QJEEM24Mab4a8: JEE Main 2024 Mathematics - Class 11 JEEM24Mab4a8 Solution: If and , where , then is equal to

To find $(g \circ g \circ g)(4),$ we first need to understand the composition of $f$ with itself, i.e., $(f \circ f)(x) = f(f(x)) = g(x).$ We can then repeatedly apply $g$ to get the given expression.

First, let's calculate $(f \circ f)(x) = g(x):$

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

$= f\left(\frac{4x+3}{6x-4}\right)$

To evaluate this expression, we substitute $\frac{4x+3}{6x-4}$ for $x$ in the function $f(x):$

$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$

Now, we simplify the expression:

$g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)}$

$= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16}$

$= \frac{34x}{34}$

$= x$

So, $g(x) = x$ for all $x$ in the domain of $g$ , which is $\mathbb{R}-\left\{\frac{2}{3}\right\}$ . It's important to note that the domain restriction is preserved through the composition because $f(x)$ has a vertical asymptote at $x = \frac{2}{3}$ which doesn't intersect the graph.

So, $g(x)$ is the identity function on its domain, which means that applying $g$ any number of times will result in the same input for $x$ in the given domain. Hence, we have:

$(g \circ g \circ g \circ g)(4) = g(g(g(g(4)))) = g(g(g(4))) = g(g(4)) = g(4) = 4$

This corresponds to option D, which is $4$ .