Question

The set of values of $k$ for which $x^2-kx+{\sin }^{-1}\left(\sin 4\right)>0$ for all real $x$ is

• A. $\left\{0\right\}$
• B. $(-2,2)$
• C. $R$
• D. $(-4,4)$

Answer 1

The correct option is D $(-4,4)$

Consider the given equation,

$x^2-kx+{\sin }^{-1}\left(\sin 4x\right)>0$

$\because {\sin }^{-1}\left(\sin 4x\right)=4$

$\therefore x^2-kx+4>0$

$\because$ Coefficient of $x^2$ = $1(>0)$

The given curve opens upwards.

If $f\left(x\right)>0$ then it should have no real roots i.e. should not cut the x-axis .

$\therefore D<0(\sqrt{D}=\sqrt{b^2-4ac}\forall a{x}^2+bx+c)$

$k^2-16<0$

$(k-4)(k+4)<0$

$k\in (-4,4)$

Hence, This is the answer.