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Show that y = log(1+x)–2x2+x,x>-1 is an increasing function on its domain. - Mathematics and Statistics

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प्रश्न

Show that y = `log (1 + x) – (2x)/(2 + x), x > - 1` is an increasing function on its domain.

योग
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उत्तर

y = `log (1 + x) – (2x)/(2 + x), x > - 1`

∴ `dy/dx =d/dx[log (1 + x) - (2x)/(2 + x)]`

= `(1)/(1 + x).d/dx(1 + x) - ((2 + x).d/dx(2x) - 2x.d/dx(2 + x))/(2 + x)^2`

= `(1)/(1 + x) xx (0 + 1) ((2 + x) xx 2 - 2x(0 + 1))/(2 + x)^2`

= `(1)/(1 + x)   - (4 + 2x - 2x)/(2 + x)^2`

= `(1)/(1 + x) - (4)/(2 + x)^2`

= `((2 + x)^2 - 4(1 + x))/((1 + x)(2 + x)^2)`

= `(4 + 4x + x^2 - 4 - 4x)/((1 + x)(2 + x)^2`

= `(x^2)/((1 + x)(2 + x)^2) >` 0 for all x > – 1

Hence, the given function is increasing function on its domain.

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अध्याय 2: Applications of Derivatives - Exercise 2.4 [पृष्ठ ९०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 2 Applications of Derivatives
Exercise 2.4 | Q 23 | पृष्ठ ९०

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