IPMAT Indore 2022 · Quantitative Ability · Question 17

Step 1 — isolate the term containing f\left( x \right).

{81}^{\,f\left( x \right)}=3-{81}^{\,x}

Step 2 — find the range of the right-hand side. For real x, {81}^{x}>0 and takes every positive value. So

3-{81}^{x}\in \left( -\infty , 3 \right)

Step 3 — impose positivity. The left side {81}^{f\left( x \right)} is an exponential, so it must be strictly positive:

0<3-{81}^{x}<3

Step 4 — solve for f\left( x \right). Taking {\log}_{81} across 0<{81}^{f\left( x \right)}<3:

f\left( x \right)<{\log}_{81}3

and {\log}_{81}3=\frac{\log 3}{\log 81}=\frac{\log 3}{4\log 3}=\frac{1}{4}=0.25.

There is no lower bound: as {81}^{x}\to{3}^{-}, the right side tends to 0 and f\left( x \right)\to -\infty.

f\left( x \right)\in \left( -\infty , 0.25 \right)

Answer: Option 3 — \left( -\infty , 0.25 \right)

Shortcut. The whole question is one observation: an exponential is strictly positive and here strictly below 3.

0<{81}^{f\left( x \right)}<3 \Rightarrow -\infty <f\left( x \right)<{\log}_{81}3=1/4

The number to spot is {\log}_{81}3=1/4, because 81={3}^{4}. Seeing 0.25 in two of the options tells you that is the bound being tested.

Then pick the interval that is open at the top and unbounded below — only option (c). Option (a) wrongly imposes a lower bound; the exponential can be arbitrarily close to 0, so f\left( x \right) runs to -\infty.