JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f(x)\) be a quadratic polynomial such that \(f(-2)\) \(+f(3)=0\). If one of the roots of \(f(x)=0\) is \(-1\), then the sum of the roots of \(f(x)=0\) is equal to
- A \(\frac{11}{3}\)
- B \(\frac{7}{3}\)
- C \(\frac{13}{3}\)
- D \(\frac{14}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{11}{3}\)
Step-by-step Solution
Detailed explanation
\(f (-2)+ f (3)=0\) \(f ( x )=( x +1)( ax + b )\) \(f (-2)+ f (3)=-1(-2 a + b )+4(3 a + b )=0\) \(2 a - b +12 a +4 b =0\) \(14 a +3 b =0\) \(\frac{- b }{ a }=\frac{14}{3}\) Sum of roots \(=\left(-1+\frac{-b}{a}\right)=-1+\frac{14}{3}=\frac{11}{3}\)
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