Question

If $\frac{2x-1}{2x^3+3x^2+x}$ is positive, then $x$ lies in the interval-

• A. $(-1.5,0.5)$
• B. $(2.5,\infty )$
• C. $(0.5,2.5)$
• D. $(-\infty ,-1.5)$

Answer 1

Answer: (A), (C), (D)

$(-1.5,0.5)$
$(0.5,2.5)$
$(-\infty ,-1.5)$

Given $\frac{2x-1}{2x^3+3x^2+x}$ is positive.

$\Rightarrow \frac{2x-1}{2x^3+3x^2+x}>0$

$\Rightarrow \frac{2x-1}{x(2x^2+3x+1)}>0$

$\Rightarrow \frac{2x-1}{x(2x+1)(x+1)}>0$

Now let us compute the signs of each term.

$2x-1=0\Rightarrow x=\frac{1}{2}$

$2x-1<0\Rightarrow x<\frac{1}{2}$

$2x-1>0\Rightarrow x>\frac{1}{2}$

$x=0\Rightarrow x=0,x<0\Rightarrow x<0,x>0\Rightarrow x>0$

$x+1=0\Rightarrow x=-1,x+1<0\Rightarrow x<-1,x+1>0\Rightarrow x>-1$

$2x+1=0\Rightarrow x=-\frac{1}{2}$

$2x+1<0\Rightarrow x<-\frac{1}{2}$

$2x+1>0\Rightarrow x>-\frac{1}{2}$

Now summarize the above in a table:

	$(-\infty ,-1)$	$-1$	$(-1,-1/2)$	$-1/2$	$(-1/2,0)$	$0$	$(0,1/2)$	$1/2$	$(1/2,\infty )$
$2x-1$	$-$	$-$	$-$	$-$	$-$	$-$	$-$	$0$	$+$
$x$	$-$	$-$	$-$	$-$	$-$	$0$	$+$	$+$	$+$
$x+1$	$-$	$0$	$+$	$+$	$+$	$+$	$+$	$+$	$+$
$2x+1$	$-$	$-$	$-$	$0$	$+$	$+$	$+$	$+$	$+$
$\frac{2x-1}{x(x+1)(2x+1)}$	$+$	$ND$	$-$	$ND$	$+$	$ND$	$-$	$0$	$+$

Now consider the above table.

Check the column $(-\infty ,-1)$, we get $\frac{2x-1}{x(x+1)(2x+1)}=\frac{-}{(-)(-)(-)}>0$ .

Now check column $(-1/2,0)$, we get $\frac{2x-1}{x(x+1)(2x+1)}=\frac{-}{(-)(+)(+)}>0$ .

and check column $(1/2,\infty )$, we get $\frac{2x-1}{x(x+1)(2x+1)}=\frac{+}{(+)(+)(+)}>0$ .

Thus $\frac{2x-1}{2x^3+3x^2+x}$ is positive at $(-\infty .-1),(-1/2,0),(1/2,\infty )$.

Thus the correct answers are $(-1.5,0.5),(0.5,2.5)and(-\infty ,-1.5)$