Q57 / 110 · VITEEE 2022 · Mathematics

If one root of the equation \((l-\mathrm{m}) \mathrm{x}^2+l \mathrm{x}+1=0\) is double of the other and if \(l\) is real then the greatest value of \(\mathrm{m}\) is \((l \neq \mathrm{m})\) :

  • A
    \(\frac{1}{3}\)
  • B
    \(\frac{8}{9}\)
  • Correct answer
    C
    \(\frac{9}{8}\)
  • D
    3

Solution

Let the roots of the equation be \(\alpha\) and \(2 \alpha\), then \(\alpha+2 \alpha=\frac{l}{\mathrm{~m}-l} \text { and } \alpha \cdot 2 \alpha=\frac{1}{l-\mathrm{m}}\) \(\Rightarrow \alpha=\frac{l}{3(\mathrm{~m}-l)} \text { and } \alpha^2=\frac{1}{2(l-\mathrm{m})} \text {. }\) Eliminating \(\alpha\), we get \(\frac{l^2}{9(\mathrm{~m}-l)^2}=\frac{1}{2(l-\mathrm{m})} \Rightarrow 2 l^2-9 l+9 \mathrm{~m}=0\) Since \(l\) is real, so \((-9)^2-4 \cdot 2 \cdot 9 \mathrm{~m} \geq 0\) \(\Rightarrow \mathrm{m} \leq \frac{9}{8}\)

Hence, the correct answer is option: 'C'.

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