IPMAT Indore 2025 · Quantitative Ability · Question 25

Step 1 — expand the determinant.

P\left( 0 \right)\,P\left( 2 \right)-P\left( 1 \right)\,P\left( 0 \right)=0 \Rightarrow P\left( 0 \right)\left[ P\left( 2 \right)-P\left( 1 \right) \right]=0

Since P\left( 0 \right)=2\ne 0, we must have

P\left( 2 \right)=P\left( 1 \right)

Step 2 — write the polynomial. Let P\left( x \right)=a{x}^{2}+bx+c. From P\left( 0 \right)=2 we get c=2.

P\left( 1 \right)=a+b+2, P\left( 2 \right)=4a+2b+2

Setting them equal:

4a+2b=a+b \Rightarrow 3a+b=0 \Rightarrow b=-3a

Step 3 — use the sum condition. With P\left( 3 \right)=9a+3b+2:

P\left( 1 \right)+P\left( 2 \right)+P\left( 3 \right)=\left( a+b+2 \right)+\left( 4a+2b+2 \right)+\left( 9a+3b+2 \right)=14a+6b+6=14

So 14a+6b=8. Substituting b=-3a:

14a-18a=8 \Rightarrow -4a=8 \Rightarrow a=-2, b=6

Step 4 — evaluate. P\left( x \right)=-2{x}^{2}+6x+2, so

P\left( 4 \right)=-32+24+2=-6

Answer: Option 3 — -6