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The set of all real values of \lambda for which the quadratic equations, \left ( \lambda ^{2}+1 \right )x^{2}-4\lambda x+2=0 always have exactly one root in the interval (0,1) is :
Option: 1 (0,2)
Option: 2 (2,4)
Option: 3 (1,3]
Option: 4 (-3,-1)

Answers (1)

best_answer

If exactly one root in (0, 1) then

\begin{array}{l} \Rightarrow \mathrm{f}(0) \cdot \mathrm{f}(1)<0 \\ \Rightarrow 2\left(\lambda^{2}-4 \lambda+3\right)<0 \\ \Rightarrow 1<\lambda<3 \end{array}

Now for \lambda=1

\begin{array}{l} 2 x^{2}-4 x+2=0 \\ (x-1)^{2}=0\\ x=1,1 \end{array}

So both roots doesn’t lie between (0, 1)

\therefore \lambda\neq1

\begin{aligned} &\text { Again for } \lambda=3\\ &\begin{aligned} & 10 x^{2}-12 x+2=0 \\ \Rightarrow & x=1, \frac{1}{5} \end{aligned} \end{aligned}

So if one root is 1 then second root lie between (0, 1) so \lambda=3

\therefore \lambda \in(1,3]

Posted by

himanshu.meshram

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