Question # 6.
(i) Find the value of k for which the function
kx + 5 when x<2
f(x) =
|x-1 when x > 2
is continuous at x = 2.
(ii) Find the value of k for which
| kx +1 when x

(i) function ( f(x) ) is continuens at ( x=2 ) ( therefore lim _{x rightarrow 2} f(x)=log _{x rightarrow 2^{+}} f(x)=f(2) )
( Rightarrow frac{d t}{x+2^{-}}(k x+5)=operatorname{lt} x-1=2 k+5 )
( Rightarrow 2 k+5=2-1=2 k+5 )
( Rightarrow 2 k+5=1 )
( Rightarrow 2 k=1-5=-4 )
( Rightarrow quad 2 k=-4 quad Rightarrow k=-2 )
Ques
(ii) function ( f(x) ) is contiat ( x=pi ) ( lim _{x rightarrow pi^{-}} f(x)=lim _{x rightarrow n^{+}} f(x)=f(n) )
( Rightarrow lim _{x rightarrow pi^{-}}(k x+1)=log _{x rightarrow n^{+}} cos x=k pi+1 )
( Rightarrow quad K T+1=cos pi=K Pi+1 )
( Rightarrow quad k Pi+1=-7 )
( Rightarrow quad K cap=-2 )
( Rightarrow quad k=frac{-2}{pi} quad underline{a}= )

# 6.
(i) Find the value of k for which the function
kx + 5 when x<2
f(x) =
|x-1 when x > 2
is continuous at x = 2.
(ii) Find the value of k for which
| kx +1 when x
is continuous at x = 1.

Solution