The set of all real values of lambda for whic | vedclass

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(B) Let $f(x) = (\lambda^{2}+1)x^{2}-4\lambda x+2$.For exactly one root to lie in $(0,1)$,we consider the condition $f(0) \cdot f(1) < 0$.$f(0) = 2$$f

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$(1,3]$

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(B) Let $f(x) = (\lambda^{2}+1)x^{2}-4\lambda x+2$.For exactly one root to lie in $(0,1)$,we consider the condition $f(0) \cdot f(1) $f(0) = 2$$f(1) = \lambda^{2}+1-4\lambda+2 = \lambda^{2}-4\lambda+3 = (\lambda-1)(\lambda-3)$So,$f(0) \cdot f(1) = 2(\lambda-1)(\lambda-3) This implies $1 Now,we check the endpoints:Case $1$: If $\lambda = 1$,the equation becomes $2x^{2}-4x+2 = 0$,which is $2(x-1)^{2} = 0$. The roots are $x=1, 1$. Neither root lies in $(0,1)$. So $\lambda 
eq 1$.Case $2$: If $\lambda = 3$,the equation becomes $10x^{2}-12x+2 = 0$,which is $2(5x^{2}-6x+1) = 0$,or $2(5x-1)(x-1) = 0$. The roots are $x = 1/5$ and $x = 1$. Since $1/5 \in (0,1)$,$\lambda = 3$ is a valid solution.Combining these,the set of values is $\lambda \in (1,3]$.

The set of all real values of $\lambda$ for which the quadratic equation $(\lambda^{2}+1)x^{2}-4\lambda x+2=0$ has exactly one root in the interval $(0,1)$ is

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