If (a, 0) is a point on a diameter of the circle `x^(2)+y^(2)=4`, then the equation `x^(2)-4x-a^(2)=0` has

Options

A. exactly one root in [-1, 0]
B. exactly one root in [2, 5]
C. distinct roots greater than -1 and less than 5
D. all of these

Answer

D

Explanation

Since (a, 0) is a point on the diameter along x-axis.
`:. -2 le a le 2 rArr 0 le a^(2)le 4`.
Let `f(x)=a^(2)-4x-a^(2)`. Clearly, discriminant of f(x)=0 is positive . So, it has real and distinct roots.
Now, `f(0) f(-1)=(-a^(2))(5-a^(2))lt 0 " " [:' 0 le a^(2) le 4]`
`rArr f(x)=0` has a root in (-1, 0]
Also, `f(2)f(5)=(-4-a^(2))(5-a^(2)) lt 0`
`rArr f(x)=0` has a root in [2, 5]
Hence, all the options are correct.