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  • Which of the option is correct for the inequality <img alt="\\\mathrm{\sqrt{-x^2+4x-3}>6-2x}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5C%5C%5Cmathrm%7B%5Csqrt%7B-x%5E2&plus

Which of the option is correct for the inequality \\\mathrm{\sqrt{-x^2+4x-3}>6-2x} ?

Option: 1

(1,3)


Option: 2

\\\mathrm{\left ( \frac{13}{5}, 3 \right )}


Option: 3

(3,∞)


Option: 4

(-∞, 3)


Answers (1)

best_answer

1. LHS should be defined, so

     -x^2+4x-3 \geq 0

    \\x^2-4x+3 \leq 0\\ (x-1)(x-3) \leq0\\ 1 \leq x \leq 3

2. When RHS < 0, then all these values will satisfy the inequation

    6 - 2x < 0 

    x > 3

3. When RHS \geq 0 (when x \leq 3), then we can square the inequation

    \\-x^2 + 4x - 3 > (6-2x)^2\\ -x^2 + 4x - 3 > 36 + 4x^2 - 24x\\ 5x^2 - 28 x + 39 < 0\\ (x-3)(5x-13) < 0 \\ 13/5 < x < 3

    Taking intersection of this result with x ≤ 3, we get 13/5 < x < 3

 

Answer is the intersection of 1 with ( 2 union 3): 13/5 < x < 3

Posted by

manish painkra

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