If y = ax^3 + 3x^2 + (2a + 1)x + 1000 is a st | vedclass

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(B) Given the function $y = ax^3 + 3x^2 + (2a + 1)x + 1000$. For $y$ to be a strictly increasing function for all $x \in \mathbb{R}$,its derivative $y

Vedclass · Accepted Answer

$a > 1$

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(B) Given the function $y = ax^3 + 3x^2 + (2a + 1)x + 1000$. For $y$ to be a strictly increasing function for all $x \in \mathbb{R}$,its derivative $y'$ must be greater than $0$ for all $x$. $y' = 3ax^2 + 6x + (2a + 1) > 0$. For a quadratic expression $Ax^2 + Bx + C > 0$ to hold for all $x$,we must have $A > 0$ and the discriminant $D Here,$A = 3a$,$B = 6$,and $C = 2a + 1$. Condition $1$: $3a > 0 \implies a > 0$. Condition $2$: $D = B^2 - 4AC $6^2 - 4(3a)(2a + 1) $36 - 12a(2a + 1) Divide by $12$: $3 - a(2a + 1) $3 - 2a^2 - a $2a^2 + a - 3 > 0$. Factoring the quadratic: $(2a + 3)(a - 1) > 0$. This inequality holds when $a > 1$ or $a Combining with Condition $1$ $(a > 0)$,we get $a > 1$.

If $y = ax^3 + 3x^2 + (2a + 1)x + 1000$ is a strictly increasing function for all values of $x$,then:

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