Let `cos^(-1) (4x^(3) -3x) = a + b cos^(-1) x`
If `x in ((1)/(2), 1]`, then the value of 'a+b pi' is

Options

A. `-1//3`
B. `-3`
C. `(1)/(3)`
D. 3.14

Answer

D

Explanation

Let `cos^(-1) x = theta " or " x = cos theta, " where " theta in [0, pi]`
`cos^(-1) (4x^(3) - 3x) = cos^(-1) (4 cos^(3) theta - 3 cos theta) = cos^(-1) (cos 3 theta) = cos^(-1) (cos alpha)`
where `alpha = 3 theta in [0, 3pi]`
Refer to the graph of `y = cos^(-1) (cos alpha), alpha in [0, 3pi]`

From the graph,
`cos^(-1) (4x^(3) -3x) = cos^(-1) (cos alpha)`
`={(alpha,0 le alpha lt pi),(2pi - alpha,pi le alpha le 2pi),(alpha -2pi,2pi lt alpha le 3pi):}`
`={(3 cos^(-1) x,0 le 3 cos^(-1) x lt pi),(2pi - 3 cos^(-1) x,pi le 3 cos^(-1) x le 2 pi),(3 cos^(-1) x - 2pi,2pi lt 3 cos^(-1) x le 3pi):}`
`={(3 cos^(-1) x,0 le 3 cos^(-1) x lt pi),(2pi - 3 cos^(-1)x,(pi le 3) cos^(-1) x le (2pi//3)),(3 cos^(-1) x - 2pi,(2pi//3) lt cos^(-1) x le pi):}`
`={(3 cos^(-1) x,(1//2) lt x le1),(2pi - 3 cos^(-1) x,(-1//2) le x le (1//2)),(3 cos^(-1) x - 2pi,-1 le x lt -(-1//2)):}`
`={(3 cos^(-1) x - 2pi,-1 le x lt (-1//2)),(2pi - 3 cos^(-1) x,(-1//2) le x le (1//2)),(3 cos^(-1) x,(1//2) lt x le 1):}`
For `x in [-1, -(1)/(2)), cos^^(-1) (4x^(3) -3x) = 3 cos^(-1) x - 2pi`
`rArr a = - 2pi and b = 3 rArr a + b pi = pi`
For `x in [-(1)/(2), (1)/(2)], cos^(-1) (4x^(3) -3x) = 2pi - 3 cos^(-1) x`
`rArr a = 2 pi and b = -3`
`rArr sin^(-1) (sin.(a)/(b)) = sin^(-1) (sin.(2pi)/(-3)) = sin^(-1) (-(sqrt3)/(2)) = -(pi)/(3)`
For `x in ((1)/(2), 1], cos^(-1) (4x^(3) - 3x) = 3 cos^(-1) x`
`rarr a = 0 and b = 3`
`:. underset(y rarr a)("lom") b cos (y) = underset(y rarr 0)("lim") 3 cos (y) = 3`