If the function `f(x) ={x+1` if `x le 1` , `2x+1` if `1 lt x le 2` and `g(x) = {x^2` if `-1 le x le 2`, `x+2 if 2 le x le 3` then the number of roots of the equation `f(g(x))=2`

Options

A. 1
B. 2
C. 4
D. None of these

Answer

B

Explanation

`f(x)={(x+1",",x le 1),(2x+1",",1lt x le 2):}`
`g(x)={(x^(2)",", -1 le x lt2),(x+2",",2le x le 3):}`
` :. f(x)={(g(x)+1",",g(x) le 1),(2g(x)+1",",1lt g(x) le 2):}`
`or f(g(x))={(x^(2)+1",",x^(2) le 1",", -1 le x lt2),(x+2+1",",x+2 le 1",",2le x le 3),(2x^(2)+1",",1 lt x^(2) le 2",", -1 le x lt2),(2(x+2)+1",",1 lt x+2 le 2",",2le x le 3):}`
`or f(g(x))={(x^(2)+1",", -1 le x le1),(2x^(2)+1",",1 lt x le sqrt(2)):}`
For `f(g(x)) =2,x^(2)+1=2 and 2x^(2)+1=2 ` i.e.,
` x=+- or x= +- (1)/(sqrt(2)).`
Therefore, `x=+-1` only. Hence, two roots.