Functional Equations Questions in English

(NONE) Given the equation: $3 f(x) + 4 f\left(\frac{1}{x}\right) = \frac{2-x}{x} \quad \dots (1)$
Substitute $x = 3$ into equation $(1)$:
$3 f(3) + 4 f\left(\frac{1}{3}\right) = \frac{2-3}{3} = -\frac{1}{3} \quad \dots (2)$
Now,substitute $x = \frac{1}{3}$ into equation $(1)$:
$3 f\left(\frac{1}{3}\right) + 4 f(3) = \frac{2 - 1/3}{1/3} = \frac{5/3}{1/3} = 5 \quad \dots (3)$
From equation $(2)$,we have $4 f\left(\frac{1}{3}\right) = -\frac{1}{3} - 3 f(3)$,so $f\left(\frac{1}{3}\right) = -\frac{1}{12} - \frac{3}{4} f(3)$.
Substitute this into equation $(3)$:
$3 \left(-\frac{1}{12} - \frac{3}{4} f(3)\right) + 4 f(3) = 5$
$-\frac{1}{4} - \frac{9}{4} f(3) + 4 f(3) = 5$
Multiply by $4$ to clear the denominator:
$-1 - 9 f(3) + 16 f(3) = 20$
$7 f(3) = 21$
$f(3) = 3$
Wait,re-evaluating the calculation:
$3 f(3) + 4 f(1/3) = -1/3$
$4 f(3) + 3 f(1/3) = 5$
Multiply first by $3$ and second by $4$:
$9 f(3) + 12 f(1/3) = -1$
$16 f(3) + 12 f(1/3) = 20$
Subtracting the first from the second:
$7 f(3) = 21 \implies f(3) = 3$.

(B) Given that $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$.
We can rewrite the expression as $f\left(3x + \frac{1}{2x}\right) = (3x)^2 + \left(\frac{1}{2x}\right)^2 + 2(3x)\left(\frac{1}{2x}\right) - 3$.
This simplifies to $f\left(3x + \frac{1}{2x}\right) = \left(3x + \frac{1}{2x}\right)^2 - 3$.
Therefore,the general form of the function is $f(t) = t^2 - 3$.
Given $f\left(x + \frac{1}{x}\right) = 1$,we substitute into the function: $\left(x + \frac{1}{x}\right)^2 - 3 = 1$.
This gives $\left(x + \frac{1}{x}\right)^2 = 4$,so $x + \frac{1}{x} = \pm 2$.
Case $1$: $x + \frac{1}{x} = 2 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1$.
Case $2$: $x + \frac{1}{x} = -2 \Rightarrow x^2 + 2x + 1 = 0 \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1$.
Thus,$x = \pm 1$.

(D) Given $f(x+y)=f(x)+f(y)$ and $f(1)=7$.
By induction,for any positive integer $r$,$f(r) = r \cdot f(1)$.
Since $f(1)=7$,we have $f(r) = 7r$.
Now,we need to calculate the sum $\sum_{r=1}^n f(r)$.
$\sum_{r=1}^n f(r) = \sum_{r=1}^n 7r = 7 \sum_{r=1}^n r$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{r=1}^n r = \frac{n(n+1)}{2}$.
Therefore,$\sum_{r=1}^n f(r) = 7 \cdot \frac{n(n+1)}{2} = \frac{7n(n+1)}{2}$.

(A) Given,$f(x) = \frac{3^x + 3^{-x}}{2}$.
We have $f(x+y) = \frac{3^{x+y} + 3^{-(x+y)}}{2}$ and $f(x-y) = \frac{3^{x-y} + 3^{-(x-y)}}{2}$.
Adding these two expressions:
$f(x+y) + f(x-y) = \frac{1}{2} [3^x \cdot 3^y + 3^{-x} \cdot 3^{-y} + 3^x \cdot 3^{-y} + 3^{-x} \cdot 3^y]$.
$f(x+y) + f(x-y) = \frac{1}{2} [3^x(3^y + 3^{-y}) + 3^{-x}(3^y + 3^{-y})]$.
$f(x+y) + f(x-y) = \frac{1}{2} (3^x + 3^{-x})(3^y + 3^{-y})$.
Since $f(x) = \frac{3^x + 3^{-x}}{2}$,we have $(3^x + 3^{-x}) = 2f(x)$ and $(3^y + 3^{-y}) = 2f(y)$.
Substituting these into the equation:
$f(x+y) + f(x-y) = \frac{1}{2} [2f(x) \cdot 2f(y)] = 2f(x)f(y)$.
Comparing this with $f(x+y) + f(x-y) = a f(x) f(y)$,we get $a = 2$.

(C) Given,$f(x+y)=f(x)+f(y)$ for all $x, y \in R$.
This is Cauchy's functional equation,and for $f: R \rightarrow R$,the solution is $f(x)=cx$.
Given $f(1)=10$,we have $c(1)=10$,so $c=10$.
Thus,$f(x)=10x$.
We need to find $\sum_{r=1}^n(f(r))^2 = \sum_{r=1}^n(10r)^2$.
$= 100 \sum_{r=1}^n r^2$.
Using the formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$= 100 \times \frac{n(n+1)(2n+1)}{6}$.
$= \frac{50}{3} n(n+1)(2n+1)$.

(B) Given $f(1)=0$ and $f(n+1)-f(n)=5n$ for all $n \in N$.
We can write the recurrence relation as $f(k+1)-f(k)=5k$.
Summing this relation from $k=1$ to $n-1$:
$\sum_{k=1}^{n-1} (f(k+1)-f(k)) = \sum_{k=1}^{n-1} 5k$.
This is a telescoping sum on the left side:
$(f(2)-f(1)) + (f(3)-f(2)) + \ldots + (f(n)-f(n-1)) = 5 \sum_{k=1}^{n-1} k$.
$f(n)-f(1) = 5 \times \frac{(n-1)n}{2}$.
Since $f(1)=0$,we have $f(n) = \frac{5}{2}(n^2-n)$.

(C) Given,$2f(x) + f\left(\frac{1}{x}\right) = 4x$ --- $(i)$
Replacing $x$ with $\frac{1}{x}$,we get:
$2f\left(\frac{1}{x}\right) + f(x) = \frac{4}{x}$ --- (ii)
Multiply equation $(i)$ by $2$:
$4f(x) + 2f\left(\frac{1}{x}\right) = 8x$ --- (iii)
Subtracting equation (ii) from equation (iii):
$(4f(x) + 2f\left(\frac{1}{x}\right)) - (f(x) + 2f\left(\frac{1}{x}\right)) = 8x - \frac{4}{x}$
$3f(x) = 8x - \frac{4}{x} = \frac{8x^2 - 4}{x}$
$f(x) = \frac{4(2x^2 - 1)}{3x}$
Now,for $S = \{x \in R : f(x) = f(-x)\}$:
$f(-x) = \frac{4(2(-x)^2 - 1)}{3(-x)} = -\frac{4(2x^2 - 1)}{3x} = -f(x)$
Setting $f(x) = f(-x)$ implies $f(x) = -f(x)$,which means $2f(x) = 0$,so $f(x) = 0$.
$\frac{4(2x^2 - 1)}{3x} = 0 \implies 2x^2 - 1 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$.
Thus,$S = \left\{-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\}$.
The number of elements in $S$ is $2$.

(B) Given,$f(x+y)=f(x) \cdot f(y)$. This is a functional equation whose solution is of the form $f(x)=a^x$.
Given $f(1)=2$,so $a^1=2 \Rightarrow a=2$. Thus,$f(x)=2^x$.
Now,consider the region enclosed by $2|x|+5|y| \leq 4$. This represents a rhombus with vertices at $(\pm 2, 0)$ and $(0, \pm 4/5)$.
The area of the rhombus is given by $4 \times \text{Area of one triangle in the first quadrant}$.
Area $= 4 \times (\frac{1}{2} \times 2 \times \frac{4}{5}) = 4 \times \frac{4}{5} = \frac{16}{5}$.
We need to express $\frac{16}{5}$ in terms of $f(1)=2^1=2$,$f(2)=2^2=4$,and $f(4)=2^4=16$.
Note that $1+f(2) = 1+4 = 5$.
Thus,the area is $\frac{16}{5} = \frac{f(4)}{1+f(2)}$.

(C) Given $f(x) = x - \frac{1}{x}$.
We know the algebraic identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$.
Applying this to $f(x)$:
$[f(x)]^3 = \left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3(x)\left(\frac{1}{x}\right)\left(x - \frac{1}{x}\right)$.
Since $f(x^3) = x^3 - \frac{1}{x^3}$ and $f(x) = x - \frac{1}{x}$,we substitute these into the equation:
$[f(x)]^3 = f(x^3) - 3(1)f(x)$.
Rearranging the terms to solve for $3f(x)$:
$3f(x) = f(x^3) - [f(x)]^3$.
Thus,the correct option is $C$.

(B) Given $f(x+y) = f(x) \cdot f(y)$ for all $x, y \in R$.
This is a functional equation of the form $f(x) = a^x$.
Given $f(2) = 9$,we have $a^2 = 9$,which implies $a = 3$ (since $f$ is non-zero).
Thus,$f(x) = 3^x$.
Therefore,$f(6) = 3^6$.

(D) Let $f(x) = ax + b$.
Given the equation $f(f(x)) = x + f(x)$.
Substituting $f(x)$ into the equation:
$a(ax + b) + b = x + (ax + b)$
$a^2x + ab + b = x + ax + b$
$a^2x + ab = x + ax$
$x(a^2 - a - 1) + ab = 0$.
For this to hold for all $x$,the coefficients must be zero:
$a^2 - a - 1 = 0$ and $ab = 0$.
Since $a^2 - a - 1 = 0$,$a$ cannot be $0$,therefore $b = 0$.
Solving $a^2 - a - 1 = 0$ using the quadratic formula:
$a = \frac{1 \pm \sqrt{5}}{2}$.
Thus,the functions are $f(x) = \left(\frac{1 + \sqrt{5}}{2}\right)x$ and $f(x) = \left(\frac{1 - \sqrt{5}}{2}\right)x$.
There are $2$ such real linear functions.

(A) Given the functional equation $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$.
This is a Jensen's functional equation,which implies that $f(x)$ is a linear function of the form $f(x) = ax + b$.
We are given $f(0) = 1$.
Substituting $x = 0$ into $f(x) = ax + b$,we get $f(0) = a(0) + b = 1$,so $b = 1$.
Now,$f(x) = ax + 1$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = a$.
We are given $f^{\prime}(0) = -1$.
Since $f^{\prime}(x) = a$ is a constant,$f^{\prime}(0) = a = -1$.
Thus,the function is $f(x) = -x + 1$.
To find $f(2)$,substitute $x = 2$ into the function:
$f(2) = -(2) + 1 = -1$.

(A) Given the functional equation $f(x)f(y) = f(x+y)$.
Since $f$ is injective,the solution to this Cauchy functional equation is of the form $f(x) = a^{kx}$ for some constant $a > 0, a \neq 1$ and $k \neq 0$.
Given that $f(x), f(y),$ and $f(z)$ are in $GP$,we have $(f(y))^2 = f(x)f(z)$.
Substituting $f(x) = a^{kx}$,we get $(a^{ky})^2 = a^{kx} \cdot a^{kz}$.
This simplifies to $a^{2ky} = a^{k(x+z)}$.
Since $a \neq 1$ and $k \neq 0$,we equate the exponents: $2ky = k(x+z)$.
Dividing by $k$,we get $2y = x+z$.
This condition implies that $x, y,$ and $z$ are in $AP$.

(A) Given that $f: R \rightarrow R$ is an injective function satisfying $f(x) f(y) = f(x+y)$ for all $x, y \in R$.
This functional equation is satisfied by the exponential function $f(x) = a^x$ for some $a > 0, a \neq 1$.
Since $f(x), f(y), f(z)$ are in $G$.$P$.,we have $(f(y))^2 = f(x) \cdot f(z)$.
Substituting $f(x) = a^x$,we get $(a^y)^2 = a^x \cdot a^z$.
This simplifies to $a^{2y} = a^{x+z}$.
Equating the exponents,we get $2y = x+z$.
This condition implies that $x, y, z$ are in $A$.$P$.

(A) Given the functional equation $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+f(0)}{3}$.
Setting $x=0$ and $y=0$,we get $f(0) = \frac{f(0)+f(0)+f(0)}{3} = f(0)$,which is always true.
Let $f(0) = c$. Then $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+c}{3}$.
Setting $y=0$,we get $f\left(\frac{x}{3}\right) = \frac{f(x)+f(0)+c}{3} = \frac{f(x)+2c}{3}$.
This implies $f(x) = 3f\left(\frac{x}{3}\right) - 2c$.
Since $f$ is differentiable at $x=0$,let $f'(0) = a$.
Using the definition of the derivative,$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$.
By differentiating the functional equation with respect to $x$,we find that $f(x)$ must be a linear function of the form $f(x) = ax+c$.
Substituting $f(x) = ax+c$ into the original equation: $a\left(\frac{x+y}{3}\right)+c = \frac{(ax+c)+(ay+c)+c}{3} = \frac{a(x+y)+3c}{3} = a\left(\frac{x+y}{3}\right)+c$.
This holds for all $x, y$. Thus,$f(x)$ is a linear function.

(D) Given $f(x) = u x^2 + v x + w$.
Since $f(x+y) = f(x) + f(y) + x y$,we substitute the expression for $f$:
$u(x+y)^2 + v(x+y) + w = (u x^2 + v x + w) + (u y^2 + v y + w) + x y$.
Expanding the left side:
$u(x^2 + 2 x y + y^2) + v x + v y + w = u x^2 + u y^2 + v x + v y + 2 w + x y$.
$u x^2 + 2 u x y + u y^2 + v x + v y + w = u x^2 + u y^2 + v x + v y + 2 w + x y$.
Comparing the coefficients of $x y$ on both sides,we get $2 u = 1$,which implies $u = \frac{1}{2}$.
Comparing the constant terms,we get $w = 2 w$,which implies $w = 0$.
Given $u + v + w = 3$,we substitute $u = \frac{1}{2}$ and $w = 0$:
$\frac{1}{2} + v + 0 = 3 \implies v = 3 - \frac{1}{2} = \frac{5}{2}$.
Thus,$f(x) = \frac{1}{2} x^2 + \frac{5}{2} x$.
Calculating $f(1)$:
$f(1) = \frac{1}{2}(1)^2 + \frac{5}{2}(1) = \frac{1}{2} + \frac{5}{2} = \frac{6}{2} = 3$.

(C) Let $u = x+2y$ and $v = x-2y$.
Adding the two equations: $u+v = (x+2y) + (x-2y) = 2x$,which implies $x = \frac{u+v}{2}$.
Subtracting the two equations: $u-v = (x+2y) - (x-2y) = 4y$,which implies $y = \frac{u-v}{4}$.
Given $f(x+2y, x-2y) = xy$,we substitute $u$ and $v$ to get $f(u, v) = \left(\frac{u+v}{2}\right) \left(\frac{u-v}{4}\right)$.
Simplifying the expression: $f(u, v) = \frac{u^2-v^2}{8}$.
Replacing $u$ with $x$ and $v$ with $y$,we get $f(x, y) = \frac{x^2-y^2}{8}$.

(C) Given: $2f(x) + 3f(-x) = 15 - 4x$ ...$(1)$
Replace $x$ with $-x$ in equation $(1)$:
$2f(-x) + 3f(x) = 15 - 4(-x) = 15 + 4x$ ...$(2)$
Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$4f(x) + 6f(-x) = 30 - 8x$ ...$(3)$
$6f(x) + 9f(-x) = 45 + 12x$ ...$(4)$
Subtract equation $(3)$ from equation $(4)$:
$(6f(x) - 4f(x)) + (9f(-x) - 6f(-x)) = (45 + 12x) - (30 - 8x)$
$2f(x) + 3f(-x) = 15 + 20x$ (This approach is complex,let's solve by substitution)
From $(1)$,$3f(-x) = 15 - 4x - 2f(x) \implies f(-x) = \frac{15 - 4x - 2f(x)}{3}$
Substitute into $(2)$: $2(\frac{15 - 4x - 2f(x)}{3}) + 3f(x) = 15 + 4x$
$30 - 8x - 4f(x) + 9f(x) = 45 + 12x$
$5f(x) = 15 + 20x$
$f(x) = 3 + 4x$
Therefore,$f(2) = 3 + 4(2) = 3 + 8 = 11$.

(C) Given $f(1)=1$ and $f(1)+2 f(2)+3 f(3)+\ldots+n f(n)=n(n+1) f(n)$ for $n \geq 2$.
Let $S_n = \sum_{k=1}^{n} k f(k)$. Then $S_n = n(n+1) f(n)$.
For $n \geq 2$,$S_n = S_{n-1} + n f(n) = n(n+1) f(n)$.
Substituting $S_{n-1} = (n-1)n f(n-1)$,we get $(n-1)n f(n-1) + n f(n) = n(n+1) f(n)$.
Dividing by $n$ $(n \geq 2)$,we get $(n-1) f(n-1) + f(n) = (n+1) f(n)$.
$(n-1) f(n-1) = n f(n) \Rightarrow f(n) = \frac{n-1}{n} f(n-1)$.
For $n=2$,$f(2) = \frac{1}{2} f(1) = \frac{1}{2}$.
For $n=3$,$f(3) = \frac{2}{3} f(2) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$.
In general,$f(n) = \frac{1}{n}$.
Thus,$f(500) = \frac{1}{500}$.

(C) Given $m = \sum_{i=1}^9 i^2 = \frac{9(9+1)(2 \times 9 + 1)}{6} = \frac{9 \times 10 \times 19}{6} = 285$.
Since $m = 285$,the expression $\frac{m}{19x}$ becomes $\frac{285}{19x} = \frac{15}{x}$.
The given equation is $3f(x) + 2f\left(\frac{15}{x}\right) = 5x$.
Replacing $x$ with $\frac{15}{x}$,we get $3f\left(\frac{15}{x}\right) + 2f(x) = 5\left(\frac{15}{x}\right) = \frac{75}{x}$.
We have a system of two equations:
$1) 3f(x) + 2f\left(\frac{15}{x}\right) = 5x$
$2) 2f(x) + 3f\left(\frac{15}{x}\right) = \frac{75}{x}$
Multiply $(1)$ by $3$ and $(2)$ by $2$:
$9f(x) + 6f\left(\frac{15}{x}\right) = 15x$
$4f(x) + 6f\left(\frac{15}{x}\right) = \frac{150}{x}$
Subtracting the second from the first:
$5f(x) = 15x - \frac{150}{x} \implies f(x) = 3x - \frac{30}{x}$.
Now,$f(5) = 3(5) - \frac{30}{5} = 15 - 6 = 9$.
$f(2) = 3(2) - \frac{30}{2} = 6 - 15 = -9$.
Therefore,$f(5) - f(2) = 9 - (-9) = 18$.

(B) Given $f(x+y)=f(x)f(y)$ with $f(1)=7$. This is a functional equation of the form $f(x)=a^x$. Since $f(1)=7$,we have $a^1=7$,so $f(x)=7^x$.
Given $g(x+y)=g(xy)$ with $g(1)=1$. Putting $y=1$,we get $g(x+1)=g(x)$. Since $g(1)=1$,it follows that $g(x)=1$ for all $x \in \mathbb{N}$.
The given summation is $\sum_{x=1}^{n} \frac{f(x)}{g(x)} = 19607$.
Substituting the functions,we get $\sum_{x=1}^{n} \frac{7^x}{1} = 19607$.
This is a geometric progression: $7 + 7^2 + \dots + 7^n = 19607$.
The sum of a $GP$ is $S_n = a\frac{r^n-1}{r-1}$. Here $a=7$ and $r=7$.
$7 \left(\frac{7^n-1}{7-1}\right) = 19607$.
$7 \left(\frac{7^n-1}{6}\right) = 19607$.
$7^n-1 = \frac{19607 \times 6}{7} = 2801 \times 6 = 16806$.
$7^n = 16807$.
Since $7^5 = 16807$,we have $n=5$.

(C) The infinite geometric series is $S = 2 + \frac{2}{3} + \frac{2}{9} + \dots = \frac{2}{1 - 1/3} = \frac{2}{2/3} = 3$.
Given equation: $\log_2(f(x)) = 3 \log_3(1 + \frac{f(x)}{f(1/x)})$.
Let $f(x) = x^2 + 1$. Given $f(6) = 6^2 + 1 = 37$,which satisfies the condition.
Checking the functional equation: $f(1/x) = \frac{1}{x^2} + 1 = \frac{1+x^2}{x^2}$.
Then $1 + \frac{f(x)}{f(1/x)} = 1 + \frac{x^2+1}{(1+x^2)/x^2} = 1 + x^2$.
The equation becomes $\log_2(x^2+1) = 3 \log_3(1+x^2)$. This is satisfied if $x^2+1 = 1$ (not possible) or if the base of the logarithm was intended to be consistent. Assuming $f(x) = x^2+1$ is the intended polynomial,the sum is $\sum_{n=1}^{10} (n^2+1) = \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} 1 = \frac{10(11)(21)}{6} + 10 = 385 + 10 = 395$.