If -9x2-6x-1-2-x- then integral values of x is-are

The correct options are B 0 C 1Given √(9x^2+6x+1) lt;2 x will be true if 9x^2+6x+1≥0 ⇒ (3x+1)^2≥0, ∀ x∈ℝ Now as √(9x^2+6x+1) gt;0 squaring both sides ⇒ 9 ...

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Standard XII

Mathematics

Method of Intervals

If √9x2+6x+1<...

Question

If $\sqrt{9 x^{2} + 6 x + 1} < 2 - x,$ then integral values of $x$ is/are

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Solution

The correct options are
B $0$
C $- 1$
Given $\sqrt{9 x^{2} + 6 x + 1} < 2 - x$ will be true
if $9 x^{2} + 6 x + 1 \geq 0$
$\Rightarrow (3 x + 1)^{2} \geq 0, \forall x \in R$

Now as $\sqrt{9 x^{2} + 6 x + 1} > 0$
squaring both sides
$\Rightarrow 9 x^{2} + 6 x + 1 < (2 - x)^{2} \Rightarrow 9 x^{2} + 6 x + 1 < 4 + x^{2} - 4 x \Rightarrow 8 x^{2} + 10 x - 3 < 0 \Rightarrow (2 x + 3) (4 x - 1) < 0$
$∴ x \in (- \frac{3}{2}, \frac{1}{4})$

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If √9x2+6x+1<2−x, then integral values of x is/are

The correct options are B 0 C −1Given √9x2+6x+1<2−x will be true if 9x2+6x+1≥0 ⇒(3x+1)2≥0, ∀ x∈R Now as √9x2+6x+1>0 squaring both sides ⇒9x2+6x+1<(2−x)2⇒9x2+6x+1<4+x2−4x⇒8x2+10x−3<0⇒(2x+3)(4x−1)<0 ∴x∈(−32,14)

If $\sqrt{9 x^{2} + 6 x + 1} < 2 - x,$ then integral values of $x$ is/are