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CAT 2025 Lesson : Functions - Sequences & Patterns

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6.3 Functions with a sequence

In these questions we establish a sequence and look for patterns.

Example 14

If f(x)=f(x−1)−f(x−2),f(1)=10f(x) = f(x - 1) - f(x - 2), f(1) = 10f(x)=f(x−1)−f(x−2),f(1)=10 and f(2)=3f(2) = 3f(2)=3, then f(100)=?f(100) = ?f(100)=?

Solution

We begin with some iterations to look for a pattern.

f(1)=10f(1) = 10f(1)=10
f(2)=3f(2) = 3f(2)=3
f(3)=3−10=−7f(3) = 3 - 10 = - 7f(3)=3−10=−7
f(4)=−7−3=−10f(4) = -7 - 3 = - 10f(4)=−7−3=−10
f(5)=−10−(−7)=−3f(5) = -10 - (-7) = - 3f(5)=−10−(−7)=−3
f(6)=−3−(−10)=7f(6) = -3 - (-10) = 7f(6)=−3−(−10)=7
f(7)=7−(−3)=10f(7) = 7 -(-3) = 10f(7)=7−(−3)=10
f(8)=10−(7)=3f(8) = 10 - (7) = 3f(8)=10−(7)=3
f(9)=3−10=−7f(9) = 3 - 10 = - 7f(9)=3−10=−7
.........

Note that f(k)=f(k+6)=f(k+12)=...f(k) = f(k + 6) = f(k + 12) = ...f(k)=f(k+6)=f(k+12)=...

∴\therefore∴ We express 100100100 as a sum of a constant and a multiple of 666

f(100)=f(4+96)=f(4)=−10f(100) = f(4 + 96) = f(4) = -10f(100)=f(4+96)=f(4)=−10

Answer: −10-10−10

6.4 Functions with a pattern

In these questions, we find the values for the next few inputs to establish a pattern. We, then, apply this pattern to find the desired output.

Example 15

A function f(x)f(x)f(x) satisfies f(1)=3600f(1) = 3600f(1)=3600 and f(1)+f(2)+.....+f(n)=n2f(n)f(1) + f(2) + ..... + f(n) = n^2 f(n)f(1)+f(2)+.....+f(n)=n2f(n) , for all positive integers n>1n \gt 1n>1 . What is the value of f(9)f(9)f(9) ?
[CAT 2007]

(111) 120120120              (222) 808080              (333) 240240240              (444) 200200200              (555) 100100100

Solution

In the equation given in the question, if we take f(n)f(n)f(n) to one side, question,

f(1)+f(2)+.....+f(n−1)+f(n)=n2f(n)f(1) + f(2) + ..... + f(n - 1 ) + f(n) = n^2 f(n)f(1)+f(2)+.....+f(n−1)+f(n)=n2f(n)

⇒ f(1)+f(2)+.....+f(n−1)=n2f(n)−f(n)f(1) + f(2) + ..... + f(n - 1 ) = n^2 f(n) - f(n)f(1)+f(2)+.....+f(n−1)=n2f(n)−f(n)

⇒ f(1)+f(2)+.....+f(n−1)=(n2−1)f(n)f(1) + f(2) + ..... + f(n - 1 ) = (n^2 - 1)f(n) f(1)+f(2)+.....+f(n−1)=(n2−1)f(n)-----(a)

In the equation given in the question, if we f(n−1)f(n - 1)f(n−1) is the last term, then

f(1)+f(2)+.....+f(n−1)=(n−1)2f(n−1)f(1) + f(2) + ..... + f(n - 1 ) = (n - 1)^2 f(n - 1) f(1)+f(2)+.....+f(n−1)=(n−1)2f(n−1)-----(b)

As LHS is the same in both (a) and (b), we equate the RHS and get the following.

(n2−1)f(n)=(n−1)2f(n−1) (n^2 - 1) f(n) = (n - 1)^2 f(n - 1)(n2−1)f(n)=(n−1)2f(n−1)

⇒ f(n)=(n−1)2f(n−1)(n+1)(n−1)f(n) = \dfrac{(n - 1)^2 f(n - 1)}{(n + 1)(n - 1)}f(n)=(n+1)(n−1)(n−1)2f(n−1)​ ⇒ f(n)=(n−1)n=1×f(n−1)f(n) = \dfrac{(n - 1)}{n = 1} \times f(n - 1)f(n)=n=1(n−1)​×f(n−1)

∴ f(2)=13×f(1)f(2) = \dfrac{1}{3} \times f(1)f(2)=31​×f(1)

f(3)=24×f(2)=24×13×f(1)f(3) = \dfrac{2}{4} \times f(2) = \dfrac{2}{4} \times \dfrac{1}{3} \times f(1)f(3)=42​×f(2)=42​×31​×f(1)

f(4)=35×f(3)=35×24×13×f(1)f(4) = \dfrac{3}{5} \times f(3) = \dfrac{3}{5} \times \dfrac{2}{4} \times \dfrac{1}{3} \times f(1)f(4)=53​×f(3)=53​×42​×31​×f(1)

The pattern is such that the numerator is the product of all natural numbers from 111 to (n−1)(n - 1)(n−1) while the denominators are the product of natural numbers from 3 to (n+1)(n + 1)(n+1)

f(9)=8×7×6×5×4×3×2×110×9×8×7×6×5×4×3×f(1)f(9) = \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3} \times f(1)f(9)=10×9×8×7×6×5×4×38×7×6×5×4×3×2×1​×f(1)

⇒ f(9)=2×110×9×3600=80f(9) = \dfrac{2 \times 1}{10 \times 9} \times 3600 = 80f(9)=10×92×1​×3600=80

Answer: (2)80(2) 80(2)80

Example 16

fff is a function for which f(1)=1f(1) = 1f(1)=1 and f(x)=2x+f(x−1)f(x) = 2x + f(x - 1)f(x)=2x+f(x−1) for each natural number x≥2x \geq 2x≥2 . Find f(31)f(31)f(31).
[XAT 2016]

(111) 869869869
(222) 929929929
(333) 951951951
(444) 991991991
(555) None of the above

Solution

f(1)f(1)f(1) is just the first term and need not be a part of the pattern. Therefore, we shall keep it as-is.
f(2)=4+f(1)f(2) = 4 + f(1)f(2)=4+f(1)
f(3)=6+f(2)=6=4+f(1)f(3) = 6 + f(2) = 6 = 4 + f(1) f(3)=6+f(2)=6=4+f(1)
f(4)=8+f(3)=8+6+4+f(1)f(4) = 8 + f(3) = 8 + 6 +4 + f(1)f(4)=8+f(3)=8+6+4+f(1)

∴ f(31)=62+60+58+......+4+f(1)f(31) = 62 + 60 + 58 + ......+ 4 + f(1)f(31)=62+60+58+......+4+f(1)

Rearranging and taking 222 common,
f(31)=2(2+3+4+........+31)+f(1)f(31) = 2(2 + 3 + 4 + ........+ 31) + f(1)f(31)=2(2+3+4+........+31)+f(1)

The 303030 terms from 222 to 313131 are in AP, with their average being 2+312=332 \dfrac{2 + 31}{2} = \dfrac{33}{2}22+31​=233​

So, their sum =30×332 = 30 \times \dfrac{33}{2} =30×233​

f(31)=2(30×332)+f(1)=990+f(1)=991 f(31) = 2(30 \times \dfrac{33}{2}) + f(1) = 990 + f(1) = 991f(31)=2(30×233​)+f(1)=990+f(1)=991

Answer: (4)991(4) 991(4)991

6.5 Product of Functions

Example 17

Let f(x)f(x)f(x) be a function satisfying f(x)f(y)=f(xy)f(x) f(y) = f(xy)f(x)f(y)=f(xy) for all real x,yx, yx,y. If f(2)=4f(2) = 4f(2)=4, then what is the value of f(12)f \left(\dfrac{1}{2}\right)f(21​)?
[CAT 2008]
(1) 000          (2) 14 \dfrac{1}{4}41​           (3)12 \dfrac{1}{2}21​          (4)111          (5) Cannot be determined

Solution

f(2)×f(1)=f(2)f(2) \times f(1) = f(2)f(2)×f(1)=f(2) ⇒ f(1)=1f(1) = 1f(1)=1

f(2)×f12=f(1)f(2) \times f \dfrac{1}{2} = f(1)f(2)×f21​=f(1) ⇒ 4×f(12)=14 \times f\left(\dfrac{1}{2} \right) = 14×f(21​)=1 ⇒ f(12)=14f\left(\dfrac{1}{2} \right) = \dfrac{1}{4}f(21​)=41​

Answer: (2) 14\dfrac{1}{4}41​

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