If the equation sin -1(x2+x+1)+ cos -1(λ x+1)=(π/2) has exactly two solution for λ ∈[a, b), then the value of (a + b) equals

Question

If the equation sin -1(x2+x+1)+ cos -1(λ x+1)=(π/2) has exactly two solution for λ ∈[a, b), then the value of (a + b) equals (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

∴ sin -1(x2+x+1)+ cos -1(λ x+1)=(π/2) If x2+x+1=λ x+1 ⇒ x 2+(1-λ) x =0 ⇒ x =0 or x =λ-1 sin -1(x2+x+1) will exists if -1 ≤ x2+x+1 ≤ 1. ⇒ x2+x ≤ 0 ⇒-1 ≤ x<0 For exactly two solution -1 ≤ λ-1<0 ⇒ 0 ≤ λ<1 ∴ a =0 text and b =1

Q. If the equation sin−1(x2+x+1)+cos−1(λx+1)=2π​ has exactly two solution for λ∈[a,b), then the value of (a+b) equals

0

1

2

3

∴sin−1(x2+x+1)+cos−1(λx+1)=2π​ If x2+x+1=λx+1 ⇒x2+(1−λ)x=0 ⇒x=0 or x=λ−1 sin−1(x2+x+1) will exists if −1≤x2+x+1≤1. ⇒x2+x≤0⇒−1≤x<0 For exactly two solution −1≤λ−1<0⇒0≤λ<1 ∴a=0 and b=1

Q. If the equation $sin^{- 1} (x^{2} + x + 1) + cos^{- 1} (λ x + 1) = \frac{π}{2}$ has exactly two solution for $λ \in [a, b)$ , then the value of $(a + b)$ equals