Sets, Relations and FunctionsMCQPYQ Nov. 19Question 1962 of 222
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If f(n)=f(n−1)+f(n−2)\displaystyle f(n) = f(n-1) + f(n-2) when n≥2\displaystyle n ≥\displaystyle \ge 2, f(0)=0\displaystyle f(0) = 0, f(1)=1\displaystyle f(1) = 1 then f(7)=?\displaystyle f(7) = ?

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option c — 13

All Options:

  • A5
  • B8
  • C13
  • DNone of these

Detailed Solution & Explanation

We are given the recurrence relation:
f(n)=f(n−1)+f(n−2)for n≥2f(n) = f(n-1) + f(n-2) \quad for \displaystyle \text{for } n ≥\displaystyle \ge 2
With initial values:
f(0)=0f(0) = 0
f(1)=1f(1) = 1
Let's calculate the terms of this sequence (which is the Fibonacci sequence) up to n=7\displaystyle n = 7:
- f(2)=f(1)+f(0)=1+0=1\displaystyle f(2) = f(1) + f(0) = 1 + 0 = 1
- f(3)=f(2)+f(1)=1+1=2\displaystyle f(3) = f(2) + f(1) = 1 + 1 = 2
- f(4)=f(3)+f(2)=2+1=3\displaystyle f(4) = f(3) + f(2) = 2 + 1 = 3
- f(5)=f(4)+f(3)=3+2=5\displaystyle f(5) = f(4) + f(3) = 3 + 2 = 5
- f(6)=f(5)+f(4)=5+3=8\displaystyle f(6) = f(5) + f(4) = 5 + 3 = 8
- f(7)=f(6)+f(5)=8+5=13\displaystyle f(7) = f(6) + f(5) = 8 + 5 = 13
The value of f(7)\displaystyle f(7) is $13$.
This corresponds directly to Option C.
*Note: The textbook answer key incorrectly lists Option D (None of these) as correct, which is a typographical error.*
Hence, **Option C** is the correct answer.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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