Let f(x) = frac{ax + b}{cx + d}. Then,f circ | vedclass

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(A) Given $f(x) = \frac{ax + b}{cx + d}$.We are given that $f \circ f(x) = x$.This implies $f(f(x)) = x$.Substituting $f(x)$ into the function:$f\left

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$d = -a$

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(A) Given $f(x) = \frac{ax + b}{cx + d}$.We are given that $f \circ f(x) = x$.This implies $f(f(x)) = x$.Substituting $f(x)$ into the function:$f\left(\frac{ax + b}{cx + d}ight) = x$$\frac{a\left(\frac{ax + b}{cx + d}ight) + b}{c\left(\frac{ax + b}{cx + d}ight) + d} = x$Multiplying numerator and denominator by $(cx + d)$:$\frac{a(ax + b) + b(cx + d)}{c(ax + b) + d(cx + d)} = x$$\frac{a^2x + ab + bcx + bd}{acx + bc + cdx + d^2} = x$$\frac{(a^2 + bc)x + (ab + bd)}{(ac + cd)x + (bc + d^2)} = x$$(a^2 + bc)x + (ab + bd) = x((ac + cd)x + (bc + d^2))$$(a^2 + bc)x + b(a + d) = (ac + cd)x^2 + (bc + d^2)x$For this to hold for all $x$,the coefficients must match. Comparing the constant term,we have $b(a + d) = 0$. If $b 
eq 0$,then $a + d = 0$,which means $d = -a$. If $d = -a$,the expression becomes $\frac{ax + b}{cx - a}$,and $f(f(x)) = \frac{a(\frac{ax+b}{cx-a}) + b}{c(\frac{ax+b}{cx-a}) - a} = \frac{a^2x + ab + bcx - ab}{acx + bc - acx + a^2} = \frac{x(a^2 + bc)}{a^2 + bc} = x$. Thus,$d = -a$ is the required condition.

Let $f(x) = \frac{ax + b}{cx + d}$. Then,$f \circ f(x) = x$ provided that

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