(D) Given the quadratic equation $f(x) = x^{2}+2ax+(3a+10) = 0$. Since the roots $\alpha$ and $\beta$ satisfy $\alpha < 1 < \beta$, the value of the f

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(D) Given the quadratic equation $f(x) = x^{2}+2ax+(3a+10) = 0$. Since the roots $\alpha$ and $\beta$ satisfy $\alpha Substituting $x=1$ into the equation: $f(1) = (1)^{2} + 2a(1) + (3a+10) $1 + 2a + 3a + 10 $5a + 11 $5a $a Therefore, the set of all possible values of $a$ is $(-\infty, -11/5)$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Quadratic expressions and Position of roots

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Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+2ax+(3a+10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:

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A
$(-\infty, -11/5) \cup (5, \infty)$
B
$(-\infty, -2) \cup (5, \infty)$
C
$(-\infty, -3)$
D
$(-\infty, -11/5)$

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4-2.Quadratic Equations and Inequations

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Quadratic expressions and Position of roots

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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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Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+2ax+(3a+10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:

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If the roots of the equation $x^2 - 2ax + a^2 + a - 3 = 0$ are real and less than $3$,then

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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 $a < -3/4$.Statement-$II$: If $f(x) = x^2 + 2(a - 3)x + 9$,then $f(6) < 0 \implies a < -3/4$.

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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 $a < -3/4$.
Statement-$II$: If $f(x) = x^2 + 2(a - 3)x + 9$,then $f(6) < 0 \implies a < -3/4$.