Question

For a function defined from set $X$ to set $Y$ where $x\in X\&y\in Y.$
It is given that $y$ is linearly dependent on $x$ and $f\left(2\right)=3,$ $f\left(0\right)=\frac{7}{3}.$ Find the function.

• A. $3y=x+7$
• B. $y=\frac{1}{3}x+\frac{7}{3}$
• C. $2y=x+7$

Answer 1

Answer: (A), (B)

$y=\frac{1}{3}x+\frac{7}{3}$

$y$ is linearly dependent on $x.$
General form of a linear funciton is $y=ax+b.$

We know, $f\left(x\right)=y.$
$\therefore f\left(x\right)=ax+b$

Now substitute $f\left(2\right)=3$ and $f\left(0\right)=\frac{7}{3}$ to get the unknown constants.
For $f\left(0\right)=\frac{7}{3},x=0$
$\Rightarrow \frac{7}{3}=a\left(0\right)+b$
$\Rightarrow b=\frac{7}{3}$
$\therefore f\left(x\right)=ax+\frac{7}{3}$

For $f\left(2\right)=3,x=2$
$\Rightarrow 3=a\left(2\right)+\frac{7}{3}$
$\Rightarrow 2a=3-\frac{7}{3}=\frac{9}{3}-\frac{7}{3}=\frac{2}{3}$
$\Rightarrow a=\frac{1}{3}$
$\therefore f\left(x\right)=\frac{1}{3}x+\frac{7}{3}$
$\Rightarrow 3y=x+7$