JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Inverse Trigonometric Functions

Given that the inverse trigonometric function assumes principal values only. Let

x, y

be any two real numbers in

[- 1, 1]

such that

cos^{- 1} x - sin^{- 1} y = α

- \frac{π}{2} \leq α \leq π

. Then, the minimum value of

x^{2} + y^{2} + 2 x y sin α

Answer: (D)

Solution

cos^{- 1} x - sin^{- 1} y = α \Rightarrow cos^{- 1} x + cos^{- 1} y = \frac{π}{2} + α

(using

sin^{- 1} y = \frac{π}{2} - cos^{- 1} y

).

Using

cos^{- 1} x + cos^{- 1} y = cos^{- 1} (x y - 1 - x^{2} 1 - y^{2})

for the sum in range, taking cosine of both sides:

x y - (1 - x^{2}) (1 - y^{2}) = cos (\frac{π}{2} + α) = - sin α

(1 - x^{2}) (1 - y^{2}) = x y + sin α

Squaring:

(1 - x^{2}) (1 - y^{2}) = x^{2} y^{2} + 2 x y sin α + sin^{2} α \Rightarrow 1 - x^{2} - y^{2} + x^{2} y^{2} = x^{2} y^{2} + 2 x y sin α + sin^{2} α

x^{2} + y^{2} + 2 x y sin α = 1 - sin^{2} α = cos^{2} α

Minimum value of

cos^{2} α

is 0 (attained at

α = \pm π /2

, within the allowed range).

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option