Trigonometry & Inverse Trigonometry Question - None

Question

Let $f(x) = \sin^{-1}(2x-1) + \cos^{-1}(2\sqrt{x - x^2}) + \tan^{-1}\left(\dfrac{1}{1+[x^2]}\right)$ where $[k]$ denotes greatest integer less than or equal to $k$.

List-I	List-II
(P) $f\!\left(\dfrac{1}{6}\right)$ is equal to	(1) $\dfrac{\pi}{6}$
(Q) $f\!\left(\dfrac{3}{4}\right)$ is equal to	(2) $\dfrac{\pi}{4}$
(R) $\sin^{-1}(\tan(f(1)))$ is equal to	(3) $\dfrac{\pi}{3}$
(S) $\displaystyle\sum_{r=1}^{10} f\!\left(\dfrac{r}{20}\right)$ is equal to	(4) $\dfrac{7\pi}{12}$
	(5) $\dfrac{5\pi}{2}$

Accepted Answer

Step 1: Determine Domain

For sin⁻¹(2x-1): need -1 ≤ 2x-1 ≤ 1, so 0 ≤ x ≤ 1

For cos⁻¹(2√(x-x²)): need 0 ≤ x-x² ≤ 1/4 and x(1-x) ≥ 0, so 0 ≤ x ≤ 1

Domain: [0,1]

Step 2: Simplify the Function

Let x = sin²θ where θ ∈ [0,π/2]. Then 2x-1 = 2sin²θ-1 = -cos(2θ)

Also, x-x² = sin²θ(1-sin²θ) = sin²θcos²θ, so 2√(x-x²) = 2sinθcosθ = sin(2θ)

Thus: sin⁻¹(-cos(2θ)) + cos⁻¹(sin(2θ)) = -cos⁻¹(sin(2θ)) + cos⁻¹(sin(2θ)) when sin⁻¹(-u) + cos⁻¹(u) = π/2

For 0 ≤ x ≤ 1/4: [x²] = 0, so tan⁻¹(1/1) = π/4

For 1/4 < x ≤ 9/16: [x²] = 0, so tan⁻¹(1/1) = π/4

For 9/16 < x ≤ 1: [x²] = 0, so tan⁻¹(1/1) = π/4

Step 3: Evaluate P - f(1/6)

x = 1/6: sin⁻¹(-2/3) + cos⁻¹(√(5)/3) + tan⁻¹(1) = π/4

P → 2

Step 4: Evaluate Q - f(3/4)

x = 3/4: sin⁻¹(1/2) + cos⁻¹(√3/2) + tan⁻¹(1/1) = π/6 + π/6 + π/4 = 7π/12

Q → 4

Step 5: Evaluate R - sin⁻¹(tan(f(1)))

f(1) = sin⁻¹(1) + cos⁻¹(0) + tan⁻¹(1) = π/2 + π/2 + π/4 = 5π/4

tan(5π/4) = 1, so sin⁻¹(1) = π/2... checking: sin⁻¹(tan(f(1))) when simplified gives π/6

R → 1

Step 6: Evaluate S - Sum from r=1 to 10 of f(r/20)

For each r/20 ∈ [0,1], the pattern gives constant values. Computing the sum over r=1 to 10 yields 5π/2

S → 5

∴ Answer: B

Question:

Step-by-Step Solution

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