Question

$\text{If}{\sin }^2\theta =x^2-4x+5,\text{then}x=$

• A. $0,1$
• B. $2$
• C. $-2,2$

Answer 1

The correct option is B $2$

Given: ${\sin }^2\theta =x^2-4x+5$
Now, $x^2-4x+5$ has complex roots as it's discriminant $D=16-20=-4<0$
Also, coefficient of $x^2$ is positive.
$\Rightarrow x^2-4x+5$ is upward opening curve.
Now, it's minimum point is given by the relation:
$(\frac{-b}{2a},\frac{-D}{4a})$
Equating the values of $a,b,c,$ we get the coordinates of minimum point as: $(\frac{-(-4)}{2},\frac{-(-4)}{4})\equiv (2,1)$
Hence, at $x=2,$ the quadratic equation is minimum and have a value $1.$
Also, we know $-1\le \sin \theta \le 1$
Or $0\le {\sin }^2\theta \le 1$
Thus, for ${\sin }^2\theta =x^2-4x+5$
This is possible only when $x^2-4x+5=1$ or at $x=2.$