Question -
Answer -
Let f(x) = (2√3 + 3) sin x + 2√3 cos x
Here, A = 2√3, B = 2√3 + 3 and C = 0
– √[(2√3)2 + (2√3 + 3)2] ≤(2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)2 + (2√3 + 3)2]
– √[12+12+9+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤√[12+12+9+12√3]
– √[33+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤√[33+12√3]
– √[15+12+6+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤√[15+12+6+12√3]
We know that (12√3 + 6 < 12√5) because the value of√5 – √3 is more than 0.5
So if we replace, (12√3 + 6 with 12√5) the aboveinequality still holds.
So by rearranging the above expression √(15+12+12√5)weget, 2√3 + √15
– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15
Hence proved.