<p>The area of the region bounded by the curve \(y = \dfrac{1}{1+(\tan x)^{1/2}}\) and the <em>x</em>-axis between the ordinates \(x = \pi/6\) and \(x = \pi/3\) is \(\pi(a/b)\). Where (a/b) in simplest form then</p>
Step-by-Step Solution
Key Concept: Use the property that for integrals of the form ∫[a to b] f(x)dx, if you substitute x = a+b-t, you can often find a symmetric relationship that simplifies the integral.
<p><strong>Step 1:</strong> Let I = ∫[π/6 to π/3] 1/(1+√(tan x)) dx</p><p><strong>Step 2:</strong> Consider the complementary integral I' = ∫[π/6 to π/3] 1/(1+√(cot x)) dx</p><p><strong>Step 3:</strong> Rewrite I': 1/(1+√(cot x)) = √(tan x)/(√(tan x) + 1). Note that 1/(1+√(tan x)) + √(tan x)/(1+√(tan x)) = 1</p><p><strong>Step 4:</strong> Therefore I + I' = ∫[π/6 to π/3] 1 dx = π/3 - π/6 = π/6</p><p><strong>Step 5:</strong> By the substitution property (x → π/2 - x), I = I', so 2I = π/6</p><p><strong>Step 6:</strong> Thus I = π/12</p><p><strong>Step 7:</strong> Comparing with π(a/b), we have a/b = 1/12, giving a = 1, b = 12 (in simplest form)</p><p><strong>Possible answers checking:</strong> A: a=1,b=12 ✓ | B: a+b=13 ✓ | C: gcd(a,b)=2 ✗ | D: b-a=11 ✓</p><p>∴ Answer: A, B, D</p>
Correct Answer: A,B,D