The value of k for which the equation x^2 - 3 | vedclass

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Question

(A) Let $f(x) = x^2 - 3x + k$. For the equation $f(x) = 0$ to have at least one root in the interval $[0, 1]$,we consider the following conditions:
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Vedclass · Accepted Answer

$0 \le k \le 2$

Vedclass · Answer

(A) Let $f(x) = x^2 - 3x + k$. For the equation $f(x) = 0$ to have at least one root in the interval $[0, 1]$,we consider the following conditions:
$1$. The product of the values at the endpoints must be less than or equal to zero: $f(0) \cdot f(1) \le 0$.
$f(0) = 0^2 - 3(0) + k = k$.
$f(1) = 1^2 - 3(1) + k = k - 2$.
So,$k(k - 2) \le 0$,which implies $0 \le k \le 2$.
$2$. The vertex of the parabola $x = -b/(2a) = 3/2$ does not lie in the interval $[0, 1]$.
$3$. Since the parabola opens upward,if the vertex is outside the interval,the only way to have a root in $[0, 1]$ is if the function changes sign at the endpoints.
Therefore,the required range for $k$ is $0 \le k \le 2$.

The value of $k$ for which the equation $x^2 - 3x + k = 0$ has at least one real root in $[0, 1]$ is

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