Question

Which of the following quadratic equations does not have both of its roots lying in the range of $y=3\sin x$?

• A. $x^2-4=0$
• B. $x^2+x-2=0$
• C. $3x^2-x=0$
• D. $x^2-2x-8=0$

Answer 1

The correct option is D $x^2-2x-8=0$

We know
$-1\le \sin x\le 1$
$\Rightarrow -3\le 3\sin x\le 3$
Range of $y=3\sin x$
$y\in [-3,3]$

Finding the roots of the quadratic equations,
$x^2-4=0$
$\Rightarrow x=\pm 2\in [-3,3]$

$x^2+x-2=0\Rightarrow (x+2)(x-1)=0$
$\Rightarrow x=1,-2\in [-3,3]$

$3x^2-x=0$
$\Rightarrow x=0,\frac{1}{3}\in [-3,3]$

$x^2-2x-8=0\Rightarrow (x-4)(x+2)=0$
$\Rightarrow x=-2,4$
$\Rightarrow x=4\notin [-3,3]$

Therefore, the roots of equation $x^2-2x-8=0$ does not lie between $[-3,3]$.