Question

$Iff:(0,\infty )\to (0,\infty )$ satisfy

$f\left(xf\right(y\left)\right)=x^2{y}^2(a\in R)$,then

${\sum }_{r=1}^n-f\left(r\right){}^n{C}_{r}is$

• A. $n{.2}^{n-1}$
• B. $n(n-1)2^{n-2}$
• C. $n{.2}^{n-1}+n(n-1)2^{n-2}$
• D. $0$

Answer 1

The correct option is C

$n{.2}^{n-1}+n(n-1)2^{n-2}$

Taking x = 1

$f\left(f\right(y\left)\right)=y^a$

$Letf\left(y\right)=\frac{1}{x}f\left(1\right)=\frac{1}{\left(f\right(y){)}^2}{y}^a..............\left(i\right)$

and Let y=1

$\left(f\right(1){)}^3=1\Rightarrow f(1)=1$

Now use y=1

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

So from $\left(1\right)\left(f\right(y){)}^2=y^a\Rightarrow y^4=y^a\Rightarrow a=4$

$\Rightarrow$ (A) is true

${\sum }_{r-1}^{n}f(r{)}^n{C}_r={\sum }_{r-1}^n{r}^2$

${}^n{C}_r$

${\sum }_{r-1}^n\left(r\right(r-1)+1)$

${}^n{C}_r$

$=n(n-1)2^{n-2}+n{.2}^{n-1}$

$\Rightarrow$ (C ) is true.

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

$\Rightarrow 2x^2{e}^x$

$\Rightarrow 3$ solutions

$\Rightarrow$ (C) is true.