If the equation x+5<1−x, then

x+5 0,x+5 x,x+5 0x2−3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4x2−3x−4>0⇒x 4. ∴ Solution set =[−5,−1)

If the equation x+5<1−x, then

x+5 0,x+5 x,x+5 0x2−3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4x2−3x−4>0⇒x 4. ∴ Solution set =[−5,−1)

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If the equation $\sqrt{x + 5} < 1 - x$ , then

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$- 5 \leq x < 1$

$5 \leq x < - 1$

$- 5 \leq x < - 1$

answer is C.

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Detailed Solution

$\begin{array}{l} \sqrt{x + 5} < 1 - x \Rightarrow x + 5 \geq 0, 1 - x > 0, x + 5 < (1 - x)^{2} \Rightarrow x \geq - 5, 1 > x, x + 5 < 1 + x^{2} - 2 x \\ \Rightarrow x \geq - 5, x < 1, x^{2} - 3 x - 4 > 0 \\ x^{2} - 3 x - 4 = 0 \Rightarrow (x - 4) (x + 1) = 0 \Rightarrow x = - 1 or 4 \\ x^{2} - 3 x - 4 > 0 \Rightarrow x < - 1 or x > 4 . ∴ Solution set = [- 5, - 1) \end{array}$