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प्रश्न
Find the values of x for which f(x) = `x/(x^2 + 1)` is (a) strictly increasing (b) decreasing.
उत्तर
f(x) = `x/(x^2 + 1)`
∴ f'(x) = `d/dx(x/(x^2 + 1))`
= `((x^2 + 1).d/dx(x) - xd/dx(x^2 + 1))/(x^2 + 1)^2`
= `((x^2 + 1)(1) - x(2x + 0))/(x^2 + 1)^2`
= `(x^2 + 1 - 2x^2)/(x^2 + 1)^2`
= `(1 - x^2)/(x^2 + 1)^2`
(a) f is strictly increasing if f'(x) > 0
i.e. if `(1 - x^2)/(x^2 + 1)^2 > 0`
i.e. if 1 – x2 > 0 ...[∵ (x2 + 1)2 > 0]
i.e. if 1 > x2
i.e. if x2 < 1
i.e. if – 1 < x < 1
∴ f is strictly increasing if – 1 < x < 1
(b) f is strictly decreasing if f'(x) < 0
i.e. if `(1 - x^2)/(x^2 + 1)^2 < 0`
i.e. if 1 – x2 < 0 ...[∵ (x2 + 1)2 > 0]
i.e. if 1 < x2
i.e. if x2 > 1
i.e. if x > 1 or x < – 1
∴ f is strictly decreasing if x < – 1 or x > 1
i.e. `x ∈( - oo, - 1) ∪ (1, oo)`.
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