JEE Main 2021 (Online) 24th February Evening ShiftQuestion 1
Let $a,b,c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a,c)$, $(2,b)$ and $(a,b)$ be $\left(\frac{10}{3},\frac{7}{3}\right)$. If $\alpha,\beta$ are the roots of the equation $ax^{2}+bx+1=0$, then the value of $\alpha^{2}+\beta^{2}-\alpha\beta$ is :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. The centroid gives two equations, the A.P. a third; then use the sum and product of roots. Formulas used. (1) Centroid $=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$(2) $\alpha^2+\beta^2-\alpha\beta=(\alpha+\beta)^2-3\alpha\beta$Step 1 - Centroid. By (1): $\frac{2a+2}{3}=\frac{10}{3}$ gives $a=4$; $\frac{c+2b}{3}=\frac73$ gives $c+2b=7$. Step 2 - A.P. $2b=a+c=4+c$. Solving: $b=\frac{11}{4}$, $c=\frac32$. Step 3 - Roots. For $4x^2+\frac{11}{4}x+1=0$: $\alpha+\beta=-\frac{11}{16}$, $\alpha\beta=\frac14$. Step 4 - Value. By (2), $$\frac{121}{256}-\frac{192}{256}=-\frac{71}{256}.$$ Answer: (B)