First slide

Functions (XII)

Question

Find the domain of the function $f (x) = \sqrt{(\frac{2}{x^{2} - x + 1} - \frac{1}{x + 1} - \frac{2 x - 1}{x^{3} + 1})}$

Moderate

A

$(- \infty, 2]$
B

$(- \infty, - 1) \cup (- 1, 2)$
C

$(- \infty, - 1) \cup (2, 3]$
D

None of these

Solution

$\begin{array}{l} f (x) = \sqrt{(\frac{2}{x^{2} - x + 1} - \frac{1}{x + 1} - \frac{2 x - 1}{x^{3} + 1})} \\ We must have \frac{2}{x^{2} - x + 1} - \frac{1}{x + 1} - \frac{2 x - 1}{x^{3} + 1} \geq 0 \\ ⇒ \frac{2 (x + 1) - (x^{2} - x + 1) - (2 x - 1)}{(x + 1) (x^{2} - x + 1)} \geq 0 \\ ⇒ \frac{- (x^{2} - x - 2)}{(x + 1) (x^{2} - x + 1)} \geq 0 \end{array}$
$\begin{array}{l} ⇒ \frac{- (x - 2) (x + 1)}{(x + 1) (x^{2} - x + 1)} \geq 0 \\ ⇒ \frac{2 - x}{x^{2} - x + 1} \geq 0, where x \neq - 1 \\ ⇒ 2 - x \geq 0, x \neq - 1 (as x^{2} - x + 1 > 0 \forall x \in R) \\ ⇒ x \leq 2, x \neq - 1 \\ Hence, domain of the function is (- \infty, - 1) \cup (- 1, 2] . \end{array}$

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