JEE Mathematics 2023 Question with Solution

Given:

$$f^1(x) = \frac{3x + 2}{2x + 3}$$

and for $n \geq 2$,

$$f^n(x) = f^1 \circ f^{n-1}(x)$$

Find: $a + b$ if

$$f^5(x) = \frac{ax + b}{bx + a}$$

with $\gcd(a,b)=1$.

Using repeated composition, the iterates are:

$$f^2(x) = f^1(f^1(x)) = \frac{13x + 12}{12x + 13}$$ $$f^3(x) = f^1(f^2(x)) = \frac{63x + 62}{62x + 63}$$

Continuing this pattern,

$$f^5(x) = \frac{1563x + 1562}{1562x + 1563}$$

Comparing with

$$f^5(x) = \frac{ax + b}{bx + a}$$

we get

$$a = 1563, \quad b = 1562$$

Therefore,

$$a+b = 1563+1562 = 3125$$

So, the required numerical value is $3125$.

Given:

$$f^1(x) = \frac{3x + 2}{2x + 3}$$

Find: the value of $a+b$.

The solution shows the coefficient pattern after successive iterations:

• $f^1(x) = \frac{3x+2}{2x+3}$
• $f^2(x) = \frac{13x+12}{12x+13}$
• $f^3(x) = \frac{63x+62}{62x+63}$
• $f^5(x) = \frac{1563x+1562}{1562x+1563}$

Thus the coefficients remain symmetric in the form

$$\frac{ax+b}{bx+a}$$

For the fifth iterate, we identify

$$a = 1563, \quad b = 1562$$

Hence,

$$a+b=3125$$

Therefore, the answer is $3125$.