JEE Main 2022 — Mathematics Algebra

Given $f(x+y)=2f(x)\cdot f(y)$f(x+y)=2f(x)⋅f(y) & $f(1)=2$f(1)=2

Now putting $x=1$x=1 & $y=1$y=1 in $f(x+y)=2f(x)\cdot f(y)$f(x+y)=2f(x)⋅f(y) we get

$$f(1+1)=2f(1)\cdot f(1)=2\times {2}^{2}={2}^{3}$$f(1+1)=2f(1)⋅f(1)=2×22=23

So $f(2)={2}^{3}$f(2)=23, Similarly

$$f(3)={2}^{5},f(4)={2}^{7}.....$$f(3)=25,f(4)=27.....

Now

$$\sum _{k=1}^{10}f(\alpha +k)=\sum _{k=1}^{10}2f(\alpha )\cdot f(k)=\frac{512}{3}({2}^{20}-1)$$k=1∑10​f(α+k)=k=1∑10​2f(α)⋅f(k)=3512​(220−1)

$$\Rightarrow 2f(\alpha )\sum _{k=1}^{10}f(k)=\frac{512}{3}({2}^{20}-1)$$⇒2f(α)k=1∑10​f(k)=3512​(220−1)

$$\Rightarrow 2f(\alpha ){f(1)+f(2)\cdots f(10)}=\frac{512}{3}({2}^{20}-1)$$⇒2f(α)f(1)+f(2)⋯f(10)=3512​(220−1)

$$\Rightarrow 2f(\alpha )(2+{2}^{3}+{2}^{5}\ldots )=\frac{512}{3}({2}^{20}-1)$$⇒2f(α)(2+23+25…)=3512​(220−1)

$$\Rightarrow 2f(\alpha )\frac{2({({2}^{2})}^{10}-1)}{{2}^{2}-1}=\frac{512}{3}({2}^{20}-1)$$⇒2f(α)22−12((22)10−1)​=3512​(220−1)

$$\Rightarrow 2\times 2f(\alpha )\times \frac{{2}^{20}-1}{3}=\frac{512}{3}({2}^{20}-1)$$⇒2×2f(α)×3220−1​=3512​(220−1)

Now comparing both side we get

$$4f(\alpha )=512\Rightarrow f(\alpha )=128$$4f(α)=512⇒f(α)=128

$\Rightarrow f(\alpha )={2}^{7}\Rightarrow \alpha =4$⇒f(α)=27⇒α=4.