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A two-branch recurrence on the integers | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

Let

f:\mathbb Z \to \mathbb R

be defined as

f\left(x+2\right) = f(x)+2\left(x+1\right)

x

is even and

f\left(x+2\right) = f(x)+1

x

is odd, and

f(1) = 1

f(2) = 5

, then

f(6) = f(41)

correct

f(7)<f(8)

correct

f(x)

is increasing

f(10)+f(11) = 59

correct

Solution

Step 1: For even

x

f\left(x+2\right)-f(x) = 2\left(x+1\right) = 2x+2

. Trying

f(x) = \tfrac{x^2}2+c

\frac{\left(x+2\right)^2}2-\frac{x^2}2 = \frac{4x+4}2 = 2x+2 .

Fitting

f(2) = 5

\tfrac42+c = 5 \Rightarrow c = 3

, so

f(x) = \frac{x^2}2+3 \qquad (x \text{ even})

Together with the recurrence,

f(2) = 5

determines

f

at every even integer, in both directions, so this is the only possibility on the evens. Step 2: For odd

x

f\left(x+2\right)-f(x) = 1

with

f(1) = 1

gives

f\left(1+2m\right) = 1+m

for every integer

m

, i.e.

f(x) = \frac{x+1}2 \qquad (x \text{ odd})

which fits

f(1) = 1

and satisfies

\tfrac{x+3}2 = \tfrac{x+1}2+1

Step 3:

f(6) = \frac{36}2+3 = 21 ,\qquad f(41) = \frac{42}2 = 21 . \quad\textbf{TRUE}

f(7) = \frac82 = 4 ,\qquad f(8) = \frac{64}2+3 = 35 ,\qquad 4<35 \quad\textbf{TRUE}

So (1) and (2) hold. Step 4:

f(8) = 35 ,\qquad f(9) = \frac{10}2 = 5

f

drops from

35

5

, so it is not increasing, and (3) is FALSE. (The two branches grow at very different rates, quadratic on the evens and linear on the odds, so the values interleave.) Step 5:

f(10) = \frac{100}2+3 = 53 ,\qquad f(11) = \frac{12}2 = 6 ,\qquad 53+6 = 59 . \quad\textbf{TRUE}

Answer: (1), (2), (4)

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