Question

The number of integral value of 'x' for $x\in (-2,10)$ if $\frac{x^2-4x-2}{x^2-x-30}>0$ is

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is B 10

$y=\frac{x^2-4x-2}{x^2-x-30}>0$

Denominator$\Rightarrow x^2-x-30=0$

$\Rightarrow x=\frac{1\pm \sqrt{1+120}}{2}$

$\Rightarrow x=6,-5$

Numerator$\Rightarrow x^2-4x-2=0$

$\Rightarrow x=\frac{4\pm \sqrt{16+8}}{2}=2\pm \sqrt{6}$

$y=\frac{(x-2-\sqrt{6})(x-2+\sqrt{6})}{(x-6)(x+5)}>0$

From the figure, $x\in (-\infty ,-5)\cup (2-\sqrt{6},2+\sqrt{6})\cup (6,\infty )$

No. of integral solutions in $x\in (-2,10)$ is 10