Trignometry - 8

Geometry Level 4

If 0 ≤ p ≤ 3 , 0 ≤ q ≤ 3 0 \leq p \leq 3 , 0 \leq q \leq 3 and the equation x 2 + 4 + 3 cos ⁡ ( p x + q ) = 2 x x^2 + 4 + 3 \cos(px+q) = 2x has atleast one solution and the value of p + q = k π 4 p+q = \dfrac{k\pi}{4} , then what is the minimum positive value of k ?


The answer is 4.

Kunal Joshi by Kunal Joshi , 24, India

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1 solution

Samarpit Swain
Jan 27, 2015

The equation can be re-arranged as, ( x − 1 ) 2 + 3 ( 1 + c o s ( p x + q ) ) = 0 (x-1)^2 +3(1+ cos(px+q))=0 => ( 1 − x ) 2 − 3 = 1 + c o s ( p x + q ) \dfrac{(1-x)^2}{-3} = 1 + cos(px+q)

Since we are dealing with the value of ( p + q ) (p+q) , we put x = 1 x=1 : 1 + c o s ( p + q ) = 0 1+cos(p+q) = 0 => c o s ( p + q ) = − 1 cos(p+q)= -1

Therefore p + q p+q = n π n\pi where n ∈ n\in I I

Hence, minimum occurs when n = 1 n=1 or k = 4 k =4 :)

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