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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` |
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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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