Question

Let $f\left(x\right)$ be a function defined as $f\left(x\right)=\frac{a}{|x|}+b.$ If $f\left(6\right)=3$ and $f(-3)=4,$ then which of the following is/are correct?

• A. $f\left(x\right)=\frac{6}{|x|}+2$
• B. Domain of $f\left(x\right)$ is $R$
• C. Domain of $f\left(x\right)$ is $R-\left\{0\right\}$
• D. Range of $f\left(x\right)$ is $(2,\infty )$

Answer 1

Answer: (A), (C), (D)

Range of $f\left(x\right)$ is $(2,\infty )$

$f\left(x\right)=\frac{a}{|x|}+b$
$f\left(6\right)=3\Rightarrow \frac{a}{6}+b=3...\left(1\right)$
$f(-3)=4\Rightarrow \frac{a}{3}+b=4...\left(2\right)$
On solving $\left(1\right)$ and $\left(2\right)$, we get
$a=6,b=2$

$\Rightarrow f\left(x\right)=\frac{6}{|x|}+2$
Clearly, $x\ne 0$
Therefore, domain of $f\left(x\right)$ is $R-\left\{0\right\}.$

Since, $\frac{6}{|x|}>0\forall x\in R-\left\{0\right\}$
$\Rightarrow f\left(x\right)>2\forall x\in R-\left\{0\right\}$
Therefore, range of $f\left(x\right)=(2,\infty )$