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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2016 · Mathematics · Quadratic Equation

If one root of the equation (1-m) x^2+1 x+1=0 is double the other...

If one root of the equation (1−m)x2+1x+1=0(1-m) x^2+1 x+1=0 is double the other and 1 is real, then what is the greatest value of mm ?

Options

  1. A−98-\frac{9}{8}
  2. B98\frac{9}{8}
  3. C−89-\frac{8}{9}
  4. D89\frac{8}{9}

Correct answer

B. 98\frac{9}{8}

Step-by-step solution

Given equation is (ℓ−m)x2+ℓx+1=0(\ell-m) x^2+\ell x+1=0 Roots are α,β\alpha, \beta. ∵\because One root is double the other. β=2α\beta=2 \alpha Sum of roots =α+β3α=−ℓℓ−mα(2α)=1(ℓ−m)⇒α2=ℓ29(ℓ−m)22α2=1ℓ−m=\alpha+\beta3 \alpha=\frac{-\ell}{\ell-m} \quad \alpha(2 \alpha)=\frac{1}{(\ell-m)}\Rightarrow \alpha^2=\frac{\ell^2}{9(\ell-m)^2} \quad 2 \alpha^2=\frac{1}{\ell-m}⇒2ℓ29(ℓ−m)2=1(ℓ−m)⇒2ℓ29(ℓ−m)=1⇒2ℓ2=9(ℓ−m)⇒2ℓ2−9ℓ+9m=0\begin{aligned} & \Rightarrow 2 \frac{\ell^2}{9(\ell-m)^2}=\frac{1}{(\ell-m)} \\ & \Rightarrow \frac{2 \ell^2}{9(\ell-m)}=1 \\ & \Rightarrow 2 \ell^2=9(\ell-m) \Rightarrow 2 \ell^2-9 \ell+9 m=0\end{aligned}$ForFor\elltoberealdiscriminantshouldbeto be real discriminant should beb^2-4 a c \geq 081-4 \times 2 \times 9 m \geq 0m \leq \frac{9}{8}$.
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