Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$ and $y$ and $f\left(1\right)=2$.If in a triangle $ABC,a=f\left(3\right),b=f\left(1\right)+f\left(3\right),c=f\left(2\right)+f\left(3\right)$, then $2A$ is equal to

• A. $C$
• B. $2C$
• C. $3C$
• D. $4C$

Answer 1

Answer: (A)

$C$

$\therefore f(x+y)=f\left(x\right)f\left(y\right)$
for $x=y=1$
$f\left(2\right)=2^2=4$
For $x=1,y=2$
$f\left(3\right)=f\left(1\right)f\left(2\right)=2^3=8$
$\therefore f\left(n\right)=2^n$
$\therefore a=f\left(3\right)=2^3=8$
$b=f\left(1\right)+f\left(3\right)=2+8=10$
and $c=f\left(2\right)+f\left(3\right)=2^2+2^3=4+8=12$
$\cos A=\frac{b^2+c^2-a^2}{2bc}$
$=\frac{{10}^2+{12}^2-8^2}{240}=\frac{3}{4}$(on simplification)
$\cos C=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{8^2+{10}^2-{12}^2}{160}=\frac{1}{8}$(on simplification)
$\cos 2A=2{\cos }^{2}A-1=2\times \frac{9}{16}-1=\frac{9}{8}-1=\frac{1}{8}=\cos C$
$\therefore \cos 2A=\cos C$
Hence$2A=C$