A real valued function f is such that f(a + b) = a + f[f(b)]- Find the...
In the given expression substitute the value of a as zero.
f(0 + b) = 0 + f(f{b))
f(f(b)) = f(b)
Therefore, we can see that f(x) = x
f(9) = 9
Hence, option 4.
A real valued function f is such that f(a + b) = a + f[f(b)]- Find the...
To find the value of f(9), we need to substitute a = 9 and b = 9 into the given function f(a b) = a f[f(b)].
Substituting a = 9 and b = 9, we get:
f(9 9) = 9 f[f(9)]
Now, let's simplify the expression step by step:
1. f(9 9) = 9 f[f(9)]
2. f(18) = 9 f[f(9)]
3. Since f(9) is still unknown, let's denote it as x:
f(18) = 9 f(x)
4. Now, let's substitute a = 18 and b = 9 into the given function f(a b) = a f[f(b)]:
f(18 9) = 18 f[f(9)]
5. Simplifying further:
f(27) = 18 f[f(9)]
6. Since f(9) is still unknown, let's denote it as x:
f(27) = 18 f(x)
7. Now, let's substitute a = 27 and b = 9 into the given function f(a b) = a f[f(b)]:
f(27 9) = 27 f[f(9)]
8. Simplifying further:
f(36) = 27 f[f(9)]
9. Since f(9) is still unknown, let's denote it as x:
f(36) = 27 f(x)
10. We can see a pattern here:
f(18) = 9 f(x)
f(27) = 18 f(x)
f(36) = 27 f(x)
11. From the pattern, we can observe that f(n) = (n-9) f(x), where n is a positive integer greater than 9.
12. Now, let's substitute n = 9 into the pattern:
f(9) = (9-9) f(x)
f(9) = 0 f(x)
13. Finally, substituting f(9) = 0 into the equation f(36) = 27 f(x):
0 = 27 f(x)
14. Since 27 f(x) equals zero, f(x) must be zero as well.
15. Therefore, f(9) = 0.
Hence, the correct answer is option 'D' - 9.