JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\mathrm{f}(\mathrm{x})\) be a quadratic polynomial such that \(\mathrm{f}(-1)+\mathrm{f}(2)=0 .\) If one of the roots of \(\mathrm{f}(\mathrm{x})=0\) is \(3,\) then its other root lies in
- A \((-3,-1)\)
- B \((1,3)\)
- C \((-1,0)\)
- D \((0,1)\)
Answer & Solution
Correct Answer
(C) \((-1,0)\)
Step-by-step Solution
Detailed explanation
\(f(x)=a(x-3)(x-\alpha)\) \(f(2)=a(\alpha-2)\) \(f(-1)=4 a(1+\alpha)\) \(f(-1)+f(2)=0 \Rightarrow a(\alpha-2+4+4 \alpha)=0\) \(a \neq 0 \Rightarrow 5 \alpha=-2\) \(\alpha=-\frac{2}{5}=-0.4\) \(\alpha \in(-1,0)\)
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