Question

If x satisfies the inequality $lo{g}_{x+3}(x^2-x)<1$ then x may belong to the interval-

• A. $(-3,-2)$
• B. $(-1,3)$
• C. $(1,3)$
• D. $(-1,0)$

Answer 1

Answer: (A)

$(-3,-2)$

Given $x$ satisfies the inequality $lo{g}_{x+3}(x^2-x)<1$

Consider $lo{g}_{x+3}(x^2-x)<1$

By definition of logarithm, we have

$x^2-x<(x+3{)}^1$

$\Rightarrow x^2-x<x+3$

$\Rightarrow x^2-2x-3<0$

Solving it as quadratic equation, $x^2-2x-3=0$, we get $(x-3)(x+1)=0$

$\Rightarrow x=3or-1$

So our solution set is $x\in (-1,3)$

This is only for quadratic though. We haven't considered the number in $x^2-x$ must be positive for the logarithm to be denied.

Hence we have, $x^2-x>0$

$\Rightarrow x^2-x=0$

$\Rightarrow x(x-1)=0$

$\Rightarrow x=0or1$

If we repeat the process with test points, we realize that the solution is $(-\infty ,0)$ and $(1,\infty )$. Therefore, we must eliminate $(0,1)$ from our solution set above.

Now, we must also guarantee that $x+3>0$.

$\Rightarrow x>-3$

This means that we can include $x>-3$ in our solution set.

The set becomes $(-3,3)$

However, if you try $x=-1.5$ (for any number in the interval $(-2,-1)$, you will realize that the result you get is not within the range of the problem, that's to say it will be greater, instead of less than $1$).

Therefore $x\in (-3,-2)\cup (-1,0)\cup (1,3)$

A graphical verification yields the same results