If x-1x-0- then least integral value of x2-2 is where x-

For any real number x x+1x≥2 if x gt;0 and x+1x≤ 2 if x lt;0 Now, from the given inequality nbsp; x+1x lt;0 we can say that nbsp;x lt;0 So, x∈( ∞,0 ...

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Standard XII

Mathematics

Inequality

If x+1x<0, th...

Question

If $x + \frac{1}{x} < 0$ , then least integral value of $x^{2} + 2$ is
(where $x \in R$ )

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Solution

For any real number $x$
$x + \frac{1}{x} \geq 2$ if $x > 0$
and $x + \frac{1}{x} \leq - 2$ if $x < 0$

Now, from the given inequality
$x + \frac{1}{x} < 0$ we can say that $x < 0$
So, $x \in (- \infty, 0)$
$\Rightarrow x^{2} \in (0, \infty) [∵ 0 is not lying in the interval] ∴ x^{2} + 2 \in (2, \infty)$

Hence, least integral value of $x^{2} + 2 = 3$

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If x+1x<0, then least integral value of x2+2 is (where x∈R)

For any real number x x+1x≥2 if x>0 and x+1x≤−2 if x<0 Now, from the given inequality x+1x<0 we can say that x<0 So, x∈(−∞,0) ⇒x2∈(0,∞) [∵0 is not lying in the interval]∴x2+2∈(2,∞) Hence, least integral value of x2+2=3

If $x + \frac{1}{x} < 0$ , then least integral value of $x^{2} + 2$ is
(where $x \in R$ )