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Question
Find the range of the following functions given by f(x) = `3/(2 - x^2)`
Solution
Let f(x) = y
y = `3/(2 - x^2)`
⇒ 2 – x2
= `3/"y"`
⇒ = `2 - 3/y`
But, we know that, x2 ≥ 0
`2 - 3/y ≥ 0`
⇒ `(2y - 3)/y ≥ 0`
⇒ y > 0 and 2y – 3 ≥ 0
⇒ y > 0 and 2y ≥ 3
⇒ y > 0 and y ≥ `3/2`
Or f(x) > 0 and f(x) ≥ `3/2`
f(x) ∈ `( – oo, 0) ∪ (3/2 , oo)`
⇒ f(x) ∈ `( – oo, 0) ∪ (3/2 , oo)`
Therefore, the range of f = `(-oo, 0) ∪ (3/2 , oo)`.
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