Prove that for any A.P. `a_1, a_2, a_3, ,` then determinant `|a_p+a_(p+m)+a_(p+2m) 2a_p+3a_(p+m)+4a_(p+2m)a_q+a_(q+m)+a_(q+2m)2a_q+3a_(q+m)+4a_(q+2m)a_r+a_(r+m)+a_(r+2m)2a_r+3a_(r+m)+4a_(r+2m)` `4a_p+9a_(p+m)+16 a_(p+2m)4a_q+9a_(q+m)+16 a_(q+2m)4a_r+9a_(r+m)+16 a_(r+2m)|=0`

Answer

1

Explanation

`f(x)=sin^(2)x+sin^(2)(x+pi//3)+cos x cos (x+pi//3)`
`rArr f(x)=sin^(2)x+(sinx cos pi//3+cos x sinpi//3)^(2)+cosx cos(x+pi//3)` ltbgt `rArr f(x)=sin^(2)x=((sinx*1)/(2)+(cosx*sqrt(3))/(2))^(2)+cosx(cosxcos pi//3-sin x sin pi//3)`
`rArr f(x)=sin^(2)x+(sin^(2)x)/(4)+(3cos^(2)x)/(4)+(2*sqrt(3))/(4)sinx cosx+(cos^(2)x)/(2)-cosxsinx*(sqrt(3))/(2)`
`=sin^(2)x+(sin^(2)x)/(4)+(3cos^(2)x)/(2)+(cos^(2)x)/(2)`
`=(5)/(4)sin^(2)x+(5)/(4)cos^(2)x=(5)/(4)`
`and gof(x)=g{f(x)}=g(5//4)=1`
Alternate Solution
`f(x)=sin^(2)x+sin^(2)(x+pi//3)+cosxcos(x+pi//3)`
`implies f'(x)=2sinx cosx+2sin(x+pi//3)cos(x+pi//3)-sinx cos(x+pi//3)-cosx sin(x+pi//3)`
`=sin2x+sin(2x+2pi//3)-[sin(x+x+pi//3)]`
`=2sin((2x+2x+2pi//3)/(2))*cos((2x-2x-2pi//3)/(2))-sin(2x+pi//3)`
`=2[sin(2x+pi//3)*cospi//3]-sin(2x+pi//3)`
`=2[sin(2x+pi//3)*(1)/(2)]-sin (2x+(pi)/(3))=0`
`rArr f(x)=c,` where c is a constant.
But `f(0)=sin^(2)0+sin^(2)(pi//3)+cos 0 cospi//3`
`=((sqrt(3))/(2))^(2)+(1)/(2)=(3)/(4)+(1)/(2)=(5)/(4)`
Therefore, `(gof)(x)=g[f(x)]=g(5//4)=1`