Question

Let $f(x+y)=f\left(x\right).f\left(y\right)$ for all $x$ and $y$ and $f\left(1\right)=2$, then area enclosed by $3|x|+2|y|\le 8$, is:

• A. $f\left(5\right)sq.units$
• B. $f\left(6\right)sq.units$
• C. $\frac{1}{3}f\left(6\right)sq.units$
• D. $f\left(4\right)sq.units$

Answer 1

Answer: (C)

$\frac{1}{3}f\left(6\right)sq.units$

Given,

$f(x+y)=f\left(x\right)f\left(y\right)$

$f\left(1\right)=2$

At $(x,y)=(1,0)$, we get

$f\left(1\right)=f\left(1\right)f\left(0\right)\Rightarrow f\left(0\right)=1$

At $(x,y)=(1,1)\Rightarrow f\left(2\right)=4$

At $(x,y)=(2,1)\Rightarrow f\left(3\right)=8$ and so on...

According to the pattern, we get

$f\left(x\right)=2^x$

Now , we need to find the area of the shaded region, which is no other than the four times the area of $\Delta AOB$.

Area of $\Delta AOB=\frac{1}{2}\times AO\times OB=\frac{1}{2}\times 4\times \frac{8}{3}=\frac{16}{3}sq.units$

Area of shaded region is, $4\times \text{Area of}\Delta AOB=4\times \frac{16}{3}=\frac{64}{3}sq.units$

The above area is also equal to $\frac{1}{3}f\left(6\right)sq.units$

Hence, option C.