The number of value of `k` for which `[x^2-(k-2)x+k^2]xx""[x^2+k x+(2k-1)]` is a perfect square is `2` b. `1` c. `0` d. none of these

Options

A. 2
B. 1
C. 0
D. none of these

Answer

2

Explanation

For given situation, `x^(2) - (k - 2)x + k^(2) = 0 and x^(2) + kx + 2k - 1 = 0`
should have both roots common or each should have equal roots. If both roots are common, then
`1/1 = (-(k - 1))/(k) = (k^(2))/(2k - 1)`
`rArr k = -k + 2 and 2k - 1 = k^(2) rArr k = 1`
If both the equations have equal roots, then
`(k - 2)^(2) - 4k^(2) = 0 and k^(2) - 4(2k - 1) = 0`
`rArr (3k - 2) (-k - 2) = 0 and k^(2) - 8k + 4 = 0`
There is no common value of k.
Therefore, k = 1 is the only possible value.