IPMAT Indore 2026 · Quantitative Ability · Question 5

Step 1 — recognise the Vandermonde determinant. For this standard matrix,

\det M=\left( b-a \right)\left( c-a \right)\left( c-b \right)

Step 2 — read off what \det M=0 means. A product is zero only if a factor is zero, so two of a, b, c must be equal. (They cannot all be zero, since abc\ne 0.)

Step 3 — try each case against a+b+c=0.

Case a=b: then 2a+c=0\Rightarrow c=-2a.

\frac{{a}^{2}+{b}^{2}}{{c}^{2}}=\frac{2{a}^{2}}{4{a}^{2}}=\frac{1}{2}

Case b=c: then a=-2b, and c=b.

\frac{{a}^{2}+{b}^{2}}{{c}^{2}}=\frac{4{b}^{2}+{b}^{2}}{{b}^{2}}=5

Case a=c: then b=-2a, and c=a.

\frac{{a}^{2}+{b}^{2}}{{c}^{2}}=\frac{{a}^{2}+4{a}^{2}}{{a}^{2}}=5

Step 4 — take the maximum.

\max =5

Answer: 5

Shortcut. Two facts do all the work: the Vandermonde form makes \det M=0 mean “two variables are equal”, and the denominator is {c}^{2}, so you want c to be the small one. That is the case where c equals one of the others and the third is -2c, giving \left( 4+1 \right)/1=5 immediately.