IPMAT Indore 2026 · Quantitative Ability · Question 23

Step 1 — rewrite as an intersection problem.

{2}^{x}={x}^{2}

so we are asking how many times the curve y={2}^{x} meets y={x}^{2}.

Step 2 — find the obvious solutions.

x=2:{2}^{2}=4={2}^{2}✓ x=4:{2}^{4}=16={4}^{2}✓

Step 3 — do not stop there — check negative x. For x<0:

{2}^{x} is small and positive, falling towards 0

{x}^{2} is positive and rising as x becomes more negative

At x=0: {2}^{0}=1>0={x}^{2}. At x=-1: {2}^{-1}=0.5<1={x}^{2}.

The difference changes sign between -1 and 0, so there is a third root there (near x\approx -0.767).

Step 4 — confirm there are no more. For x>4 the exponential grows faster and stays above {x}^{2} forever, and between the roots the curves cross cleanly.

\text{exactly three real solutions}

Answer: Option 2 — exactly three real solutions

Shortcut. Everyone finds x=2 and x=4 and picks “two”. The mark is in remembering that {x}^{2} is defined for negative x while {2}^{x} stays positive — guaranteeing one crossing on the left. Sketch the two curves; never solve this one algebraically.