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 the interval $(0,1)$, the product of the values of the function a

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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 the interval $(0,1)$, the product of the values of the function at the endpoints must be negative, i.e., $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, $2 \cdot (\lambda-1)(\lambda-3) This implies $1 Now, we check the boundary cases: Case $1$: If $\lambda = 1$, the equation becomes $2x^{2}-4x+2 = 0$, which simplifies to $2(x-1)^{2} = 0$. The roots are $x=1, 1$. Since neither root lies strictly inside $(0,1)$, $\lambda = 1$ is not included. Case $2$: If $\lambda = 3$, the equation becomes $10x^{2}-12x+2 = 0$, which simplifies to $2(5x^{2}-6x+1) = 0$, or $2(5x-1)(x-1) = 0$. The roots are $x = 1/5$ and $x = 1$. Since $1/5$ lies in $(0,1)$, $\lambda = 3$ is included. Thus, 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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