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Question: Answered & Verified by Expert

If fx=e1+1x-ae1x+1:x0       b:x=0 (where a and b are arbitrary constants) is continuous at x=0, then the value of a2 is equal to

(use e=2.7)

MathematicsContinuity and DifferentiabilityJEE Main
Solution:
1219 Upvotes Verified Answer
The correct answer is: 7.29
At x=0,
LHL=limh0+f0-h=limh0+e1-1h-ae-1h+1=0-a0+1=-a
RHL=limh0+f0+h=limh0+e1+1h-ae1h+1=limh0e-e-1h×a1+e-1h=e
and f0=b
fx is continuous at x=0
-a=e=ba=-e
a2=e2

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