Find-the-maximum-and-minimum-value-of-1-sinx-1-sin3x-1-sin2x-

Question Number 114765 by dw last updated on 21/Sep/20

F i n d t h e m a x i m u m . a n d m i n i m u m v a l u e o f ⌊ 1 + s i n x ⌋ + ⌊ 1 + s i n 3 x ⌋ + ⌊ 1 + s i n 2 x ⌋

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

let

f (x) = ⌊ 1 + \sin x ⌋ + ⌊ 1 + \sin 2 x ⌋ + ⌊ 1 + \sin 3 x ⌋

Now,

- 1 ⩽ \sin nx ⩽ 1

0 ⩽ ⌊ 1 + \sin nx ⌋ ⩽ 2

∴ the \min . value of f (x) = 0

now for the \max . value

the period of f (x) = L . C . M (2 π, \frac{2 π}{2}, \frac{2 π}{3}) = 2 π

we know that the fundamental period of

\sin x \in [- \frac{π}{2}, \frac{π}{2}]

∴ the \max . value of

f (x) = \max . {f (\frac{π}{2}), f (\frac{π}{4}), f (\frac{π}{6})}

{∵ sinx is an increasing function in [- \frac{π}{2}, \frac{π}{2}]}

f (\frac{π}{2}) = ⌊ 1 + \sin \frac{π}{2} ⌋ + ⌊ 1 + \sin \frac{2 π}{2} ⌋ + ⌊ 1 + \sin \frac{3 π}{2} ⌋

= 2 + 1 + 0 = 3

f (\frac{π}{4}) = ⌊ 1 + \sin \frac{π}{4} ⌋ + ⌊ 1 + \sin \frac{2 π}{4} ⌋ + ⌊ 1 + \sin \frac{3 π}{4} ⌋

= 1 + 2 + 1 = 4

f (\frac{π}{6}) = ⌊ 1 + \sin \frac{π}{6} ⌋ + ⌊ 1 + \sin \frac{2 π}{6} ⌋ + ⌊ 1 + \sin \frac{3 π}{6} ⌋

= 1 + 1 + 2 = 4

∴ the \max . value of f (x) = 4

Commented by dw last updated on 21/Sep/20

T h a n k y o u S i r

Commented by PRITHWISH SEN 2 last updated on 21/Sep/20

welcome