Question

The number of integral values of $k$ such that, the inclination of the tangent at any point on the curve $y=x^3-k{x}^2+x+1$ lies in $(0,\frac{\pi }{2}),$ is equal to

Answer 1

$y=x^3-k{x}^2+x+1\frac{dy}{dx}=3x^2-2kx+1$
$\because$ the inclination of the tangent lies in $(0,\frac{\pi }{2}),$
its slope is positive.
$\Rightarrow \frac{dy}{dx}>0,\forall x\in R$
$\Rightarrow 3x^2-2kx+1>0,\forall x\in R$
$\Rightarrow (-2k{)}^2-4(3\left)\right(1)<0$
$\{\because a>0,D<0\}$
$\Rightarrow k^2-3<0$
$\Rightarrow k\in (-\sqrt{3},\sqrt{3})$
So, the number of integral values of $k$ is $3$
$(k=-1,0,1)$