If f:[−2,2]→R is defined by f(x)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩√1+ex−√1−exxfor−2≤x<0x+3x+1for0≤x≤2 is continuous on[−2,2], then e=
A
2√3
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B
3
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C
32
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D
3√2
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Solution
The correct option is D3 limx→0−f(x) =limx→0−√1+ex−√1−exx =limx→0−1+ex−(1−ex)(√1+ex+√1−ex)x =limx→0−2ex(√1+ex+√1−ex)x =limx→0−2e(√1+ex+√1−ex) =2e2 =e ............(i) limx→0+f(x) =limx→0+x+3x+1 =3...............(ii) For the function to be continuous LHL=RHL ∴e=3.