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 is continuous at x = 1.

JEE/Engineering Exams

Maths

Solution

255

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(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}= )