If both the roots of the quadratic equation x | vedclass

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(A) Let $f(x) = x^2 - mx + 4$. For the roots $\alpha, \beta$ to be real,distinct,and lie in $[1, 5]$,the following conditions must be satisfied:$(1)$

Vedclass · Accepted Answer

$(4, 5)$

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(A) Let $f(x) = x^2 - mx + 4$. For the roots $\alpha, \beta$ to be real,distinct,and lie in $[1, 5]$,the following conditions must be satisfied:$(1)$ Discriminant $D > 0$:$D = (-m)^2 - 4(1)(4) = m^2 - 16 > 0$$m^2 > 16 \Rightarrow m \in (-\infty, -4) \cup (4, \infty)$$(2)$ $f(1) > 0$ (since roots are distinct and lie in $[1, 5]$,$f(1)$ cannot be $0$ as $1$ is not a root):$f(1) = 1 - m + 4 = 5 - m > 0 \Rightarrow m $(3)$ $f(5) > 0$:$f(5) = 25 - 5m + 4 = 29 - 5m > 0 \Rightarrow m $(4)$ Vertex location $1 $1 Taking the intersection of all conditions:$m \in (4, \infty) \cap (-\infty, 5) \cap (-\infty, 5.8) \cap (2, 10) = (4, 5)$Thus,$m \in (4, 5)$. Option $A$ is correct.

If both the roots of the quadratic equation $x^2 - mx + 4 = 0$ are real and distinct and they lie in the interval $[1, 5]$,then $m$ lies in the interval.

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