Q.

Let y=f(x)=sin3(π3(cos(π32(-4x3+5x2+1)32))). Then, at x=1,                [2023]

1 2y'+3 π2y=0  
2 2 y'-3π2y=0  
3 2y'+3π2y=0  
4 y'+3π2y=0  

Ans.

(3)

Let -4x3+5x2+1=t, then

y=sin3(π3(cos(π32 t3/2)))

dtdx=-12x2+10x

∴  dydx=3sin2(π3(cos(π32 t3/2)))·cos(π3cos(π32 t3/2))·π3(-sin(π32 t3/2))

        π32·32t1/2·dtdx

⇒dydx=-3sin2(π3(cos(π32(-4x3+5x2+1)3/2))·cos(π3cos(π32(-4x3+5x2+1)3/2))·π3sin(π32(-4x3+5x2+1)3/2)·π22(-4x3+5x2+1)1/2(-12x2+10x)

At x=1, t=-4+5+1=2

dydx(x=1)=-3sin2(π3cos(π32 23/2))·cos(π3cos(π32 23/2))·π3sin(π32 23/2)·π22 21/2·(-2)

=π2sin2(π3(cos2π3))·cos(π3cos2π3)·sin(2π3)

=π2sin2(-π3·12)·cos(-π3·12)32

=32π2·14·32=316π2

at x=1, y=sin3(π3(cosπ32(2)3/2))=sin3(π3(cos2π3))

=sin3(-π3×12)=-18

⇒ 2y'+3π2y=2×316π2+3π2×(-18)=0