Let f(x)=x^3+ax^2+bx+5 ^2x be an increasing function on R. Then, … - Vidyadip

MCQ

Let $\text{f}(\text{x})=\text{x}^3+\text{a}\text{x}^2+\text{b}\text{x}+5\sin^2\text{x}$ be an increasing function on $R$. Then, $a$ and $b$ satisfy :

A
$a^2 - 3b - 15 > 0$
B
$a^2 - 3b + 15 > 0$
✓
$a^2 - 3b + 15 < 0$
D
$a < 0$ and $b > 0$

✓

Answer

Correct option: C.

$a^2 - 3b + 15 < 0$

$\text{f}(\text{x})=\text{x}^3+\text{a}\text{x}^2+\text{b}\text{x}+5\sin^2\text{x}$
$\text{f}'(\text{x})=3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x})$
Given, $f(x)$ is increasing on $R.$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\text{R}$
$\Rightarrow3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x})>0,\forall\ \text{x}\in\text{R}$
Since, the quadratic function is $> 0,$ its discriminant is $< 0.$
$\Rightarrow(2\text{a})^2-4(3)(\text{b}+5\sin2\text{x})<0$
$\Rightarrow4\text{a}^2-12\text{b}-60\sin2\text{x}<0$
$\Rightarrow\text{a}^2-3\text{b}-15\sin2\text{x}<0$
We know that the minimum value of $\sin2\text{x}$ is $-1.$
$\therefore\ \text{a}^2-3\text{b}-15<0$

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