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Question Number 66225 by Rio Michael last updated on 11/Aug/19

Given that     f(x)= { ((−x + 1,  x≤ 3_ )),((kx −8,    x >3)) :}  is continuous then  f(5) =   A   2  B   0  C  −2  D  −1

$${Given}\:{that}\:\:\:\:\:{f}\left({x}\right)=\begin{cases}{−{x}\:+\:\mathrm{1},\:\:{x}\leqslant\:\mathrm{3}_{} }\\{{kx}\:−\mathrm{8},\:\:\:\:{x}\:>\mathrm{3}}\end{cases} \\ $$ $${is}\:{continuous}\:{then}\:\:{f}\left(\mathrm{5}\right)\:=\: \\ $$ $${A}\:\:\:\mathrm{2} \\ $$ $${B}\:\:\:\mathrm{0} \\ $$ $${C}\:\:−\mathrm{2} \\ $$ $${D}\:\:−\mathrm{1} \\ $$ $$ \\ $$

Commented byPrithwish sen last updated on 11/Aug/19

A.  2

$$\mathrm{A}.\:\:\mathrm{2} \\ $$

Commented bymathmax by abdo last updated on 11/Aug/19

f(3) =lim_(x→3^+ ) f(x) ⇒3k−8=−2 ⇒3k=6 ⇒k=2  ⇒f(x)=1−x if x≤3 and f(x)=2x−8 if x>3  ⇒f(5)=2×5−8 =2

$${f}\left(\mathrm{3}\right)\:={lim}_{{x}\rightarrow\mathrm{3}^{+} } {f}\left({x}\right)\:\Rightarrow\mathrm{3}{k}−\mathrm{8}=−\mathrm{2}\:\Rightarrow\mathrm{3}{k}=\mathrm{6}\:\Rightarrow{k}=\mathrm{2} \\ $$ $$\Rightarrow{f}\left({x}\right)=\mathrm{1}−{x}\:{if}\:{x}\leqslant\mathrm{3}\:{and}\:{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{8}\:{if}\:{x}>\mathrm{3} \\ $$ $$\Rightarrow{f}\left(\mathrm{5}\right)=\mathrm{2}×\mathrm{5}−\mathrm{8}\:=\mathrm{2} \\ $$

Answered by som(math1967) last updated on 11/Aug/19

3k−8=−3+1⇒k=(6/3)=2  f(5)=2×5−8=2

$$\mathrm{3}{k}−\mathrm{8}=−\mathrm{3}+\mathrm{1}\Rightarrow{k}=\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2} \\ $$ $${f}\left(\mathrm{5}\right)=\mathrm{2}×\mathrm{5}−\mathrm{8}=\mathrm{2} \\ $$

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