2IIM Quant Notes  Functions  Part (1/2)

rajesh_balasubramanian
Director, 2IIM Online CAT Preparation  IIT Madras  IIM Bangalore  CAT 100th percentile  CAT 2011, 2012 and 2014.
How many onto functions can be defined from the set A = {1, 2, 3, 4} to {a, b, c}?
81
79
36
45First let us think of the number of potential functions possible. Each element in A has three options in the codomain. So, the number of possible functions = 3^4 = 81.
Now, within these, let us think about functions that are not onto. These can be under two scenarios.Scenario 1: Elements in A being mapped on to exactly two of the elements in B (There will be one element in the codomain without a preimage).
Let us assume that elements are mapped into A and B.
Number of ways in which this can be done = 2^4 – 2 = 14
2^4 because the number of options for each element is 2. Each can be mapped on to either A or B
2 because these 24 selections would include the possibility that all elements are mapped on to A or all elements being mapped on to B. These two need to be deducted.The elements could be mapped on B & C only or C & A only.
So, total number of possible outcomes = 14 * 3 = 42.Scenario 2: Elements in A being mapped to exactly one of the elements in B. (Two elements in B without preimage). There are three possible functions under this scenario. All elements mapped to a, or all elements mapped to b or all elements mapped to c.
Total number of onto functions = Total number of functions – Number of functions where one element from the codoamin remains without a preimage  Number of functions where 2 elements from the codoamin remain without a preimage.
⇒ Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36
f(x + y) = f(x)f(y) for all x, y, f(4) = +3 what is f(8)?
1/3
1/9
9
6f(x + 0) = f(x) f(0)
f(0) = 1
f(4 + 4) = f(0)
f(4 + 4) = f(4) f(4)
1 = +3 x f(4)
f(4) =13
f(8) = f(4 + (4)) = f(4) f(4)
f(8) = 1/3 * 1/3 = 1/9Find the maximum value of f(x); if f(x) is defined as the Min {(x – 1)^2 + 2, (x – 2)^2 + 1}
1
2
0
3First let us find the range where Min ((x – 1)^2 + 2, (x – 2)^2 + 1) is – (x – 1)^2 + 2.
In other words, in which range is – (x – 1)^2 + 2 < (x – 2)^2 + 1.
–(x^2 – 2x +1) + 2 < x2 – 4x + 4 + 1
0 < 2x^2 – 6x + 4
x^2 – 3x + 2 > 0
(x – 1) (x – 2) > 0
=> x > 2 or x < 1So, for x ∈ (1, 2) , f(x) = (x – 2)^2 + 1
And f(x) = –(x – 1)^2 + 2 elsewhere.Let us also compute f(1) and f(2)
f(1) = 2, f(2) = 1
For x ∈ (∞, 1), f(x) = –(x – 1)^2 + 2f(1) = 2
For x ∈ (1, 2), f(x) = (x – 2)^2 + 1f(2) = 1
For x ∈ (2, ∞), f(x) = –(x – 1)^2 + 2
For x < 1 and x > 2, f(x) is (square) + 2 and so is less than 2.
When x lies between 1 and 2, the maximum value it can take is 2. f(1) = 2 is the highest value f(x) can take.As a simple rule of thumb, the best way to approach this question is to solve the two expressions. This gives us the meeting points of the two curves. One of the two meeting points should be the maximum value.
Consider functions f(x) = x^2 + 2x, g(x) = √(x+1) and h(x) = g(f(x)). What are the domain and range of h(x)?
h(x) = g(f(x)) = g(x^2 + 2x) = √(x^2 + 2x + 1) = √(x+1)^2 = x + 1
This bit is very important, and often overlooked.
√9 = 3, not ±3
If x^2 = 9, then x can be ±3, but √9 is only +3.So, √(x^2) = x, not +x, not ±x
Domain of x + 1 = ( ∞, +∞), x can take any value.
As far as the range is concerned, x + 1 cannot be negative. So, range = [0, ∞)Correct Answer: Domain: ( ∞, +∞), Range [0, ∞]
[x] = greatest integer less than or equal to x. If x lies between 3 & 5, what is the probability that [x^2] = [x]^2?
Roughly 0.64
Roughly 0.5
Roughly 0.14
Roughly 0.36Let us take a few examples.
[3^2] = [3]^2
[3.5^2] = 12, [3.5]^2 = 9
[4^2] = 16, [4]^2 = 16
For x ∈ (3, 5). [x]^2 can only take value 9, 16 and 25.Let us see when [x^2] will be 9, 16 or 25.
If [x^2] = 9,
x^2 ∈ [9, 10)
=> x ∈ [3, √10)[x^2] = 16
x^2 ∈ [16, 17)
=> x ∈ [4, √17)In the given range [x^2] = 25 only when x = 5
So [x^2] = [x]^2 when x ∈ [3, √10] or [4, √10) or 5.
Probability = (√10 − 3 + √17−4)/2
(3.16 − 3 + 4.12−4)/2 = 0.28/2 = 0.14Give the domain and range of the following functions:
f(x) = x^2 + 1
g(x) = log(x + 1)
h(x) = 2^x
f(x) = 1/(x+1)
p(x) = x + 1
q(x) = [2x], where [x] gives the greatest integer less than or equal to xf(x) = x^2 + 1
Domain = All real numbers (x can take any value)
Range [1, ∞). Minimum value of x^2 is 0.g(x) = log (x + 1)
Domain = Log of a negative number is not defined so (x + 1) > 0 or x > 1
Domain ( 1, ∞)
Range = (∞, +∞)Note: Log is one of those beautiful functions that is defined from a restricted domain to all real numbers. Log 0 is also not defined. Log is defined only for positive numbers
h(x) = 2^x Domain  All real numbers.
Range = (0, ∞)
The exponent function is the mirror image of the log function.f(x) = 1/(x+1)
Domain = All real numbers except 1
Range = All real numbers except 0.p(x) = x + 1
Domain = All real numbers
Range = [0, ∞) Modulus cannot be negativeq(x) = [2x], where [x] gives the greatest integer less than or equal to x
Domain = All real numbers
Range = All integersThe range is NOT the set of even numbers. [2x] can be odd. [2 * 0.6] = 1. It is very important to think fractions when you are substituting values.
How many elements are present in the domain of (9–x)C(x+1)?
5
6
4
7For nCr to be defined, we should have r greater than or equal to zero and, n greater than or equal to r.
Therefore 9 – x ≥ x + 1 or 4 ≥ x and
also x + 1 > 0 or x > –1
Therefore x takes on the values {–1,0,1,2,3,4}
Therefore domain = {–1,0,1,2,3,4} and range = {10C0,9C1,8C2,7C3,6C4,5C5}
There are 6 elements in the domain.
Answer choice (b).f(x + y) = f(x)f(y) for all x, y, f(4) = + 3 what is f(–8)?
1/3
1/9
9
6f(x + 0) = f(x) f(0)
f(0) = 1
f(4 + – 4) = f(0)
f(4 + – 4) = f(4) f(–4)
1 = +3 x f(–4)
f(4) = 1/3
f(– 8 ) = f(– 4 + (– 4)) = f(– 4) f(– 4)
f(– 8 ) = 1/3 x 1/3 = 1/9If f(x – 3) = 2x^3 + p – qx and f(x^2 – 4) = x^2 – 8q + 6p, then what is the value of p – q?
5
10
6
Cannot determinef(x – 3) = 2x^3 + p – qx
Let x = 3, then f(0) = 54 + p – 3q  (1)
f(x^2 – 4) = x^2 – 8q + 6p
Let x^2 = 4, then f(0) = 4 – 8q + 6p  (2)From (1) and (2)
54 + p – 3q = 4 – 8q + 6p
50 = 5p – 5q
p  q = 10Given that x is real and f(x) = f(x + 1) + f(x – 1). Determine the value of ‘a’ that will satisfy f(x) + f(x + a) = 0?
1
2
1
3f(x) = f(x + 1) + f(x – 1)
Let f(x) = p and f(x – 1) = qf(x + 1) = f(x + 1 + 1) + f(x + 1 – 1) (Put x = x + 1 here!)
= f(x + 2) + f(x)
p – q = f(x + 2) + p
Or f(x + 2) = qf(x + 2) = f(x + 2 + 1) + f(x + 2 – 1) (Put x = x + 2 here!)
= f(x + 3) + f(x + 1)
– q = f(x + 3) + p – q
Or f(x + 3) = pAt this point we notice that f(x) = p and f(x+3) = p
f(x) + f(x + 3) = 0 (This is the condition to be satisfied to determine a)
Hence ‘a’ = 3x is a real number such that f(x) = 1/x when x > 0 and f(x) = 1/(x + 1) otherwise. Also
$f^n$ (x) = f($f ^{ n 1}$ (x)). What is f(3) +$f^2$ (3) +$f^3$ (3) +$f^4$ (3)?
2/3
14/3
0
3f(x) = 1/x when x > 0; f(x) = 1/(x + 1) otherwise
f(3) = 1/3
f(3) = 1/2$f^2$ (3) = f(f(3) = f(1/3) = 3$f^2$ (3) = f(f(3) = f(1/2) = 1/(1/2 + 1) = 2f(3) +
$f^2$ (3) +$f^3$ (3) +$f^4$ (3)
= = 1/3 + f($f^1$ (3)) + f($f^2$ (3)) + f($f^3$ (3))
= = 1/3 + f(1/2) + f(3) + f(f($f^2$ (3)))
= 1/3 + 2 + 1/3 + f(f(2))
= 8/3 + f(1/2)
= 8/3 + 2
= 14/3Which of the following functions are identical?
f(x) = x^3/x^2
g(x) = (√x)^2
h(x) = x
a) f(x) and g(x)
b) f(x) and h(x)
c) All 3 are identical
d) None of these are identicalFor functions to be identical, their domains should be equal
f(x) – x can’t be zero
g(x) – x can’t be negative
h(x) – x can take all possible values
Therefore, Answer – (d)