Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

Let f : [0, 2] → R be the function defined by

f(x) = (3 - sin(2πx)) sin(πx - π/4) - sin(3πx + π/4).

if α, β ∈ [0, 2] are such that {x ∈ [0, 2] : f(x) ≥ 0 = [α, β] then the value of β - α is ...

solution : function seems so lengthy.

let (πx - π/4) = y

then, f(y) = [3 - sin(2y + π/2)] siny - sin(3π/4 + 3y + π/4)

= (3 - cos2y)siny + sin3y

as for x ∈ [0, 2] , f(x) ≥ 0

so f(y) ≥ 0

⇒(3 - cos2y ) siny + sin3y ≥ 0

⇒(2 + 1 - cos2y) siny + 3siny - 4sin³y ≥ 0

⇒(2 + 2sin²y) siny + 3siny - 4sin³y ≥ 0

⇒siny [2 + 2sin²y + 3 - 4sin²y] ≥ 0

⇒siny (5 - 2 sin²y) ≥ 0

siny ≥ 0 , so y ∈ [ 0, π] and (5 - 2sin²y) ≥ 0 for all y belongs to R

implies that , siny ≥ 0, y ∈ [ 0, π]

so, 0 ≤ πx - π/4 ≤ π

⇒π/4 ≤ πx ≤ 5π/4

⇒1/4 ≤ x ≤ 5/4

i.e, x ∈ [1/4, 5/4] = [α, β]

so α = 1/4 and β = 5/4

then, β - α = 5/4 - 1/4 = 1

Therefore the value of β - α is 1