Question

The number of integral values of $x$ that satisfy the inequality $\frac{4x^2}{(1-\sqrt{2x+1}{)}^2}<2x+9$, is

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is B $5$

Taking $2x+1=a^2$
$\Rightarrow x=\frac{a^2-1}{2}$, where $a\ge 0$.

$\frac{4{\left(\frac{a^2-1}{2}\right)}^2}{{(1-a)}^2}<2\left(\frac{a^2-1}{2}\right)+9$
$\Rightarrow \frac{(a^2-1{)}^2}{(1-a{)}^2}<a^2+8$
$\Rightarrow (a+1{)}^2<a^2+8;a\ne 1$
$\Rightarrow a^2+2a+1<a^2+8$
$\Rightarrow a<\frac{7}{2}\text{and}a\ne 1\Rightarrow 0\le a^2<\frac{49}{4}$
$\Rightarrow -\frac{1}{2}\le x<\frac{45}{8}$
As $a\ne 1\Rightarrow x\ne 0$

Therefore, $x=1,2,3,4,5$