What is the value of k if the area bounded by the curve `y = sin kx, y= 0, x = pi//k, x = pi//(3k) ` is `3` sq units ?

Options

A. `1//2`
B. `1`
C. `3//2`
D. `2`

Answer

A

Explanation

Given `x = pi//k, x = (pi)/(3k)`
and `y = sin kx`
Let the required area be A
So, A = 3 (given )
Therefore Area, `a = int_(pi//3k)^(pi//k) sin kx dx`
`rArr 3 = -[(coskx)/(k)]_(pi//3k)^(pi//k)`
`rArr 3 = - 1/k[cos pi - cos' (pi)/(3)]`
`rArr 3 = - 1/k [-1 - 1/2]`
( `:' cos pi = cos (pi//2 + pi//2) = - sin pi//2` and `cos 60^(@) = 1/2`)
`rArr 3 = 3/(2k) rArr k = (1)/(2)`