If both roots of the equation `x^2-2ax + a^2-1=0` lie between `(-2,2)` then a lies in the interval

Options

A. `R`
B. `(-1,1)`
C. `(-2,2)`
D. `(-3,-1)uu(1,3)`

Answer

B

Explanation

Let `f(x)=x^(2)-2ax+a^(2)-1`
Now four cases arise:
Case I `Dge0`

`implies(-2a)^(2)-4.1(a^(2)-1)ge0`
`implies 4ge0` ltbr.gt `:.a epsilonR`
Case II `f(-2)gt0`
`implies4+4a+a^(2)-1gt0`
`impliesa^(2)+4a+3gt0`
`implies(a+1)(a+3)gt0`
`:.a epsilon(-oo,-3)uu(-1,oo)`
Case III `f(2)gt0`
`implies4-4a+a^(2)-1gt0`
`impliesa^(2)-4a+3gt0`
`implies(a-1)(a-3)gt0`
`:.a epsilon(-oo,1)uu(3,oo)`
Case IV `-2ltx` coordinate of vertex `lt2`
`implies-2lt2alt2`
`:.a epsilon(-1,1)`
Hence `[a]=-1,0`