$\textbf{Edit:}$ To do this question quickly in exam, you can do something like informal induction:
$A)$ Let, $f(2^n - 1) = 2^n - 1$ is true for all natural $n \geq 1$, it means $f(2^{n-1} - 1) = 2^{n-1} - 1$ is also true for $n \geq 2$.
Now,
$f(2^n-1) = f((2^n - 1 -1)+1) = f(2^n -2 + 1) = f(2(2^{n-1}-1)+1)$
$ = 2f(2^{n-1}-1)+1 = 2(2^{n-1}-1)+1 = 2^n - 1$
Similarly,
$B)$ $f(2^n) = f(2\times2^{n-1}) = 2f(2^{n-1}) - 1 = 2\times 1 -1 = 1$
$C)$ $f(5.2^n) = f(2(5\times2^{n-1})) = 2f(5\times 2^{n-1})-1$
$ = 2 \times (2^n + 1) - 1 = 2^{n+1} + 1$
$D)$ $f(2^n + 1) = f(2\times 2^{n-1} + 1) = 2f(2^{n-1}) + 1 = 2 \times 1 + 1 = 3$
Hence, A,B and C are correct options.
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Given: $f(2n) = 2f(n)-1$ and $f(2n+1)=2f(n) + 1$ for $n \geq 1$
Now, I shall check options one by one.
$\textbf{Option (B)}:$
$f(2^n) = 2f(\frac{2^n}{2}) -1 $ $[\because 2^n$ is an even term for $n\geq 1$ and since for even terms recurrence is defined as $f(2n) = 2f(n)-1$ which can be written as $f(n) = 2f(\frac{n}{2})-1$ So, I can write $f(2^n)$ as $2f(\frac{2^n}{2}) -1 $]
So, $f(2^n) = 2f(2^{n-1}) -1$
$=2[2f(2^{n-2})-1]-1= 2^2f(2^{n-2})-2-1$
Like this I can write:
$f(2^n) = 2^kf(2^{n-k})-2^{k-1}-2^{k-2}-...-1$
Put $n-k=0 \implies k=n$
So, $f(2^n) = 2^n*1 -(1+2+...+2^{n-1}) = 2^n – \left(\frac{1*(2^n-1)}{2-1} \right) = 2^n – 2^n +1 = 1$
Hence, $f(2^n) = 1$
$\textbf{Option (A)}:$
Since, $2^n-1$ is an odd term for $n \geq 1$ So, I can get the value of $f(2^n-1)$ from the recurrence which is defined for odd terms i.e. $f(2n+1)=2f(n) + 1$ which can be written as $f(n)= f(\frac{n-1}{2}) + 1$
So, $f(2^n -1) = 2f(\frac{2^n-1-1}{2}) = 2f(\frac{2^n-2}{2}) + 1 = 2f(2^{n-1}-1) +1 $
So, $f(2^n -1) = 2f(2^{n-1}-1) +1 $
$= 2[2f(2^{n-2} – 1)+1]+1 = 2^2f(2^{n-2}-1)+2 + 1$
Like this I can write as:
$f(2^n – 1) = 2^kf(2^{n-k}-1) + 2^{k-1} +2^{k-2}+...+1$
Put $2^{n-k}-1 = 1 \implies 2^{n-k} = 2 \implies n-k=1 \implies k=n-1$
So,
$f(2^n – 1) = 2^{n-1}*1 + 2^{n-2} + 2^{n-3}+...+1 = 1+2+….+2^{n-1}$
$= \left(\frac{2^n-1}{2-1}\right) = 2^n – 1$
So, $f(2^n – 1) = 2^n – 1$
$\textbf{Option (C)}:$
For $n \geq 1,$ $5 \times 2^n$ will give an even value, So, I can use recurrence which is defined for even terms i.e. $f(2n) = 2f(n)-1$ which can be written as $f(n) = 2f(\frac{n}{2})-1$ So, I can write $f(5 \times 2^n)$ as $2f(\frac{5\times2^n}{2}) -1 $]
So,
$f(5 \times 2^n) = 2f(5 \times 2^{n-1}) – 1$
$= 2 [ 2f(5 \times 2^{n-2}) – 1] -1 = 2^2 f(5 \times 2^{n-2}) – 2 -1$
Like this I can write:
$f(5 \times 2^n) = 2^kf(5\times 2^{n-k}) – 2^{k-1} – 2^{k-2}-...-2^0$
Since $f(5 \times 2^1) = f(10) = 2f(5)-1 = 2[2f(2)+1]-1 = 2[2+1]-1 =5$
So, put $5\times 2^{n-k} = 10 \implies 2^{n-k} = 2^1 \implies n-k = 1 \implies k=n-1$
So,
$f(5 \times 2^n) = 2^{n-1} \times 5 – 2^{n-2} – 2^{n-3} -...-1 = 5\times 2^{n-1} – [1+2+...+2^{n-2}]$
$f(5 \times 2^n) = 5 \times 2^{n-1} – \left(\frac{2^{n-1}-1}{2-1}\right) = 5 \times 2^{n-1} – 2^{n-1} + 1 = 2^{n-1}(5-1) + 1 $
$f(5 \times 2^n)= 2^{n-1}\times 2^2 + 1 = 2^{n+1} + 1$
$\textbf{Option (D)}:$
Since, $2^n + 1$ is an odd value for $n \geq 1$, So, I can get the values for $f(2^n + 1)$ from the recurrence which is defined for odd values of $f.$ i.e. $f(2n+1)=2f(n) + 1$ which can be written as $f(n)= f(\frac{n-1}{2}) + 1$
So,
$f(2^n +1 ) = 2f (\frac{2^n +1 -1}{2})+1 = 2 f(2^{n-1})+1 = 2\times1 +1 = 3$
Hence, $\textbf{Answer: A,B,C}$ (By putting values of $n$ and evaluating $f(.)$, we can eliminate some option also here.)