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Difference of Extreme Values of abc for Vertex (4,2) | JEE

JEE Maths question with a full step-by-step solution.

Question
If the parabola y=ax2+bx+cy = ax^2 + bx + c has vertex at (4,2)(4, 2) and a∈[1,3]a \in [1, 3], then the difference between the extreme values of abcabc is:
A36003600
B144144
C34563456correct
Dcannot be found
Solution
Step 1: The vertex gives −b2a=4⇒b=−8a-\dfrac{b}{2a} = 4 \Rightarrow b = -8a and 4ac−b24a=2⇒c=16a+2\dfrac{4ac - b^2}{4a} = 2 \Rightarrow c = 16a + 2. Step 2: Then abc=a(−8a)(16a+2)=−16(8a3+a2)abc = a(-8a)(16a + 2) = -16(8a^3 + a^2), which decreases on [1,3][1, 3], so the extremes are at the endpoints:
a=1: −144;a=3: −3600⇒difference=3456.a = 1:\ -144; \qquad a = 3:\ -3600 \Rightarrow \text{difference} = 3456.
 Correct answer: (3)
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