The number of values of k for which the equat | vedclass

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(A) Let $f(x) = x^2 - 3x + k$. For the roots $\alpha, \beta$ to be real,distinct,and lie in the interval $(0, 1)$,the following conditions must be sat

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(A) Let $f(x) = x^2 - 3x + k$. For the roots $\alpha, \beta$ to be real,distinct,and lie in the interval $(0, 1)$,the following conditions must be satisfied:$1$. Discriminant $D > 0$: $D = (-3)^2 - 4(1)(k) = 9 - 4k > 0 \implies k $2$. $f(0) > 0$: $0^2 - 3(0) + k > 0 \implies k > 0$.$3$. $f(1) > 0$: $1^2 - 3(1) + k > 0 \implies 1 - 3 + k > 0 \implies k > 2$.$4$. The vertex of the parabola $x = -b/(2a)$ must lie in $(0, 1)$: $x = 3/2 = 1.5$. Since $1.5$ is not in the interval $(0, 1)$,it is impossible for both roots to lie in $(0, 1)$.Alternatively,for roots to lie in $(0, 1)$,the sum of roots $\alpha + \beta$ must satisfy $0 < \alpha + \beta < 2$. Here,$\alpha + \beta = 3$,which is not less than $2$. Therefore,no such $k$ exists.

The number of values of $k$ for which the equation $x^2 - 3x + k = 0$ has two real and distinct roots lying in the interval $(0, 1)$ is:

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