1.

For `f(x)=(kcosx)/(pi-2x)`, if `x!=pi/2`, `3`, if `x=pi/2`, then find the value of `k` so that `f` is continous at `x=pi/2`

Answer» `f(x)=(kcosx)/(pi-2x)`
if `x!=pi/2`
`=k/2(cosx)/(pi/2-x)=k/2(sin(pi/2-x)/(pi/2-n))`
`=lim_(x->pi/2)f(x)=lim_(x->pi/2)k/2sin(pi/2-n)/(pi/2-x)`
`=k/2 lim_(x->pi/2)(sin(pi/2-x))/(pi/2-x)`
`=k/2*1=k/2`
if f is continous at x=`pi/2`
`lim_(x->pi/2)f(x)=f(pi/2)`
`k/2-3,k=6`.


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