SATHEE: Theory of Equations 1 Question 25

Theory of Equations 1 Question 25

26. The set of all real numbers $x$ for which $x^{2} - | x + 2 | + x > 0$ is

(2002, 1M)

(a) $(- \infty, - 2) \cup (2, \infty)$

(b) $(- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$

(d) $(\sqrt{2}, \infty)$

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Answer:

Correct Answer: 26. (b)

Solution:

Given, $x^{2} - | x + 2 | + x > 0$

Case I When $x + 2 \geq 0$

$\begin{array}{lc} ∴ & x^{2} - x - 2 + x > 0 \Rightarrow x^{2} - 2 > 0 \\ \Rightarrow & x < - \sqrt{2} or x > \sqrt{2} \\ \Rightarrow & x \in [- 2, - \sqrt{2}) \cup (\sqrt{2}, \infty) \end{array}$

Case II When $x + 2 < 0$

$\begin{array}{ll} ∴ & x^{2} + x + 2 + x > 0 \\ \Rightarrow & x^{2} + 2 x + 2 > 0 \\ \Rightarrow & (x + 1)^{2} + 1 > 0 \end{array}$

which is true for all $x$ .

$∴ x \leq - 2 or x \in (- \infty, - 2)$

From Eqs. (ii) and (iii), we get

$x \in (- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$

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Theory of Equations 1 Question 25

26. The set of all real numbers x for which x2−|x+2|+x>0 is

Answer:

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26. The set of all real numbers $x$ for which $x^{2} - | x + 2 | + x > 0$ is