If 9x2+6x+1 < (2−x) then x∈

Given inequation 9x2+6x+1 0 and 9x2+6x+1 0 ⇒ x<2Case(iii): 9x2+6x+1< (2−x)2⇒ 9x2+6x+1< 4+x2−4x⇒ 8x2+10x−3<0⇒ 8x2+12x−2x−3<0⇒ 4x(2x+3)−1(2x+3)<0⇒ (2x+3)(4x−1)<0∴ −32<x<14∴ x∈ (−32, 14)

If 9x2+6x+1 < (2−x) then x∈

Given inequation 9x2+6x+1 0 and 9x2+6x+1 0 ⇒ x<2Case(iii): 9x2+6x+1< (2−x)2⇒ 9x2+6x+1< 4+x2−4x⇒ 8x2+10x−3<0⇒ 8x2+12x−2x−3<0⇒ 4x(2x+3)−1(2x+3)<0⇒ (2x+3)(4x−1)<0∴ −32<x<14∴ x∈ (−32, 14)

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$I f \sqrt{9 x^{2} + 6 x + 1} < (2 - x) t h e n x \in$

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$(- \frac{3}{2}, \frac{1}{4})$

$(- \frac{3}{2}, \frac{1}{4}]$

$[- \frac{3}{2}, \frac{1}{4})$

$(- \infty, \frac{1}{4})$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n i n e q u a t i o n \sqrt{9 x^{2} + 6 x + 1} < (2 - x) \\ \Rightarrow 9 x^{2} + 6 x + 1 \geq 0, 2 - x > 0 a n d 9 x^{2} + 6 x + 1 < (2 - x) \\ C a s e (i) : 9 x^{2} + 6 x + 1 \geq 0 \\ \Rightarrow {(3 x + 1)}^{2} \geq 0 \Rightarrow \forall x \in R \\ C a s e (i i) : 2 - x > 0 \Rightarrow x < 2 \\ C a s e (i i i) : 9 x^{2} + 6 x + 1 < {(2 - x)}^{2} \\ \Rightarrow 9 x^{2} + 6 x + 1 < 4 + x^{2} - 4 x \\ \Rightarrow 8 x^{2} + 10 x - 3 < 0 \\ \Rightarrow 8 x^{2} + 12 x - 2 x - 3 < 0 \\ \Rightarrow 4 x (2 x + 3) - 1 (2 x + 3) < 0 \\ \Rightarrow (2 x + 3) (4 x - 1) < 0 \\ ∴ \frac{- 3}{2} < x < \frac{1}{4} \\ ∴ x \in (\frac{- 3}{2}, \frac{1}{4}) \end{array}$