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Question

If f:[2,2]R is defined by f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪1+ex1exxfor2x<0x+3x+1for0x2
is continuous on[2,2], then e=

A
23
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B
3
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C
32
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D
32
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Solution

The correct option is D 3
limx0f(x)
=limx01+ex1exx
=limx01+ex(1ex)(1+ex+1ex)x
=limx02ex(1+ex+1ex)x
=limx02e(1+ex+1ex)
=2e2
=e ............(i)
limx0+f(x)
=limx0+x+3x+1
=3...............(ii)
For the function to be continuous
LHL=RHL
e=3.

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