If f(x) ={`x^2-1, 0 lt x lt 2` , `2x+3 , 2 le x lt 3`then the quadratic equation whose roots are `lim_(x->2^+)f(x)` and `lim_(x->2^-)f(x)` is

Options

A. `x^(2)-6x+9=0`
B. `x^(2)-7x+8=0`
C. `x^(2)-14x+49=0`
D. `x^(2)-10x+21=0`

Answer

d

Explanation

Given, `f(x)={:{(x^(2)-1",",,0 lt xlt 2),(2x+3",",,2le x lt 3):}`,
`therefore lim_(xto2^(-))f(x)=lim_(xto2^(-1))(x^(2)+1)`
`lim_(hto0)[(2-h)^(2)]=lim_(hto0)(4+h^(2)-4h-1)`
`=lim_(hto0)(h^(2)-4h+3)=3`
and `lim_(xto2^(+))f(x) = lim_(xto2^(+))(2x+3)`
`=lim_(hto0)[2(2+h)+3]=lim_(hto0)(4+2h+3)=7`
So, the quadratic equation whose roots are 3 and 7 is `x^(2)-(3+7)x+ 3 xx 7=0` i.e.,
`x^(2)-10x+21=0`