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Question

Let fx=x-4x-4+a,x<4 a+b ,x=4x-4x-4+b,x>4.
Then, f (x) is continuous at x = 4 when
(a) a = 0, b = 0
(b) a = 1, b = 1
(c) a = −1, b = 1
(d) a = 1, b = −1.

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Solution

(d) a = 1, b = −1.

Given: fx=x-4x-4+a, if x<4a+b, if x=4x-4x-4+b, if x>4

We have
(LHL at x = 4) = limx4-fx=limh0f4-h

=limh04-h-44-h-4+a=limh0-h-h+a=a-1


(RHL at x = 4) = limx4+fx=limh0f4+h

=limh04+h-44+h-4+b=limh0hh+b=b+1

Also,
f4=a+b

If f(x) is continuous at x = 4, then

limx4-fx=limx4+fx=f4

a-1=b+1=a+b
a-1=a+b and b+1=a+b
b=-1 and a=1

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