IPMAT Indore 2024 · Quantitative Ability · Question 8

Step 1 — simplify f. Split on the sign of x:

f\left( x \right)=|x+\left| x \right||=\left\{\begin{matrix}\left| 2x \right|=2x, & x\ge 0 \\ \left| x-x \right|=0, & x<0\end{matrix}\right.

So f\left( x \right)=2x for x\ge 0 and 0 otherwise.

Step 2 — use the given equation. Write t=f\left( a \right). Then g\left( f\left( a \right) \right)=1/t, so

t+\frac{1}{t}=\frac{13}{6}\Rightarrow 6{t}^{2}-13t+6=0 \Rightarrow \left( 2t-3 \right)\left( 3t-2 \right)=0

t=\frac{3}{2}\text{or}t=\frac{2}{3}

Note t\ne 0, so a>0 and t=f\left( a \right)=2a.

Step 3 — recover a.

2a=\frac{3}{2}\Rightarrow a=\frac{3}{4}, 2a=\frac{2}{3}\Rightarrow a=\frac{1}{3}

Step 4 — evaluate f\left( g\left( a \right) \right)=f\left( 1/a \right). Both values of a are positive, so 1/a>0 and f\left( 1/a \right)=2/a:

a=\frac{3}{4}\Rightarrow \frac{2}{3/4}=\frac{8}{3}, a=\frac{1}{3}\Rightarrow \frac{2}{1/3}=6

The maximum is 6.

Answer: 6

Shortcut. Two compressions. First, |x+\left| x \right|| is just 2{x}^{+} — twice the positive part. Second, t+1/t=13/6 is the classic reciprocal pair 3/2,2/3 (spot it from 13=9+4 over 6). Then f\left( g\left( a \right) \right)=2/a=4/t, which is largest when t is smallest: 4\div 2/3=6. No need to compute both branches.