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CAT 2019 Slot 1 — QA Question 19

Basics (Functions)Easy

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8 f(m + 1) − f(m) = 2, then m equals

Answer & solution

Answer: 10

Solution

Easy

ff has two rules depending on parity, so split on whether mm is odd or even. In each case m+1m+1 flips parity, so pick the matching rule for f(m+1)f(m+1) and f(m)f(m). One case yields no positive integer; the other does.

1

Case mm odd. Then m+1m+1 is even, so f(m+1)=(m+1)(m+2)f(m+1)=(m+1)(m+2) and f(m)=m+3f(m)=m+3.

8(m+1)(m+2)−(m+3)=2⇒ 8m2+24m+16−m−3=2⇒ 8m2+23m+11=0\begin{aligned} &8(m+1)(m+2)-(m+3)=2\\ &\Rightarrow\ 8m^2+24m+16-m-3=2\\ &\Rightarrow\ 8m^2+23m+11=0 \end{aligned}
2

Reject case 1. Sum of roots −238<0-\tfrac{23}{8}<0 and product 118>0\tfrac{11}{8}>0, so both roots are negative.

no positive integer root⇒ discard\begin{aligned} &\text{no positive integer root}\quad\Rightarrow\ \text{discard} \end{aligned}
3

Case mm even. Then m+1m+1 is odd, so f(m+1)=(m+1)+3=m+4f(m+1)=(m+1)+3=m+4 and f(m)=m(m+1)f(m)=m(m+1).

8(m+4)−m(m+1)=2⇒ 8m+32−m2−m=2⇒ m2−7m−30=0\begin{aligned} &8(m+4)-m(m+1)=2\\ &\Rightarrow\ 8m+32-m^2-m=2\\ &\Rightarrow\ m^2-7m-30=0 \end{aligned}
4

Solve and pick the valid root. Factor the quadratic from step 3.

(m−10)(m+3)=0⇒ m=10 or m=−3⇒ m=10(positive integer)\begin{aligned} &(m-10)(m+3)=0\\ &\Rightarrow\ m=10\ \text{or}\ m=-3\\ &\Rightarrow\ m=10\quad\text{(positive integer)} \end{aligned}
m=10m=10
CAT 2019 Slot 1 QA Q19: For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive — Solution | TheCATExam