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```Solved Examples on Continuity

Illustration 1: The function

f(x) = [log(1+ax) – log(1-bx)]/ x

is not defined at x = 0. The value which should be assigned to f at x=0 so that it is continuous at 0 is (1983)

1. a-b                                                              2. a+b

3. log a + log b                                                 4. None of these

Solution: We know for the function f to be continuous we must have

f(0) = limx→0 f(x)

= limx→0[log(1+ax) – log(1-bx)]/ x

= limx→0a[log(1+ax)] /ax+ b log(1-bx)]/ -bx

= a.1 + b.1

= a+b (since f(0) = a + b) Illustration 2: The function f(x) = [x]2 – [x2] is discontinuous at (1999)

1. all integers

2. all integers except 0 and 1

3. all integers except 0

4. all integers except 1

Solution: This question is based on greatest integer function. We know that all integers are critical points for the greatest integer function.

Case 1: When x ∈ I

f(x) = [x]2 – [x2] = x2 –x2 = 0

Case 2: when x does not belong to I

If 0<x<1, then [x] = 0

And 0<x2<1, then [x2] = 0

Next, if 1 ≤ x2< 2

Then 1≤ x < √2

So, [x] = 1 and [x2] = 1

Therefore, f(x) = [x]2 – [x2] = 0 if 1 ≤ x <√2

Hence, f(x) = 0 if 0≤ x < √2

This shows that f(x) is continuous at x =1.

Hence f(x) is discontinuous in (-∞, 0) ∪[√2, ∞) on many other points.

Illustration 3: let f(x) = x+a, if x < 0

= |x-1|, if x ≥ 0

and g(x) = x+1, if x < 0

= (x-1)2 + b, if x ≥ 0

Where a and b are non-negative real numbers. Determine the composite function gof. If (gof)(x) is continuous for all real x determine the values of a and b. (2002)

Solution: (gof)(x) = f(x) + 1, if f(x) < 0

= (f(x) -1)2 + b, if f(x)≥ 0

Hence ,(gof)(x) = x+a+1, if x < 0

= (x+a-1)2 + b, if –a ≤ x <0

= (|x-1| - 1)2 + b, if x ≥ 0

As (g0f)(x) is continuous at x =-a,

(gof)(-a) = (gof)(-a+) = (gof)(-a-)

So, 1+b = 1+b = 1 hence, this gives b = 0

Also, gof is continuous at x = 0

So, gof(0) = gof(0+)= gof(0-)

So, b = b = (a-1)2+b
Hence, a = 1.
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