Trigonometry & Inverse Trigonometry
Domain and Range of Inverse Trigonometric Functions
Grade 12
Question:
<p><strong>Ex. 34.</strong> <strong>Statement I:</strong> If α, β are roots of <em>6x</em>² + 11<em>x</em> + 3 = 0, then cos α exists but not cos⁻¹β (α > 0).</p><p><strong>Statement II:</strong> Domain of cos⁻¹<em>x</em> is [−1, 1].</p>
<p>(a) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I</p>
<p>(b) Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I</p>
<p>(c) Statement I is correct but Statement II is incorrect</p>
<p>(d) Statement II is correct but Statement I is incorrect</p>
Step-by-Step Solution
Key Concept: The domain of cos⁻¹x is restricted to [−1, 1]. If a value lies outside this interval, its inverse cosine does not exist.
<p><strong>Solution:</strong></p><p>Given: 6<em>x</em>² + 11<em>x</em> + 3 = 0</p><p>⟹ 6<em>x</em>² + 9<em>x</em> + 2<em>x</em> + 3 = 0</p><p>⟹ 3<em>x</em>(2<em>x</em> + 3) + 1(2<em>x</em> + 3) = 0</p><p>⟹ (2<em>x</em> + 3)(3<em>x</em> + 1) = 0</p><p>⟹ <em>x</em> = −3/2, −1/3</p><p>Since −1 > −3/2 or −1/3, we have α = −1/3, β = −3/2</p><p>Now, cos α = cos(−1/3) exists (as −1/3 is just a number).</p><p>For cos⁻¹β = cos⁻¹(−3/2): Since β = −3/2 ∉ [−1, 1], cos⁻¹β does not exist.</p><p>Statement I is true (cos α exists but not cos⁻¹β).</p><p>Statement II is true (Domain of cos⁻¹<em>x</em> is [−1, 1]).</p><p>Statement II correctly explains Statement I because cos⁻¹β does not exist precisely because β is outside the domain [−1, 1].</p><p>∴ Answer is (a).</p>
Correct Answer: a