Question

If `e^((cos^2x + cos^4x + cos^6x + ...∞)log_e2` satisfies the equation t² – 9t + 8 = 0, then the value of `(2sinx)/(sinx + sqrt(3)cosx)(0 < x ,< π/2)` is ______.

पर्याय

• `3/2`

• `2sqrt(3)`

• `1/2`

• `sqrt(3)`

Answer 1

If `e^((cos^2x + cos^4x + cos^6x + ...∞)log_e2` satisfies the equation t² – 9t + 8 = 0, then the value of `(2sinx)/(sinx + sqrt(3)cosx)(0 < x ,< π/2)` is `underlinebb(1/2)`. 

Explanation:

Given: `e^((cos^2x + cos^4x + cos^6x + .......)log_e2` satisfies the equation t² – 9t + 8 = 0.

Sum of G.P. with first term = a, common ratio = r for infinite terms = `a/(1 - r)`.

In case of cos²x + cos⁴x + cos⁶x.....

a = cos²x, r = `(cos^4x)/(cos^2x)` = cos²x

cos²x + cos⁴x + cos⁶x..... = `(cos^2x)/(1 - cos^2x)`

= `(cos^2x)/(sin^2x)`

= cot²x

`e^((cos^2x + cos^4x + .......)log_e2` = `e^(cot^2x.log_e2)`

= `e^(log_e2^(cos^2x)`  ...(∵ mlogx = logx^m)

= `2^(cot^2x)`  ...`(∵ a^(loga^x) = x)`

`2^(cot^2x)` satisfies t² – 9t + 8 = 0

⇒ t² – t – 8t + 8 = 0 

⇒ t(t – 1) – 8(t – 1) = 0

⇒ t = 1 or 8

⇒ Either, `2^(cot^2x)` = 1 or `2^(cot^2x)` = 8 

⇒ `2^(cot^2x)`  = 2⁰ or `2^(cot^2x)` = 2³ 

⇒ cot²x = 0 or cot²x = 3

∵ `0 < x < π/2`

∴ cotx = `sqrt(3)` ⇒ x = `π/6`

 Now, `(2sinx)/(sinx + sqrt(3)cosx)` = `2/(1 + sqrt(3) xx cot  π/6` = `1/2`.