Correct Answer - Option 2 : 2 ≤ a ≤ 6
Concept:
sin θ is defined as
-1 ≤ sin θ ≤ 1
solution of a quadratic equation is given by
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Calculation:
cos2x + a sin x = 2a - 7
∵ cos 2x = 1 - 2 sin2x
⇒ 1 - 2 sin2x + a sin x = 2a - 7
⇒ 2 sin2x - a sin x + 2a - 8 = 0
Using the formula
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(⇒ \ sin\ x = {a \pm \sqrt{a^2-4.2.(2a\ -\ 8)} \over 2.2}\)
\(⇒ \ sin\ x = {a \pm \sqrt{a^2-16a+64} \over 4}\)
∵ (a - b)2 = a2 - 2ab + b2
\(⇒ \ sin\ x = {a \pm \sqrt{(a\ -\ 8)^2} \over 4}\)
\(⇒ \ sin\ x = {a \pm\ {(a\ -\ 8)} \over 4}\)
\(⇒ \ sin\ x = {a \pm\ {(a\ -\ 8)} \over 4}\)
⇒ sin x = 2, \(\frac{a\ -\ 4}{2}\)
We know that
-1 ≤ sin x ≤ 1
Therefore, sin x = 2 is not possible
\(-1\ \le\ \frac{a\ -\ 4}{2}\ \le\ 1\)
⇒ 2 ≤ a ≤ 6
