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1- sin' x3 cos xif x <Letf(x) =If f(x) is continuous at x =find a and b.NEUENb (1-sin x)| (A-2752Tencendo 1 |
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Answer» hii dudGiven:f(x)=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩1−sin3x3cos2x,ifx<π2a,ifx=π2b(1−sinx)(π−2x)2,ifx>π2fx=1-sin3x3cos2x,ifx<π2a,ifx=π2b1-sinxπ-2x2,ifx>π2 We have(LHLatx=π2π2) =limx→π2−f(x)=limh→0f(π2−h)limx→π2-fx=limh→0fπ2-h =limh→0(1−sin3(π2−h)3cos2(π2−h))=limh→0(1−cos3h3sin2h)=13limh→0((1−cosh)(1+cos2h+cosh)(1−cosh)(1+cosh))=13limh→0((1+cos2h+cosh)(1+cosh))=13(1+1+11+1)=12=limh→01-sin3π2-h3cos2π2-h=limh→01-cos3h3sin2h=13limh→01-cosh1+cos2h+cosh1-cosh1+cosh=13limh→01+cos2h+cosh1+cosh=131+1+11+1=12 (RHLatx=π2π2) =limx→π2+f(x)=limh→0f(π2+h)limx→π2+fx=limh→0fπ2+h=limh→0(b[1−sin(π2+h)][π−2(π2+h)]2)=limh→0(b(1−cosh)[−2h]2)=limh→0(2bsin2h24h2)=limh→0⎛⎝2bsin2h216h24⎞⎠=b8limh→0(sinh2h2)2=b8×1=b8=limh→0b1-sinπ2+hπ-2π2+h2=limh→0b1-cosh-2h2=limh→02bsin2h24h2=limh→02bsin2h216h24=b8limh→0sinh2h22=b8×1=b8 Also,f(π2)=afπ2=a Iff(x) is continuous atx=π2π2, thenlimx→π2−f(x)=limx→π2+f(x)=f(π2)limx→π2-fx=limx→π2+fx=fπ2 ⇒12=b8=a⇒12=b8=a ⇒a=12andb=4 |
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