1.

If the function `f(x) = {(kx + 5 ", when " x le 2),(x -1 ", when " x gt 2):}` is continuous at x = 2 then k = ?

Answer» we have `f(2) = ( k xx 2 +5) = ( 2 k +5)`
`lim_(x to 2x) f(x)= lim_(h to 0) f( 2 +h)`
` lim_( h to 0) {( 2 +h) -1} = lim_(h to 0) ( 1+h) =1`
` lim_(x to 2-) f(x) = lim_( h to 0) f ( 2-h) `
` lim_(h to 0) {" k ( 2-h) +5} = lim_( h to 0) {( 2 k + 5) -kh } = ( 2k +5)`
Now ` lim_(x to 2) f(x) " exists only when " 2k + 5 =1 ` i.e when k = -2 .
when k = -2 we have ` lim_( x to 2) f (x) = f( 2) =1 `
Hence f (x) is continuous at x=2 when k =-2


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