JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f(x)=a x^{2}+b x+c\) 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\(...\)
- A \(-4\)
- B \(\frac{13}{2}\)
- C \(\frac{23}{2}\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(f (0)+3+\lambda+4=14\) \(\therefore f (0)=7-\lambda= c\) \(f (1)= a + b + c =3\) \(f (3)=9 a +3 b + c =4 \quad\)...(ii) \(f (-2)=4 a -2 b + c =\lambda \quad\)...(iii) \((ii) - (iii)\) \(a+b=\frac{4-\lambda}{5}\) put in equation (i) \(\frac{4-\lambda}{5}+7-\lambda=3\)…
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