The number of ordered pairs (α, β), where α, β ∈(-π, π) satisfying cos (α-β)=1 and cos (α+β)=(1/e) is

Question

The number of ordered pairs (α, β), where α, β ∈(-π, π) satisfying cos (α-β)=1 and cos (α+β)=(1/e) is (A) 0 (B) 1 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

Since, cos(α-β)=1 ⇒ α-β=2nπ But -2π < α-β < 2π [as α, β ∈ (-π, π)] ∴ α-β=0 ...(i) Given, co (α+β)=(1/e) ⇒ cos 2α=(1/e) < 1, which is true for four values of α. [as -2π < 2α < 2π]

Q. The number of ordered pairs $(α, β),$ where $α, β \in (- π, π)$ satisfying cos $(α - β) = 1 an d cos (α + β) = \frac{1}{e}$ is

0

1

2

4

$S in ce, cos (α - β) = 1$
$\Rightarrow α - β = 2 nπ$
But $- 2 π < α - β < 2 π [a s α, β \in (- π, π)]$
$∴ α - β = 0... (i)$
Given, $co (α + β) = \frac{1}{e}$
$\Rightarrow cos 2 α = \frac{1}{e} < 1, w hi c hi s t r u e f or f o u r v a l u eso f α .$
$[a s - 2 π < 2 α < 2 π]$

Q. The number of ordered pairs (α,β), where α,β∈(−π,π) satisfying cos (α−β)=1andcos(α+β)=e1​ is

0

1

2

4

Since,cos(α−β)=1 ⇒α−β=2nπ But −2π<α−β<2π[asα,β∈(−π,π)] ∴α−β=0...(i) Given, co(α+β)=e1​ ⇒cos2α=e1​<1,whichistrueforfourvaluesofα. [as−2π<2α<2π]

Q. The number of ordered pairs $(α, β),$ where $α, β \in (- π, π)$ satisfying cos $(α - β) = 1 an d cos (α + β) = \frac{1}{e}$ is