1.Relation and Function
hard

माना द्विघात बहुपद $f ( x )$ इस प्रकार है कि $f (-2)+ f (3)=0$ है। यदि $f ( x )=0$ का एक मूल $-1$ है, तो $f ( x )=0$ के मूलों का योगफल है :

A

$\frac{11}{3}$

B

$\frac{7}{3}$

C

$\frac{13}{3}$

D

$\frac{14}{3}$

(JEE MAIN-2022)

Solution

$f (-2)+ f (3)=0$

$f ( x )=( x +1)( ax + b )$

$f (-2)+ f (3)=-1(-2 a + b )+4(3 a + b )=0$

$2 a – b +12 a +4 b =0$

$14 a +3 b =0$

$\frac{- b }{ a }=\frac{14}{3}$

Sum of roots $=\left(-1+\frac{-b}{a}\right)=-1+\frac{14}{3}=\frac{11}{3}$

Standard 12
Mathematics

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