## How many integer values ​​of the parameter (a in left( { – 10; + infty } right)) for the function (y = left| {{x^3} + left( {a + 2} right)x + 9 – {a^2}} right|) covariates on the interval (left( {0;1} right))? – Math book

How many integer values ​​of the parameter $$a \in \left( { – 10; + \infty } \right)$$ so that the function $$y = \left| {{x^3} + \left( {a + 2} \right)x + 9 – {a^2}} \right|$$ covariates over the interval $$\left( {0;1} \right)$$?

A. twelfth.

B. 11.

C. 6.

D. 5.

Select REMOVE

Consider $$f\left( x \right) = {x^3} + \left( {a + 2} \right)x + 9 – {a^2}$$

$$f’\left( x \right) = 3{x^2} + a + 2$$

To $$y = \left| {f\left( x \right)} \right|$$ covariates over the interval $$\left( {0;1} \right)$$

TH1:$$\left\{ \begin{array}{l}f’\left( x \right) \ge 0,\forall x \in \left( {0;1} \right)\\f\left( 0 \right) \ge 0\end{array} \right.$$

$$\Leftrightarrow \left\{ \begin{array}{l}3{x^2} + a + 2 \ge 0,\forall x \in \left( {0;1} \right)\\9 – {a^2} \ge 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a \ge \mathop {Max}\limits_{\left( {0;1} \ right)} \left( { – 3{x^2} – 2} \right)\\9 – {a^2} \ge 0\end{array} \right \Leftrightarrow \left\{ \begin{array }{l}a \ge – 2\\ – 3 \le a \le 3\end{array} \right \Rightarrow a \in \left[ { – 2;3} \right]$$

$$a = \left\{ { – 2; – 1;0;1;2;3;} \right\}$$ → 6 values

TH2:$$\left\{ \begin{array}{l}f’\left( x \right) \le ,\forall x \in \left( {0;1} \right)\\f\left( 0 \ right) \le 0\end{array} \right.$$

$$\Leftrightarrow \left\{ \begin{array}{l}3{x^2} + a + 2 \le 0,\forall x \in \left( {0;1} \right)\\9 – {a^2} \le 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a \le \mathop {Min}\limits_{\left( {0;1} \ right)} \left( { – 3{x^2} – 2} \right)\\9 – {a^2} \le 0\end{array} \right \Leftrightarrow \left\{ \begin{array }{l}a \le – 5\\\left[\begin{array}{l}a\ge3\\a\le–3\end{array}\right\end{array}\right\Rightarrowa\le–5$$[\begin{array}{l}a\ge3\\a\le –3\end{array}\right\end{array}\right\Rightarrowa\le –5\)

Combined with the problem condition $$a = \left\{ { – 9; – 8; – 7; – 6; – 5} \right\}$$ → 5 values

So there are 11 satisfying values.