If the range of function ` f(x) = (x + 1)/(k+x^(2))` contains the interval
[-0,1] , then values of k can be equal to

Options

A. 0
B. 0.5
C. 1.25
D. 1.5

Answer

1,2,3

Explanation

` y in [0.1]`
where ` y = (x +1)/(k+x^(2))`
y = 0 when x = 0- 1
If ` y ne 0 , yx^(2) - x + ky - 1 = 0, x in `R
`rArr D ge 0 `
`rArr 1-4y (ky -1) ge 0 `
` rArr 4ky^(2) - 4y - 1 le 0 `
`AA y in [0,1]` (Given)
` f(0) le 0 and f(1) le 0 `
`rArr -1 lt 0 and 4k - 4 - 1 lt 0 `
` rArr k lt (5)/(4) `
` therefore k in (-infty, (5)/(4)]`
Also, ` k gt 0 ` , otherwise graph of ` 4ky^(2) - 4y - 1` is concave
downward and is negative for unbounded interval values of y.
Hence , ` k in [ 0, (5)/(4)]`