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**Illustration 1: The function ****f(x) = [log(1+ax) â€“ 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) = lim_{xâ†’0} f(x)

= lim_{xâ†’0}[log(1+ax) â€“ log(1-bx)]/ x

= lim_{xâ†’0}a[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} â€“ [x^{2}] is discontinuous at (1999)**

f(x) = [x]

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

And 0<x

Next, if 1 â‰¤ x

Then 1â‰¤ x < âˆš2

So, [x] = 1 and [x

Therefore, f(x) = [x]

Hence, f(x) = 0 if 0â‰¤ x < âˆš2

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

Hence f(x) is discontinuous in (-âˆž, 0) âˆª[âˆš2, âˆž) on many other points.

**Illustration 3: let f(x) = x+a, if x < 0****= |x-1|, if x â‰¥ 0****and g(x) = x+1, if x < 0****= (x-1) ^{2} + b, if x â‰¥ 0**

= (f(x) -1)

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

= (x+a-1)

= (|x-1| - 1)

As (gof)(x) is continuous at x =-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

So, b = b = (a-1)

Hence, a = 1.

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