JEE Mathematics 2023 Question with Solution

Given: $f(x) = ax - 3$ and $g(x) = x^b + c$, where $a, b, c \in \mathbb{N}$.

Find: $(f \circ g)(ac) + (g \circ f)(b)$.

From the given inverse,

$$(f \circ g)^{-1}(x) = \left(\frac{x - 7}{2}\right)^{1/3}$$

so the function itself is

$$(f \circ g)(x) = 2x^3 + 7.$$

Now,

$$(f \circ g)(x) = f(g(x)) = a(x^b + c) - 3.$$

Comparing with

$$2x^3 + 7,$$

we get

$$a = 2, \quad b = 3, \quad c = 5.$$

Therefore,

$$ac = 2 \cdot 5 = 10.$$

Now,

$$(f \circ g)(ac) = (f \circ g)(10) = 2(10)^3 + 7 = 2007.$$

Also,

$$(g \circ f)(x) = g(f(x)) = (2x - 3)^3 + 5.$$

Hence,

$$(g \circ f)(b) = (g \circ f)(3) = (2 \cdot 3 - 3)^3 + 5 = 3^3 + 5 = 32.$$

So,

$$(f \circ g)(ac) + (g \circ f)(b) = 2007 + 32 = 2039.$$

Therefore, the required value is $2039$.

Given: $(f \circ g)^{-1}(x) = \left(\frac{x - 7}{2}\right)^{1/3}$.

Find: the value of $(f \circ g)(ac) + (g \circ f)(b)$.

If

$$y = (f \circ g)(x),$$

then from the inverse relation,

$$x = \left(\frac{y - 7}{2}\right)^{1/3}.$$

Cubing both sides,

$$x^3 = \frac{y - 7}{2}.$$

So,

$$y - 7 = 2x^3$$

and hence

$$y = 2x^3 + 7.$$

Thus,

$$(f \circ g)(x) = 2x^3 + 7.$$

But also,

$$(f \circ g)(x) = f(g(x)) = a(x^b + c) - 3 = ax^b + ac - 3.$$

Matching powers and constants with

$$2x^3 + 7,$$

we obtain

$$a = 2, \quad b = 3,$$

and

$$ac - 3 = 7 \Rightarrow ac = 10.$$

Since $a = 2$, this gives

$$c = 5.$$

Now evaluate the two terms.

First,

$$(f \circ g)(ac) = (f \circ g)(10) = 2(10)^3 + 7 = 2007.$$

Next,

$$(g \circ f)(x) = g(ax - 3) = (ax - 3)^b + c = (2x - 3)^3 + 5.$$

So,

$$(g \circ f)(b) = (g \circ f)(3) = (6 - 3)^3 + 5 = 27 + 5 = 32.$$

Therefore,

$$(f \circ g)(ac) + (g \circ f)(b) = 2007 + 32 = 2039.$$

The required answer is $2039$.