Let S be the set of all α ∈ R such that the equation, cos 2x + αsinx = 2α - 7 has a solution. Then S is equal to :

Question

Let S be the set of all α ∈ R such that the equation, cos 2x + αsinx = 2α - 7 has a solution. Then S is equal to : (A) [2,6] (B) [3,7] (C) R (D) [1,4]

Tardigrade Admin · Accepted Answer

cos2x + α sinx = 2 α - 7 ⇒ 2 sin2 x - α sin x + 2 α - 8 = 0 sin2 x - (α/2) sinx + α - 4 = 0 ⇒ sin x = 2 (rejected) or sin x = (α - 4/2) ⇒ |(α - 4/2)| le 1 ⇒ α ∈ [2,6]

Q. Let $S$ be the set of all $α \in R$ such that the equation, cos $2 x + α$ sinx = 2 $α - 7$ has a solution. Then $S$ is equal to :

$[2, 6]$

$[3, 7]$

$R$

$[1, 4]$

$cos 2 x + α s in x = 2 α - 7$
$\Rightarrow 2 s i n^{2} x - α s in x + 2 α - 8 = 0$
$s i n^{2} x - \frac{α}{2} s in x + α - 4 = 0$
$\Rightarrow s in x = 2 (re j ec t e d) or s in x = \frac{α - 4}{2}$
$\Rightarrow ∣ \frac{α - 4}{2} ∣ \leq 1$
$\Rightarrow α \in [2, 6]$

Q. Let S be the set of all α∈R such that the equation, cos 2x+αsinx = 2α−7 has a solution. Then S is equal to :

[2,6]

[3,7]

R

[1,4]

cos2x+αsinx=2α−7 ⇒2sin2x−αsinx+2α−8=0 sin2x−2α​sinx+α−4=0 ⇒sinx=2(rejected)orsinx=2α−4​ ⇒∣2α−4​∣≤1 ⇒α∈[2,6]

Q. Let $S$ be the set of all $α \in R$ such that the equation, cos $2 x + α$ sinx = 2 $α - 7$ has a solution. Then $S$ is equal to :

$[2, 6]$

$[3, 7]$

$R$

$[1, 4]$

$cos 2 x + α s in x = 2 α - 7$
$\Rightarrow 2 s i n^{2} x - α s in x + 2 α - 8 = 0$
$s i n^{2} x - \frac{α}{2} s in x + α - 4 = 0$
$\Rightarrow s in x = 2 (re j ec t e d) or s in x = \frac{α - 4}{2}$
$\Rightarrow ∣ \frac{α - 4}{2} ∣ \leq 1$
$\Rightarrow α \in [2, 6]$