JAMIA 2021 — Mathematics PYQ
JAMIA | Mathematics | 2021If cos−1α+cos−1β+cos−1γ=3π, then α(β+γ)+β(γ+α)+γ(α+β) equals
Choose the correct answer:
- A.
0
- B.
1
- C.
6
(Correct Answer) - D.
12
6
Explanation
Given Equation
cos−1α+cos−1β+cos−1γ=3π
1. Range of Inverse Cosine
The range of the function f(x)=cos−1x is [0,π].
This means the maximum value for any cos−1x is π.
For the sum of three such terms to be 3π:
cos−1α=π
cos−1β=π
cos−1γ=π
2. Finding the Values of α,β,γ
Since cosπ=−1:
α=cosπ=−1
β=cosπ=−1
γ=cosπ=−1
3. Evaluating the Expression
The expression is:
E=α(β+γ)+β(γ+α)+γ(α+β)
Substitute α=−1,β=−1,γ=−1:
E=(−1)(−1−1)+(−1)(−1−1)+(−1)(−1−1)
E=(−1)(−2)+(−1)(−2)+(−1)(−2)
E=2+2+2
E=6
Final Answer:
The value of the expression is 6.
Explanation
Given Equation
cos−1α+cos−1β+cos−1γ=3π
1. Range of Inverse Cosine
The range of the function f(x)=cos−1x is [0,π].
This means the maximum value for any cos−1x is π.
For the sum of three such terms to be 3π:
cos−1α=π
cos−1β=π
cos−1γ=π
2. Finding the Values of α,β,γ
Since cosπ=−1:
α=cosπ=−1
β=cosπ=−1
γ=cosπ=−1
3. Evaluating the Expression
The expression is:
E=α(β+γ)+β(γ+α)+γ(α+β)
Substitute α=−1,β=−1,γ=−1:
E=(−1)(−1−1)+(−1)(−1−1)+(−1)(−1−1)
E=(−1)(−2)+(−1)(−2)+(−1)(−2)
E=2+2+2
E=6
Final Answer:
The value of the expression is 6.