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Question

Function whose jump (non-negative difference of LHL and RHL) of discontinuity is greater than or equal to one, is/are-

A
f(x)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪(e1/x+1)(e1/x1),x<0(1cosx)x,x>0
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B
g(x)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪x1/31x1/21;x>1lnx(x1);12<x<1
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C
u(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪sin12xtan13x;x(0,12]|sinx|x;x<0
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D
w(x)={log3(x+2);x>2log1/2(x2+5);x<2
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Solution

The correct options are
A f(x)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪(e1/x+1)(e1/x1),x<0(1cosx)x,x>0
C u(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪sin12xtan13x;x(0,12]|sinx|x;x<0
D w(x)={log3(x+2);x>2log1/2(x2+5);x<2
We will check by options
Option A:
f(x)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪(e1/x+1)(e1/x1),x<0(1cosx)x,x>0
LHL=limx0f(x)=limh0f(0h)
=limh0(e1/h+1)(e1/h1)
=0+101=1
RHL=limx0+f(x)=limh0f(0+h)
=limh0(1cosh)h
=limh02sin2h/2h
=limh0h42sin2h/2h2h2
=0×1=0
Difference between LHL and RHL =0(1)=1
Hence, jump of discontinuity is 1.
Option B:
f(x)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪x1/31x1/21;x>1lnx(x1);12<x<1
RHL=limx1+f(x)=limh0f(1+h)
=limh0(1+h)1/31(1+h)1/21
=limh0(1+h319h2+....)1(1+12h18h2.....)1
=limh0h(1319h+....)h(1218h.....)
RHL=23
LHL=limx1f(x)=limh0f(1h)
=limh0ln(1h)1h1
limh0ln(1+(h))h
=1 (limx0ln(1+x)x=1)
Difference between LHL and RHL 123=13<1
Here, jump of discontinuity for f(x) is not 1 or greater than 1.
Option C:
f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪sin12xtan13x;x(0,12]|sinx|x;x<0
LHL=limx0f(x)=limh0f(0h)
=limh0sin|0h|h
=limh0|sinh|h
=limh0sinhh=1
RHL=limx0+f(x)=limh0f(0+h)
=limh0sin12htan13h
=limh02hsin12h2h3htan13h3h
RHL=23
Difference between LHL and RHL 23(1)=53>1
Hence, jump of discontinuity for f(x) is greater than 1.
Option D:
f(x)={log3(x+2);x>2log1/2(x2+5);x<2
RHL=limx2+f(x)=limh0f(2+h)
=limh0log3(h+4)
RHL=log34=1.26
LHL=limx2f(x)=limh0f(2h)
=limh0log1/2(h24h+9)
LHL=log1/29=0.653
Difference between LHL and RHL 1.26(0.653)>1
Hence, jump of discontinuity for f(x) is greater than 1.

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