Question

Column I		Column II
A	The function $f(x)=\frac{3 x-2}{x+4}$ has an inverse that can be written in the form $f^{-1}(x)=\frac{x+b}{c x+d}$. The value of $(b+c+d)$ is	P	even
B	Let $f(x)=\frac{x}{1+x}$ and let $g(x)=\frac{r x}{1-x}$.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 $f(x)=2 x+1$ 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 $f ( x )=2 x +1$
$f ( f ( x ))=2(2 x +1)+1=4 x +3, f ( f ( f ( x )))=8 x +7 ; f ( f ( f ( f ( x ))))=16 x +15$
Now L.H.S $=30 x +26$
So $30 x +26=116 \Rightarrow 30 x =90$
Hence $x=3$.