TITA Based Questions: Functions | Quantitative Aptitude (Quant) - CAT PDF Download

Ans: 540

Set A={2,3,5,7,11,13} so |A|=6
Set B={1, 8, 27} so |B|=3
Without any restrictions, each element in A can map to any of the 3 elements in B. Thus, the total number of functions is: $3\mathrm{<sup\; style="font-size:\; 12px\; !important">6</sup>}=729$
Excluding Functions That Miss One Element in B: 
If a function does not map to an element in B, there are 2 elements in B left for mapping. 
The total number of such functions (for each specific element not mapped) is: $2\mathrm{<sup\; style="font-size:\; 12px\; !important">6</sup>}=64$
Since there are 3 elements in B, the total number of such functions is:3 x 64 = 192
Adding Back Functions That Miss Two Elements in B: If a function misses two elements in B, there is only 1 element left for mapping. 
The total number of such functions is: 1⁶ = 1. 
Since there are ³C₂ ways to choose which two elements are missed, the total number of such functions is: 3
Using the inclusion-exclusion principle, the number of functions where all elements of B are mapped by at least one element of A is:
729 - 192 + 3 = 540. 

Ans: 4095
Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question. 
f(16000) can be written as f(2⁸ ×  5⁴)
Now, we can try to find these individual values:
For any prime p: f(p)=1
f(p²) = f(p)f(p)+f(p)+f(p) = 1 + 1 + 1 = 3
f(p³)=f(p2)f(p)+f(p²)+f(p) = 3 + 3 + 1 = 7
This way, we can find the function output for any prime number raised to a power. 
We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8
f(p⁴) = 7 + 7 + 1 = 15
f(p⁵) = 15 + 15 + 1 = 31
f(p⁶) = 31 + 31 + 1 = 63
f(p⁷) = 63 + 63 + 1 = 127
f(p⁸) = 127 + 127 + 1 = 255

Using these values in the original expression of f(2⁸ × 5⁴) = f(2⁸)f(5⁴)+f(2⁸)+f(5⁴) we get
f(2⁸ × 5⁴) = (255 ×  15) + 255 + 15 = 4095

Ans: 45
From the inequality and nature of x, and y, we get the given diagram:

We need to find the area of the quadrilateral ABDE = area of rectangle ABCD + area of triangle CDE
⇒ Area of ABCD = (7-3)*5 = 20 units, and the area of triangle CDE = (1/2)*10*5 = 25 units.
Hence, the area of the quadrilateral ABDE = (20+25) = 45 units.

Ans: 3
Given that f(3x + 2y, 2x - 5y) = 19x.
Let us assume the function f(a,b) is a linear combination of a and b.
⇒ f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x
⇒ 3m + 2n = 19 and 2m - 5n = 0
Solving we get m = 5 and n = 2
⇒ f(a,b) = 5a+2b
⇒ f(x,2x) = 5x + 2(2x) = 9x = 27 
=> x = 3.

Ans: 50
x > = a, so |x - a| = x - a
x < 100, so |x - 100| = 100-x
f(x) = (x-a) + (100-x) + |x-a-50| =100
or, |x-a-50| = a
From the graph we can can see that when x = a then
|x - a - 50| = a
or, a = 50
Similarly when x = a + 100
|x - a - 50| = a
or, a = 50
So value of a is 50 when f(x) is 100.

Ans: 24
f(x) ≥ 0 for all real numbers x, so D <=0
Since f(2)=0 therefore x = 2 is a root of f(x)
Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D = 0
Therefore f(x) = a(x − 2)²
 f(4)=6 
or, a(x − 2)²
a = 3/2
= 24

Ans: 12
f(x)+f(x−1)=1 ...... (1) 
f(x² − x) = 5 ...... (2) 
g(x) = x²  
Substituting x = 1 in (1) and (2), we get
f(0) = 5
f(1) + f(0) = 1
f(1) = 1 - 5 = -4
f(2) + f(1) = 1
f(2) = 1 + 4 = 5
f(n) = 5 if n is even and f(n) = -4 if n is odd
f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

Ans: 4
f(x)= x² +ax+b 
f(x+1) = x² +2x+1+ax+a+b 
f(x−1) = x² −2x+1+ax−a+b 
g(x)=f(x+1)−f(x−1)=4x+2a 
Now g(20)=72 from this we get a=−4 ; f(x) = x² −4x +b 
For this expression to be greater than zero it has to be a perfect square which is possible for b≥ 4 
Hence the smallest value of 'b' is 4.

Ans: 3 
The area of the region contained by the lines ∣x∣ − y ≤ 1, y ≥ 0 and y ≤ 1 is the white region. 
Total area = Area of rectangle + 2 x Area of triangle

Hence, 3 is the correct answer.

Ans: 4
Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that
x(2)+x(3)+x(4)+........x(10)= 47*9=423...................(1)
Similarly x(1)+x(2)+x(3)+x(4)................+x(9)= 42*9=378...............(2)
Subtracting both the equations we get x(10)-x(1)=45
Now, the sum of the 10 observations from equation (1) is 423+x(1)
Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5
Hence difference in average will be 46.5-42.5 = 4 which is the correct answer