JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q64

MathematicsSingle correct+4 / -1Question 64 of 90

Let $f(x)=ax^2+bx+c$ be such that $f(1)=3$ , $f(-2)=\lambda$ and $f(3)=4$ . If $f(0)+f(1)+f(-2)+f(3)=14$ , then $\lambda$ is equal to :

-4

\dfrac{13}2

\dfrac{23}2

Answer and worked solution

Official answer: D

Equations. $a+b+c=3$ , $9a+3b+c=4$ , $c+\lambda=7$ and $\lambda=4a-2b+c$ .

Solve. $a=\frac{9-2\lambda}6$ , and substituting gives $\lambda=4$ .

(D)

Topics:Quadratic expressions

JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 64