Question

Consider a real valued function $f:R\to R$ satisfying $f\left(\frac{2x+3y}{5}\right)=\frac{2f\left(x\right)+3f\left(y\right)}{5}\forall x,y\in R$ and $f\left(0\right)=2;f^{\prime }\left(0\right)=1$
Minimum distance of a point on graph of $y=f\left(x\right)$ from origin is less than or equal to

• A. $\sqrt{2}$ units
• B. $\frac{1}{2}$ units
• C. $\frac{1}{\sqrt{5}}$ units
• D. $\frac{1}{\sqrt{7}}$ units

Answer 1

The correct option is A $\sqrt{2}$ units

$f\left(\frac{2x+3y}{5}\right)=\frac{2f\left(x\right)+3f\left(y\right)}{5}\forall x,yϵR$

Given $f\left(0\right)=2$

$f\left(\frac{2}{5}\right)=\frac{2}{5}f\left(1\right)$

$f\left(x\right)=kx+2$

$f\left(0\right)=2$

$f^{\prime }\left(0\right)=k=1$

Therefore, $f\left(x\right)=x+2$

(Image)

Let P be the minimum distance from origin to $f\left(x\right)$

$PQ=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2}$

$PQ=2\sqrt{2}$

And $PM=\sqrt{2}$

Now, minimum distance $P=\sqrt{4-2}$

$P=\sqrt{2}$ Answer