Question

If the tangent at each point of the curve $y=\frac{2}{3}{x}^3-2a{x}^2+2x+5$ makes an acute angle with the positive direction of x-axis, then

• A. $a\ge 1$
• B. $-1\le a\le 1$
• C. $a\le -1$
• D. none of these

Answer 1

The correct option is B $-1\le a\le 1$

We have, $y=\frac{2}{3}{x}^3-2a{x}^2+2x+5$

$\Rightarrow \frac{dy}{dx}=2x^2-4ax+2.$

Since, the tangent makes an acute angle with the positive direction of x-axis, therefore,

$\frac{dy}{dx}\ge 0\Rightarrow 2x^2-4ax+2\ge 0$ for all $x$

$\Rightarrow 16a^2-16\le 0$

$(\because$ Disc. $={\left(4a\right)}^2-4\left(2\right)\left(2\right)\le 0)$

$\Rightarrow a^2-1\le 0$ i.e., $(a-1)(a+1)\le 0$

$\Rightarrow -1\le a\le 1.$