Question

If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______. 

Options

• x² – 6x + 9 = 0

• x² – 7x + 8 = 0

• x² – 14x + 49 = 0

• x² – 10x + 21 = 0

Answer 1

If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is x² – 10x + 21 = 0. 

Explanation:

Given `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`

∴ `lim_(x -> 2^-) f(x) = lim_(x -> 2^-) (x^2 - 1)`

`lim_(h -> 0) [(2 - h)^2 - 1] =  lim_(h -> 0) (4 + h^2 - 4h - 1)`

= `lim_(h -> 0) (h^2 - 4h + 3)`

= 3

And `lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (2x + 3)`

= `lim_(h -> 0) [2(2 - h) + 3]`

= 7

Therefore, 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