Question

Let $f\left(x\right)$ be a function defined as $f\left(x\right)=a|x|+b.$ If $f\left(6\right)=5$ and $f(-3)=4,$ then which of the following is/are correct?

• A. $f\left(x\right)=3|x|+\frac{1}{3}$
• B. $ab=1$
• C. Domain of $f\left(x\right)$ is $R$
• D. Range of $f\left(x\right)=[3,\infty )$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $ab=1$
C Domain of $f\left(x\right)$ is $R$
D Range of $f\left(x\right)=[3,\infty )$

$f\left(x\right)=a|x|+b$
$f\left(6\right)=5\Rightarrow 6a+b=5...\left(1\right)$
$f(-3)=4\Rightarrow 3a+b=4...\left(2\right)$
On solving $\left(1\right)$ and $\left(2\right)$, we get
$a=\frac{1}{3},b=3$

$\Rightarrow f\left(x\right)=\frac{|x|}{3}+3$
Clearly, domain of $f\left(x\right)$ is $R$

Since, $|x|\ge 0\forall x\in R$
$\Rightarrow \frac{|x|}{3}+3\ge 3$
Therefore, range of $f\left(x\right)=[3,\infty )$