Number of integral values of \( x \) satisfying the inequation \( \fra

1.	Number of integral values of \( x \) satisfying the inequation \( \frac{x}{x+2} \leq \frac{1}{\|x\|} \) is (a) 2 (b) 8 (c) 4
Answer» \(\frac{x}{x+2}\leq\frac1{\|x\|}\) ⇒ \(\frac{x\|x\|}{x+2}\leq1\) (Multiplying both sides by positive value \|x\|) I₆ x = -3, then \(\frac{x\|x\|}{x+2}=\frac{-3\times3}{-3+2}=\frac{-9}{-1}=9>1\) (Not satisfies) (\(\therefore\) I₆ x < -2, then x\|x\| < 0 & x + 2 < 0 ⇒ \(\frac{x\|x\|}{x+2}>1\)) x = -2 is not possible x = -1 then \(\frac{x\|x\|}{x+2} = \frac{-1\times1}{-1+2}=\frac{-1}1=-1<1\) (Satisfies) x = 0 then \(\frac{x\|x\|}{x+2} = 0<1\) (Satisfies) x = 1 then \(\frac{x\|x\|}{x+2} = \frac13<1\) (Satisfies) x = 2 then \(\frac{x\|x\|}{x+2} = \frac{2\times2}{2+2}=\frac44=1\) (Satisfies) I₆ x > 2 then x\|x\| = x² > x + 2 ⇒ \(\frac{x\|x\|}{x+2} >1\) (Not satisfies) \(\therefore\) Number of integral values which satisfies given inequality are 4.

Number of integral values of \( x \) satisfying the inequation \( \frac{x}{x+2} \leq \frac{1}{|x|} \) is (a) 2 (b) 8 (c) 4

Answer»

\(\frac{x}{x+2}\leq\frac1{|x|}\)

⇒ \(\frac{x|x|}{x+2}\leq1\) (Multiplying both sides by positive value |x|)

I₆ x = -3, then \(\frac{x|x|}{x+2}=\frac{-3\times3}{-3+2}=\frac{-9}{-1}=9>1\) (Not satisfies)

(\(\therefore\) I₆ x < -2, then x|x| < 0 & x + 2 < 0 ⇒ \(\frac{x|x|}{x+2}>1\))

x = -2 is not possible

x = -1 then \(\frac{x|x|}{x+2} = \frac{-1\times1}{-1+2}=\frac{-1}1=-1<1\) (Satisfies)

x = 0 then \(\frac{x|x|}{x+2} = 0<1\) (Satisfies)

x = 1 then \(\frac{x|x|}{x+2} = \frac13<1\) (Satisfies)

x = 2 then \(\frac{x|x|}{x+2} = \frac{2\times2}{2+2}=\frac44=1\) (Satisfies)

I₆ x > 2 then x|x| = x² > x + 2

⇒ \(\frac{x|x|}{x+2} >1\) (Not satisfies)

\(\therefore\) Number of integral values which satisfies given inequality are 4.

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