Question

The interval(s) of $x$ which satisfy the inequality $\frac{(1-3x{)}^7(1+5x{)}^4(x+5{)}^6}{(x^2-1)(x-6{)}^3(1-2x{)}^4}>0$ is/are

• A. $(-5,-1)$
• B. $(-\frac{1}{5},\frac{1}{3})$
• C. $(1,6)$
• D. $(-\infty ,-5)$

Answer 1

Answer: (B), (C)

The correct options are
B $(-\frac{1}{5},\frac{1}{3})$
C $(1,6)$

$\frac{(1-3x{)}^7(1+5x{)}^4(x+5{)}^6}{(x^2-1)(x-6{)}^3(1-2x{)}^4}>0$
Making coefficient of $x$ positive
$\Rightarrow \frac{(3x-1{)}^7(1+5x{)}^4(x+5{)}^6}{(x^2-1)(x-6{)}^3(2x-1{)}^4}<0$
Sign changes only for odd powers
$\Rightarrow \frac{(3x-1{)}^7(1+5x{)}^4(x+5{)}^6}{(x-1)(x+1)(x-6{)}^3(2x-1{)}^4}<0$

Critical points are
$\{-5,-1,-\frac{1}{5},\frac{1}{3},\frac{1}{2},1,6\}$


$\therefore x\in (-1,\frac{1}{3})\cup (1,6)-\{-\frac{1}{5}\}$