CAT 2019 Quant Question with Solution 24

Correct Answer: 3

$f(a+1)+f(a+2)+\ldots+f(a+n)=16\left(2^{n}-1\right)$

$\Rightarrow f(a)f(1)+f(a)f(2)+...+f(a)f(n)=16\left( {{2}^{n}}-1 \right)$

$\Rightarrow f(a)\left( f(1)+f(2)+...+f(n) \right)=16\left( {{2}^{n}}-1 \right)$

Take n=1,

$\Rightarrow f(a)f(1)=16\left( {{2}^{1}}-1 \right)=16$

$\Rightarrow f(a)\times 2=16\Rightarrow f(a)=8$

Therefore,

$\begin{align} & f(a)\left( f(1)+f(2)+...+f(n) \right)=16\left( {{2}^{n}}-1 \right) \\ & \Rightarrow f(1)+f(2)+...+f(n)=2\left( {{2}^{n}}-1 \right) \\ \end{align}$

If n=2, then $f(1)+f(2)=2\left( {{2}^{2}}-1 \right)=6$

$\Rightarrow f(2)=6-f(1)=6-2=4$

If n=3, then $f(1)+f(2)+f(3)=2\left( {{2}^{3}}-1 \right)=14$

$\Rightarrow f(3)=6-f(1)-f(2)=14-2-4=8=f(a)$

Hence a =3.