\{ x {R} : {{6+x-x²}{2x+5} ≥ {{6+x-x²}{x+4} \} =

Correct Answer is : [- 2 , - 1] ∪ {3}

(- ∞ , - 4] ∪ ≤ft[-5/2, -1]

Correct Answer is : [- 2 , - 1] ∪ {3}

Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 13 of 160

Marks: +1, -0

\{ x \in \mathbb{R} : \frac{\sqrt{6+x-x^2}}{2x+5} \ge \frac{\sqrt{6+x-x^2}}{x+4} \}

Solution:

It is given that,

\; \frac{\sqrt{6+x-x^{2}}}{2x+5} \ge \; \frac{\sqrt{6+x-x^{2}}}{x+4}

Rewrite the above equation.

\; \frac{\sqrt{6+x-x^{2}}}{2x+5} \left( \; \frac{1}{2x+5} - \; \frac{1}{x+4} \right) \ge 0

\sqrt{6+x-x^{2}} \left( \; \frac{x+4-2x-5}{(2x+5)(x+4)} \right) \ge 0

\sqrt{6+x-x^{2}} \left( \; \frac{-(x+1)}{(2x+5)(x+4)} \right) \ge 0

(x+1)(2x+5)(x+4) \le 0

Now,

x \in (-\infty, -4) \cup \left[ -\frac{5}{2}, -1 \right] \; \; \; \; \; \cdots (I)

For expression to be exist,

6+x-x^{2} \ge 0

x^{2}-x-6 \le 0

(x-3)(x+2) \le 0

x \in [-2,3] \; \; \; \; \; \cdots (II)

From equation (I) and (II),

x \in [-2,-1] \cup \{3\}

Go to Question:

Prev Question

Next Question