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Question Number 73567 by Rio Michael last updated on 13/Nov/19

Determine the value of a and b fir which the function f,defined by  f(x) =  { ((−2sinx,  x < −(π/2))),((asinx + b,−(π/2) ≤ x < (π/2))),((cosx,              x > (π/2))) :}  is continouos

$${Determine}\:{the}\:{value}\:{of}\:{a}\:{and}\:{b}\:{fir}\:{which}\:{the}\:{function}\:{f},{defined}\:{by} \\ $$ $${f}\left({x}\right)\:=\:\begin{cases}{−\mathrm{2}{sinx},\:\:{x}\:<\:−\frac{\pi}{\mathrm{2}}}\\{{asinx}\:+\:{b},−\frac{\pi}{\mathrm{2}}\:\leqslant\:{x}\:<\:\frac{\pi}{\mathrm{2}}}\\{{cosx},\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:>\:\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$ $${is}\:{continouos} \\ $$

Commented byRio Michael last updated on 13/Nov/19

thank you

$${thank}\:{you} \\ $$

Commented bykaivan.ahmadi last updated on 13/Nov/19

lim_(x→−(π^− /2)) f(x)=−2sin(−(π/2))=2  lim_(x→−(π^+ /2) ) f(x)=lim_(x→−(π^+ /2)) (asinx+b)=−a+b  ⇒−a+b=2  and similarly for x=(π/2) we have  a+b=0  ⇒ { ((−a+b=2)),((a+b=0)) :}⇒a=−1,b=1.

$${lim}_{{x}\rightarrow−\frac{\pi^{−} }{\mathrm{2}}} {f}\left({x}\right)=−\mathrm{2}{sin}\left(−\frac{\pi}{\mathrm{2}}\right)=\mathrm{2} \\ $$ $${lim}_{{x}\rightarrow−\frac{\pi^{+} }{\mathrm{2}}\:} {f}\left({x}\right)={lim}_{{x}\rightarrow−\frac{\pi^{+} }{\mathrm{2}}} \left({asinx}+{b}\right)=−{a}+{b} \\ $$ $$\Rightarrow−{a}+{b}=\mathrm{2} \\ $$ $${and}\:{similarly}\:{for}\:{x}=\frac{\pi}{\mathrm{2}}\:{we}\:{have} \\ $$ $${a}+{b}=\mathrm{0} \\ $$ $$\Rightarrow\begin{cases}{−{a}+{b}=\mathrm{2}}\\{{a}+{b}=\mathrm{0}}\end{cases}\Rightarrow{a}=−\mathrm{1},{b}=\mathrm{1}. \\ $$

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