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Standard XII

Mathematics

Sum of n Terms

Let f be a fu...

Question

Let $f$ be a function defined on the set of all positive integers such that $f (x y) = f (x) + f (y)$ for all positive integers $x, y$ . If $f (12) = 24$ and $f (8) = 15$ , the value of $f (48)$ is

31

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33

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34

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Solution

The correct option is D $34$
$Let x = 2, y = 2 f (4) = f (2) + f (2) \Rightarrow f (4) = 2 f (2) \Rightarrow f (2 \times 4) = f (2) + f (4) = 15 \Rightarrow 3 f (2) = 15 \Rightarrow f (2) = 5 Now, f (48) = f (12 \times 4) = f (12) + f (4) = 24 + 2 \times 5 = 34$

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Similar questions

Q. Let f be a function defined on the set of all positive integers such that

f (x y) = f (x) + f (y)

for all positive integers

x, y

. If

f (12) = 24

and

f (8) = 15

, the value of

f (48)

Q. Let

f : Z \to Z

be a function satisfying

f (0) \neq 0, f (1) = 0

and

f (x y) + f (x) f (y) = f (x) + f (y)

;

(\begin{matrix} f (x - y) - f (0) \end{matrix}) f (x) f (y) = 0

,
for all

x, y, \in Z

, simultaneously.
(a) Find the set of all possible values of the function

f

.
(b) If

f (10) \neq 0 a n d f (2) = 0

, find the set of all integers

n

such that

f (n) \neq 0

Q. Let

f

be a function satisfying

f (x y) = \frac{f (x)}{y}

for all positive real numbers

x

and

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f (30) = 20

, then the value of

f (40)

Q. Let

f : N \to N

(

N

being the set of positive integers) be a function defined by

f (x) =

the biggest positive integer obtained by reshuffling the digits of

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f

Let $f : R^{+} \to R^{+}$ be a differentiable function satisfying $f (x y) = \frac{f (x)}{y} + \frac{f (y)}{x}$ for all $x, y ϵ R^{+}$ . Also $f (1) = 0, f^{'} (1) = 1$ . If M is the greatest value of $f (x)$ then $[m + e]$ is ___ (where [.] represents Greatest Integer Function).

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Let f be a function defined on the set of all positive integers such that f(xy)=f(x)+f(y) for all positive integers x,y. If f(12)=24 and f(8)=15, the value of f(48) is

The correct option is D 34 Let x=2,y=2f(4)=f(2)+f(2)⇒f(4)=2f(2)⇒f(2×4)=f(2)+f(4)=15⇒3f(2)=15⇒f(2)=5Now, f(48)=f(12×4)=f(12)+f(4)=24+2×5=34

Let $f$ be a function defined on the set of all positive integers such that $f (x y) = f (x) + f (y)$ for all positive integers $x, y$ . If $f (12) = 24$ and $f (8) = 15$ , the value of $f (48)$ is

The correct option is D $34$
$Let x = 2, y = 2 f (4) = f (2) + f (2) \Rightarrow f (4) = 2 f (2) \Rightarrow f (2 \times 4) = f (2) + f (4) = 15 \Rightarrow 3 f (2) = 15 \Rightarrow f (2) = 5 Now, f (48) = f (12 \times 4) = f (12) + f (4) = 24 + 2 \times 5 = 34$