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

Answer

A

Explanation

The given equation will be a perfect square, if either the equations
`x^(2) - (a-2) x+a^(2) = 0 and x^(2) + ax + (2a -1) = 0`
have both roots common or each has equal roots.
Thus, we have the following case:
CASE I When two equations have both roots common In this case, we have
`(1)/(1)=(-(a-2))/(a)=(a^(2))/(2a-1)`
`rArr" "a = - a+2 and 2a - 1 = a^(2)`
`rArr" "a=1 and a^(2) - 2a + 1 = 0 rArr a = 1`.
CASE II When two equations have equal roots
`(a-2)^(2) - 4a^(2) = 0 and a^(2) - 4(2a - 1) = 0`
`rArr" "-3a^(2) - 4a + 4 = 0 and a^(2) - 8a + 4 = 0`
`rArr" "3a^(2) + 4a - 4 = 0 and a^(2) - 8a + 4 = 0`
`rArr" "(3a - 2) (a+2)=0 and a^(2) - 8a + 4 = 0`
There is no value of a satisfying these two equations simultaneouslly.
Hence, for a = 1, the given expression is a perfect square.