NTA Abhyas JEE Main2020MathematicsFunctionsPractice
The number of real solution(s) of the equation s i n - 1 x 2 - 5 x + 5 + c o s - 1 4 x - x 2 - 3 = π is/are
Options
- Aone
- Btwo
- Czero
- Dinfinite
Correct answer
A. one
Step-by-step solution
Since, x 2 - 5 x + 5 ≥ 0 ⇒ 0 ≤ s i n - 1 x 2 - 5 x + 5 ≤ π 2 Since, 4 x - x 2 - 3 ≥ 0 ⇒ 0 ≤ c o s - 1 4 x - x 2 - 3 ≤ π 2 ∴ For LHS to be π s i n - 1 x 2 - 5 x + 5 = π 2 and c o s - 1 4 x - x 2 - 3 = π 2 So, x 2 - 5 x + 5 = 1 & 4 x - x 2 - 3 = 0 ⇒ x 2 - 5 x + 5 = 1 & x 2 - 4 x + 3 = 0 ⇒ x = 1 , 4 & x = 1,3 Hence, the common value is x = 1