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CAT 2025 Lesson : Functions - Box & Composite Functions

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6.Special Types of Functions

6.1 Greatest/Smallest Integer Functions

These are also called step functions as they look like steps when graphically plotted.

Greatest integer function of xx, also denoted as [x]\bm{[x]}, returns the greatest integer less than or equal to xx.

For instance, [1] = 1 , [1.2] =1 , [1.99] = 1,[2] = 2[1] \space = \space 1 \space , \space [1.2] \space = 1 \space , \space [1.99] \space = \space 1, [2] \space = \space 2

This is also called a floor function.
Least integer function of xx, returns the least integer greater than or equal to xx.

For instance, LIF(11) = 1 , = \space 1 \space , \space LIF(1.21.2) = 2 ,  \space =\space 2 \space , \space LIF(1.991.99)  = 2\space = \space 2,
LIF(22)  = 2\space = \space 2

This is also called a ceiling function.

Example 12

If [X] denotes the greatest integer less than or equal to X, then

[13]+[13+199]+[13+299]+[13+9899]=\left[ \dfrac{1}{3} \right] + \left[ \dfrac{1}{3} + \dfrac{1}{99} \right] + \left[ \dfrac{1}{3} + \dfrac{2}{99} \right] + \left[ \dfrac{1}{3} + \dfrac{98}{99} \right] =
[XAT 2008]

(11)33      (22)34      (33)66      (44)67      (55)98

Solution

[13+099]=0\left[ \dfrac{1}{3} + \dfrac{0}{99} \right] = 0, [13+199]=0\left[ \dfrac{1}{3} + \dfrac{1}{99}\right] = 0 and so on till [13+6599]\left[ \dfrac{1}{3} + \dfrac{65}{99} \right]

From [13+6699]=1\left[ \dfrac{1}{3} + \dfrac{66}{99} \right] = 1 onwards, all the terms equal 11.

Between 6666 and 9898 (both inclusive), there are total of 98−66+1=3398 - 66 + 1 = 33 terms, each of whose value is 11.

∴\therefore Total of these 3333 terms is 3333.

Answer: (1) 3333

6.2 Nested & Composite Functions

In nested functions, we will given a function f(x)f(x) and asked to find f(f(x))f(f(x)). This is a function within a function.

In composite functions, we will be given 22 or more functions, say f(x)f(x) and g(x)g(x). We will then be asked to find f(g(x))f(g(x)), etc.

Example 13

If f(x)=x2,g(x)=10−xf(x) = x^2, g(x) = 10 - x and h(x)=10xh(x) = \dfrac{10}{x}, then

(I) What is f(g(x))?f(g(x))?
(II) What is g(g(x))?g(g(x))?
(III) What is h(g(f(x)))h(g(f(x)))?

Solution

Case I: f(g(x))=f(10−x)f(g(x)) = f(10 - x)

=(10−x)2= (10 - x)^2 =100−20x+x2= \bm{100 - 20x + x^2}

Case II: g(g(x))=g(10−x)g(g(x)) = g(10 - x)

=10−(10−x)= 10 - (10 - x) =x=x

Case III: h(g(f(x)))=h(g(x2))h(g(f(x))) = h(g(x^2)) =h(10−x2)= h(10 - x^2)

=1010−x2\bm{= \dfrac{10}{10 - x^2}}


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