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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
Solution
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
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