If exactly one root of the equation x^2 + (a | vedclass

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(B) Let $f(x) = x^2 + (a - 1)x + 2a$. For exactly one root to lie in $(0, 3)$,we check two cases:Case $1$: $f(0) \cdot f(3) < 0$$f(0) = 2a$$f(3) = 9 +

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$(-\infty, 0] \cup (6, \infty)$

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(B) Let $f(x) = x^2 + (a - 1)x + 2a$. For exactly one root to lie in $(0, 3)$,we check two cases:Case $1$: $f(0) \cdot f(3) $f(0) = 2a$$f(3) = 9 + 3(a - 1) + 2a = 9 + 3a - 3 + 2a = 5a + 6$$2a(5a + 6) Case $2$: One root is at the boundary.If $x = 0$ is a root,$f(0) = 0$ $\Rightarrow 2a = 0$ $\Rightarrow a = 0$. The equation becomes $x^2 - x = 0$,roots are $0, 1$. Since $1 \in (0, 3)$,$a = 0$ is included.If $x = 3$ is a root,$f(3) = 0$ $\Rightarrow 5a + 6 = 0$ $\Rightarrow a = -1.2$. The equation becomes $x^2 - 2.2x - 2.4 = 0$,roots are $3, -0.8$. Since $3$ is not in $(0, 3)$,$a = -1.2$ is excluded.Combining these,the set of values is $a \in (-1.2, 0]$.Note: The provided options appear to have a typo in the equation coefficients. Based on the standard interpretation of such problems,the correct range is $(-1.2, 0]$. Given the options provided,if we assume the equation was $x^2 - (a+1)x + 2a = 0$,the range would be $(-\infty, 0] \cup (6, \infty)$.

If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

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Statement-$I$: If the roots $\alpha, \beta$ of the equation $x^2 + 2(a - 3)x + 9 = 0$,$a \in R$ satisfy $\alpha < 6 < \beta$,then $a < -3/4$.
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