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JEE Main Mock Test Series 2020 & Previous Year Papers

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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)+ 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 (gof)(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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