Question

Which of the following statement(s) is/are always holds true for $k<0$

• A. $a>b\Rightarrow \frac{a}{k}<\frac{b}{k},a,b<0$
• B. $a>b\Rightarrow a^k>b^k,a,b<0$
• C. Maximum value of $a^k+a^{-k}$ is $-2,a<0$
• D. $a>b\Rightarrow -ka<-kb,a,b>0$

Answer 1

The correct option is A $a>b\Rightarrow \frac{a}{k}<\frac{b}{k},a,b<0$

If $a>b,$ then multiplying or dividing with the term $k(<0)$ changes the sign of the expression irrespective of previous condition.

When $a>b\text{where}a,b<0$, then $-a<-b,\text{where}-a,-b>0$
$\Rightarrow (-a{)}^k>(-b{)}^k$
Now, $a^k>b^k$ only when $k$ is even, and when $k$ is odd, then $a^k<b^k$

$a^k+a^{-k}$ is of form $x+\frac{1}{x}.$
We know that, $x+\frac{1}{x}\le -2$ when $x<0$ and $x+\frac{1}{x}\ge 2$ when $x>0.$
So, when $k$ is even, $a^k>0$ i.e. $x>0.$
Hence, in this case $a^k+a^{-k}\ge 2$
Eg: $a=-1,k=-2$
$\Rightarrow a^k+a^{-k}=2$

Here $k<0\Rightarrow -k>0$ multiplying with positive value does not changes the sign.