\)</span></p>
<p>(c) <span>\(\{c - b, b + c\}\)</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The maximum and minimum values of A sin x + B cos x are ±√(A² + B²). Here the amplitude is b, and adding constant c shifts both extrema by c.
<p><strong>Step 1:</strong> Consider the expression <span>$a\sin x + b\sqrt{1 - a^2}\cos x + c$</span></p><p><strong>Step 2:</strong> Find the amplitude of the trigonometric part:</p><p><span>$\sqrt{a^2 + b^2(1 - a^2)} = \sqrt{a^2 + b^2 - a^2b^2}$</span></p><p><strong>Step 3:</strong> Simplify:</p><p><span>$= \sqrt{b^2(a^2 + 1 - a^2)} = \sqrt{b^2} = b$</span></p><p><strong>Step 4:</strong> The expression <span>$a\sin x + b\sqrt{1 - a^2}\cos x$</span> ranges from <span>$-b$</span> to <span>$b$</span>.</p><p><strong>Step 5:</strong> Adding constant <span>$c$</span>:</p><p>Minimum value = <span>$-b + c = c - b$</span></p><p>Maximum value = <span>$b + c$</span></p><p>∴ Answer is (c) <span>$\{c - b, b + c\}$</span></p>
Correct Answer: C