Question

Let f and g be two functions defined on R such that $f(x+y)=f\left(x\right)g\left(y\right)+g\left(x\right)f\left(y\right)$ for each $x,yϵR$ Let
$\Delta =\left|\begin{array}{ccc}f(\theta +\alpha )& f\left(\alpha \right)& g\left(\alpha \right)\\ f(\theta +\beta )& f\left(\beta \right)& g\left(\beta \right)\\ f(\theta +\gamma )& f\left(\gamma \right)& g\left(\gamma \right)\end{array}\right|$ then $\Delta$ equals

• A. $f\left(\alpha \right)f\left(\beta \right)f\left(\gamma \right)$
• B. $g\left(\alpha \right)g\left(\beta \right)g\left(\gamma \right)$
• C. $f\left(\alpha \right)g\left(\alpha \right)+f\left(\beta \right)g\left(\beta \right)+f\left(\gamma \right)g\left(\gamma \right)$
• D. $0$

Answer 1

Answer: (D)

$0$

$\Delta =\left|\begin{array}{ccc}f(\theta +\alpha )& f\left(\alpha \right)& g\left(\alpha \right)\\ f(\theta +\beta )& f\left(\beta \right)& g\left(\beta \right)\\ f(\theta +\gamma )& f\left(\gamma \right)& g\left(\gamma \right)\end{array}\right|$
Applying $C_1\to C_1-g\left(\theta \right)C_2-f\left(\theta \right)C_3$
and using $f(x+y)=f\left(x\right)g\left(y\right)+g\left(x\right)f\left(y\right)$, we get
$\Delta =\left|\begin{array}{ccc}0& f\left(\alpha \right)& g\left(\alpha \right)\\ 0& f\left(\beta \right)& g\left(\beta \right)\\ 0& f\left(\gamma \right)& g\left(\gamma \right)\end{array}\right|=0$
Hence, option 'D' is correct.