MHTCET 16 Apr 2026 Shift 1 Paper

Concept:The composition $(f \circ f)(x)$ reduces to the identity function, so repeated composition leaves the input unchanged.Explanation:We are given $f(x)=\frac{4x+3}{6x-4}$, where $x \neq \frac{2}{3}$.First, find $g(x)=(f \circ f)(x)=f(f(x))$.Substitute $f(x)$ into itself:$$f(f(x))=\frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}$$Simplify the numerator:$$4\left(\frac{4x+3}{6x-4}\right)+3=\frac{16x+12+18x-12}{6x-4}=\frac{34x}{6x-4}$$Simplify the denominator:$$6\left(\frac{4x+3}{6x-4}\right)-4=\frac{24x+18-24x+16}{6x-4}=\frac{34}{6x-4}$$Hence,$$f(f(x))=\frac{\frac{34x}{6x-4}}{\frac{34}{6x-4}}=x$$So $g(x)=x$, which is the identity function.Since applying the identity function repeatedly does not change the input,$$(g \circ g \circ g \circ g \circ g)(3)=3$$Answer: $3$, which is Option A.