Answer 1

f(x) = `{((1-sin^3x)/(3cos^2x),`

For continuity,

`lim_(x->(pi^-)/2)f(x)=lim_(x->(pi^+)/2)f(x)=f(pi/2)`

`lim_(x->(pi^-)/2)f(x)=lim_(x->(pi^-)/2)((1-sin^3x)/(3cos^2x))=lim_(x->(pi^-)/2)((1-sinx)(1+sin^2x+sinx))/(3[1-sin^2x])`

`lim_(x->pi/2)f(x)=lim_(x-pi/2)(1+sin^2x+sinx)/(3(1+sinx))=(1+1+1)/(3(2))=1/2`

Let `pi/2-x=theta=>x=pi/2-theta`

`lim_(x->pi^+)=lim_(theta->0)q[(1-sin(pi/2-theta))/(20)^2]=q/4lim_(theta->0)(1-costheta)/theta^2`

`=q/4lim_(theta->0)(2sin^2`

Now, `lim_(x->pi/2)f(x)=lim_(x->pi^+)f(x)=f(pi/2)`

`=>1/2=p=q/8`

`=>p=1/2 `