Limits, Continuity & Differentiability
Application of Derivatives to Root Analysis
Grade 12
Question:
<p>Let \(f(x) = \sin x - ax - b\). Then, \(f(x) = 0\) has</p>
<p>(a) only one real root which is positive, if \(a > 1, b \geq 0\)</p>
<p>(b) only one real root which is negative, if \(a > 1, b < 0\)</p>
<p>(c) only one real root which is negative, if \(a \leq -1, b \geq 0\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Analyze monotonicity of f(x) using derivative. When |a| > 1, the function is monotonic, ensuring exactly one real root whose sign depends on the value of f(0) = -b.
<p><strong>Solution:</strong> Given $f(x) = \sin x - ax - b$</p><p>$f'(x) = \cos x - a$</p><p><strong>Case 1:</strong> If $a > 1$, then $f'(x) = \cos x - a < 0$ always (since $\cos x \leq 1$). So $f(x)$ is entirely decreasing.</p><p>Therefore, $f(x) = 0$ has only one real root, which is positive if $f(0) \geq 0$ (i.e., $-b \geq 0$ or $b \leq 0$) and negative if $f(0) < 0$ (i.e., $b > 0$).</p><p><strong>Case 2:</strong> When $a \leq -1$, then $f(x)$ is entirely increasing. Therefore, $f(x)$ has only one real root which is negative if $f(0) \geq 0$ and positive if $f(0) < 0$.</p><p>∴ Answers are (a), (b) and (c).</p>
Correct Answer: A,B,C