If $(a,0)$ lies on the diameter of circle $x^2+y^2=4$ then $x^2-4x-a^2=0$ has

Question

A) Exactly one real root in $(-1,0]$

B) Exactly one real root in $[2,5]$

C) Distinct roots greater than 1

D) Distinct roots less than 5

$$x=\frac{4\pm \sqrt {16+4a^2}}{2}$$

$$x=2\pm \sqrt{4+a^2}$$

Also $a\in [-2,2]$

So $a^2\in [0,4]$

Then $x=4,0$ and $x=2(1\pm \sqrt 2)$

Which account for the extrema and minima

These 4 values satisfy A, C and D, but not B. However, the given answer is all 4. How is B correct?

Answer 1

As you point out $-2 \le a \le -2$ so.

$\sqrt{4 + 0} \le\sqrt{4 + a^2} \le \sqrt {4+4}$

So $2 \le 2+ \sqrt{4 + a^2}\le 2 + \sqrt 8$

So $2 \le 2 + \sqrt{4+a^2} \le 2 + \sqrt 8 < 2 + \sqrt 9 = 2+3 =5$.

So the root $2 + \sqrt{4+a^2} \in [2, 5)\subset [2,5]$.

Answer 2

Since $\sqrt {4+a^2}>0, \forall a\in[-2,2] $ so the equation always have 2 distinct roots. We shall look at the root $2+\sqrt{4+a^2}$.

We know that $\sqrt x$ is an increasing function .

Hence $0\le a^2 \le 4\implies 0< 4+a^2\le 8<9\implies 0<\sqrt {4+a^2}<3\implies 2<2+\sqrt{4+a^2}<5$