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Solve the given inequality for real x: x + x2 + x3 < 11 - Mathematics

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प्रश्न

Solve the given inequality for real x: x + `x/2` + `x/3` < 11

योग
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उत्तर

`x  + x/2` + `x/3` <11`

= `x(1 + 1/2 + 1/3) < 11`

= `(6x + 3x + 2x)/6 < 11`

 = `(11x)/6 < 11`

= `(11x)/(6 xx 11) < 11/11`

= `x/6 < 1`

= x < 6

Thus, all real numbers x, which are less than 6, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, 6).

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Linear Inequalities - Exercise 6.1 [पृष्ठ १२२]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 6 Linear Inequalities
Exercise 6.1 | Q 9 | पृष्ठ १२२
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