If the inequality -3- x 2-2 x -10- 0 holds good- then x -A- -0- -B- -2-14-C- - 0-D- -

The correct option is D ϕGiven √( 3x^2 + 2x + 10)≥ 0 to be true if 3x^2 + 2x + 10≥0 ⇒3x^2 + 2x + 10≤0 ⇒ x^2+2x+10 lt;0 ⇒ (x+1)^2+9 lt;0 which is not possi ...

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Standard XI

Mathematics

Square Root of a Complex Number

If the inequa...

Question

If the inequality $\sqrt{\frac{- 3}{x^{2} + 2 x + 10}} \geq 0$ holds good, then $x \in$

[0, \infty)

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(- \infty, 0]

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ϕ

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[2, 14]

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Solution

The correct option is C $ϕ$
Given $\sqrt{\frac{- 3}{x^{2} + 2 x + 10}} \geq 0$ to be true
if $\frac{- 3}{x^{2} + 2 x + 10} \geq 0$
$\Rightarrow \frac{3}{x^{2} + 2 x + 10} \leq 0 \Rightarrow x^{2} + 2 x + 10 < 0 \Rightarrow (x + 1)^{2} + 9 < 0$
which is not possible for any $x \in R$

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If the inequality √−3x2+2x+10≥0 holds good, then x∈

The correct option is C ϕGiven √−3x2+2x+10≥0 to be true if −3x2+2x+10≥0 ⇒3x2+2x+10≤0⇒x2+2x+10<0⇒(x+1)2+9<0 which is not possible for any x∈R

If the inequality $\sqrt{\frac{- 3}{x^{2} + 2 x + 10}} \geq 0$ holds good, then $x \in$

The correct option is C $ϕ$
Given $\sqrt{\frac{- 3}{x^{2} + 2 x + 10}} \geq 0$ to be true
if $\frac{- 3}{x^{2} + 2 x + 10} \geq 0$
$\Rightarrow \frac{3}{x^{2} + 2 x + 10} \leq 0 \Rightarrow x^{2} + 2 x + 10 < 0 \Rightarrow (x + 1)^{2} + 9 < 0$
which is not possible for any $x \in R$