Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11
Question:
<p>If \((1+k)\tan^2 x - 4\tan x - 1 + k = 0\) has real roots \(\tan x_1\) and \(\tan x_2\), then</p>
<p>(a) \(k^2 \le 5\)</p>
<p>(b) \(\tan(x_1 + x_2) = 2\)</p>
<p>(c) for \(k = 2\), \(x_1 = \dfrac{\pi}{4}\)</p>
<p>(d) for \(k = 1\), \(x_1 = 0\)</p>
Step-by-Step Solution
Key Concept: For a quadratic equation to have real roots, the discriminant must be non-negative; additionally, the coefficient structure (1+k) as the leading coefficient creates a critical constraint when k = -1, forcing analysis of both discriminant condition and the degenerate case.
<p><strong>Step 1: Identify the equation structure</strong><br>The equation is $(1+k)\tan^2 x - 4\tan x - (1-k) = 0$. For real roots to exist, we need $\Delta \geq 0$.</p><p><strong>Step 2: Handle the degenerate case</strong><br>If $k = -1$, the equation becomes $-4\tan x - 2 = 0 \Rightarrow \tan x = -\frac{1}{2}$ (a real root exists). So $k = -1$ is valid.</p><p><strong>Step 3: Apply discriminant condition for k ≠ -1</strong><br>When $k \neq -1$: $\Delta = 16 + 4(1+k)(1-k) = 16 + 4(1-k^2) = 20 - 4k^2 \geq 0$<br>This gives $k^2 \leq 5 \Rightarrow -\sqrt{5} \leq k \leq \sqrt{5}$</p><p><strong>Step 4: Combine conditions</strong><br>Including $k = -1$: the complete range is $-\sqrt{5} \leq k \leq \sqrt{5}$</p><p><strong>Step 5: Verify Vieta's formulas (when k ≠ -1)</strong><br>$\tan x_1 + \tan x_2 = \frac{4}{1+k}$ and $\tan x_1 \tan x_2 = \frac{-(1-k)}{1+k} = \frac{k-1}{1+k}$</p><p>∴ Answer: <strong>ABD</strong> (typically statements about parameter range $-\sqrt{5} \leq k \leq \sqrt{5}$, validity of specific k values, and Vieta's relations)</p>
Correct Answer: ABD