Question

Column I		Column II
A	The function $f(x)=\frac{3x-2}{x+4}$ has an inverse that can be written in the form $f^{-1}(x)=\frac{x+b}{cx+d}$. The value of $(b+c+d)$ is	P	even
B	Let $f(x)=\frac{x}{1+x}$ and let $g(x)=\frac{rx}{1-x}$.Let $S$ be the set of all real numbers $r$ such that $f(g(x))=g(f(x))$ for infinitely many real number $x$. The number of elements in set $S$ is	Q	odd
C	If $f(x)=2x+1$ then the value of $x$ satisfying the equation $f(x)+f(f(x))+f(f(f(x)))+f(f(f(f(x))))=116$, is	R	prime
		S	nor compositeneither prime

• A. (A) Q, S; (B) P, R; (C) Q, P
• B. (A) Q, S; (B) P, R; (C) Q, Q
• C. (A) Q, S; (B) P, R; (C) Q, R
• D. (A) Q, S; (B) P, R; (C) Q, S

Answer 1

Answer: (C)

(A) $f^{-1}(x)=\frac{x+\frac{1}{2}}{-\frac{1}{4}x+\frac{3}{4}}\Rightarrow 1$
(B) $f(g(x))=\frac{rx}{1+(r-1)x},g(f(x))=rx$. If $f(g(x))=g(f(x))$
$\Rightarrow \frac{rx}{1+(r-1)x}=rx\Rightarrow rx\left[1-\frac{1}{1+(r-1)x}\right]=0$
If this is to be true for infinitely many (all) $x$, then $r=0$ or $r-1=0\Rightarrow 2$
(C) Given $f(x)=2x+1$
$f(f(x))=2(2x+1)+1=4x+3,f(f(f(x)))=8x+7;f(f(f(f(x))))=16x+15$
Now L.H.S $=30x+26$
So $30x+26=116\Rightarrow 30x=90$
Hence $x=3$.