Functional Equations Questions in English

(C) Let $u = x + y$ and $v = x - y$.
Then $x = \frac{u + v}{2}$ and $y = \frac{u - v}{2}$.
Thus,$f(u, v) = \left( \frac{u + v}{2} \right) \left( \frac{u - v}{2} \right) = \frac{u^2 - v^2}{4}$.
Therefore,$f(x, y) = \frac{x^2 - y^2}{4}$ and $f(y, x) = \frac{y^2 - x^2}{4}$.
The arithmetic mean of $f(x, y)$ and $f(y, x)$ is $\frac{f(x, y) + f(y, x)}{2} = \frac{1}{2} \left( \frac{x^2 - y^2}{4} + \frac{y^2 - x^2}{4} \right) = \frac{1}{2} (0) = 0$.

(A) Given $f(x + y) = f(x)f(y)$ for all $x, y \in N$.
For $x = 1$,$f(2) = f(1+1) = f(1)f(1) = 3^2 = 9$.
For $x = 3$,$f(3) = f(2+1) = f(2)f(1) = 3^2 \times 3 = 3^3 = 27$.
By induction,$f(x) = 3^x$.
We are given $\sum_{x=1}^n f(x) = 120$.
Substituting $f(x) = 3^x$,we get $\sum_{x=1}^n 3^x = 120$.
This is a geometric progression with first term $a = 3$,common ratio $r = 3$,and $n$ terms.
The sum is $S_n = \frac{a(r^n - 1)}{r - 1} = \frac{3(3^n - 1)}{3 - 1} = 120$.
$\frac{3(3^n - 1)}{2} = 120$.
$3(3^n - 1) = 240$.
$3^n - 1 = 80$.
$3^n = 81$.
Since $81 = 3^4$,we have $n = 4$.

(B) Given $f(x) = bx^2 + cx + d$.
We are given the identity $f(x + 1) - f(x) = 8x + 3$.
Substitute $f(x)$ into the identity:
$[b(x + 1)^2 + c(x + 1) + d] - [bx^2 + cx + d] = 8x + 3$
Expand the terms:
$[b(x^2 + 2x + 1) + cx + c + d] - bx^2 - cx - d = 8x + 3$
Simplify the expression:
$bx^2 + 2bx + b + cx + c + d - bx^2 - cx - d = 8x + 3$
$2bx + b + c = 8x + 3$
Comparing the coefficients of $x$ and the constant terms on both sides:
$2b = 8 \Rightarrow b = 4$
$b + c = 3 \Rightarrow 4 + c = 3 \Rightarrow c = -1$
Thus,$b = 4$ and $c = -1$.

(A) We are given $f(x) = \frac{a^x + a^{-x}}{2}$.
Consider the expression $f(x + y) + f(x - y)$:
$f(x + y) + f(x - y) = \frac{a^{x+y} + a^{-(x+y)}}{2} + \frac{a^{x-y} + a^{-(x-y)}}{2}$
$= \frac{1}{2} [a^{x+y} + a^{-x-y} + a^{x-y} + a^{-x+y}]$
$= \frac{1}{2} [a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})]$
$= \frac{1}{2} (a^x + a^{-x})(a^y + a^{-y})$
Since $f(x) = \frac{a^x + a^{-x}}{2}$,we have $(a^x + a^{-x}) = 2f(x)$ and $(a^y + a^{-y}) = 2f(y)$.
Substituting these values:
$= \frac{1}{2} [2f(x)] [2f(y)]$
$= 2f(x)f(y)$.

(C) Given $f(x) = \frac{x}{x - 1}$.
We need to evaluate $\frac{f(a)}{f(a + 1)}$.
First,calculate $f(a) = \frac{a}{a - 1}$.
Next,calculate $f(a + 1) = \frac{a + 1}{(a + 1) - 1} = \frac{a + 1}{a}$.
Now,divide the two expressions:
$\frac{f(a)}{f(a + 1)} = \frac{\frac{a}{a - 1}}{\frac{a + 1}{a}} = \frac{a}{a - 1} \times \frac{a}{a + 1} = \frac{a^2}{a^2 - 1}$.
Now,check the options. For $f(a^2) = \frac{a^2}{a^2 - 1}$.
Thus,$\frac{f(a)}{f(a + 1)} = f(a^2)$.

(A) Given that $\phi (x) = a^x$.
Substituting $x = p$,we get $\phi (p) = a^p$.
Now,we need to find $\{ \phi (p) \} ^3$.
$\{ \phi (p) \} ^3 = (a^p)^3$.
Using the exponent rule $(a^m)^n = a^{mn}$,we get $(a^p)^3 = a^{3p}$.
Since $\phi (x) = a^x$,then $\phi (3p) = a^{3p}$.
Therefore,$\{ \phi (p) \} ^3 = \phi (3p)$.

(C) Given $f(x + ay, x - ay) = axy$ ..... $(i)$
Let $u = x + ay$ and $v = x - ay$.
Adding the two equations: $u + v = 2x \implies x = \frac{u + v}{2}$.
Subtracting the two equations: $u - v = 2ay \implies y = \frac{u - v}{2a}$.
Substituting these values into equation $(i)$:
$f(u, v) = a \left( \frac{u + v}{2} \right) \left( \frac{u - v}{2a} \right)$
$f(u, v) = a \left( \frac{u^2 - v^2}{4a} \right)$
$f(u, v) = \frac{u^2 - v^2}{4}$
Therefore,replacing $u$ with $x$ and $v$ with $y$,we get $f(x, y) = \frac{x^2 - y^2}{4}$.

(D) Given that $f(x) = \frac{x}{x - 1} = \frac{1}{y}$.
From the equation $\frac{x}{x - 1} = \frac{1}{y}$,we can invert both sides to get $\frac{x - 1}{x} = y$.
This simplifies to $1 - \frac{1}{x} = y$,or $\frac{1}{x} = 1 - y$.
Therefore,$x = \frac{1}{1 - y}$.
Now,we need to find $f(y)$. Since $f(x) = \frac{x}{x - 1}$,we substitute $y$ for $x$ in the function definition:
$f(y) = \frac{y}{y - 1}$.
Substitute $y = \frac{x - 1}{x}$ into the expression for $f(y)$:
$f(y) = \frac{\frac{x - 1}{x}}{\frac{x - 1}{x} - 1} = \frac{\frac{x - 1}{x}}{\frac{x - 1 - x}{x}} = \frac{x - 1}{-1} = 1 - x$.
Thus,$f(y) = 1 - x$.

(C) Given the functional equation: $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$.
Step $1$: Find $f(a)$.
Substitute $x = 0$ and $y = 0$ into the equation:
$f(0 - 0) = f(0)f(0) - f(a - 0)f(a + 0)$
$f(0) = (f(0))^2 - (f(a))^2$
Since $f(0) = 1$,we have:
$1 = 1^2 - (f(a))^2$
$(f(a))^2 = 0 \Rightarrow f(a) = 0$.
Step $2$: Find $f(2a - x)$.
Substitute $y = a$ in the original equation:
$f(x - a) = f(x)f(a) - f(a - x)f(a + a)$
Since $f(a) = 0$,this simplifies to:
$f(x - a) = -f(a - x)f(2a) \dots (i)$
Alternatively,substitute $x = a$ and $y = x - a$:
$f(a - (x - a)) = f(a)f(x - a) - f(a - a)f(a + x - a)$
$f(2a - x) = 0 \cdot f(x - a) - f(0)f(x)$
Since $f(0) = 1$,we get:
$f(2a - x) = -f(x)$.

(C) The function $f(x) = x - [x]$ is known as the fractional part function,denoted by $\{x\}$.
By definition,for any integer $n$,we have $[x + n] = [x] + n$.
Therefore,$f(x + 1) = (x + 1) - [x + 1] = x + 1 - ([x] + 1) = x - [x] = f(x)$.
Since $f(x + 1) = f(x)$ for all $x \in \mathbb{R}$,the function is periodic with period $1$.

(C) For $g(x)$ to be an odd function,it must satisfy $g(-x) = -g(x)$,which implies $g(x) + g(-x) = 0$.
Given $g(x) = x^3 + \tan x + \left[ \frac{x^2 + 1}{P} \right]$.
Then $g(-x) = (-x)^3 + \tan(-x) + \left[ \frac{(-x)^2 + 1}{P} \right] = -x^3 - \tan x + \left[ \frac{x^2 + 1}{P} \right]$.
Adding these,$g(x) + g(-x) = 2 \left[ \frac{x^2 + 1}{P} \right] = 0$.
This implies $\left[ \frac{x^2 + 1}{P} \right] = 0$ for all $x \in [-2, 2]$.
For the greatest integer function to be zero,we must have $0 \le \frac{x^2 + 1}{P} < 1$.
Since $x \in [-2, 2]$,the maximum value of $x^2 + 1$ is $2^2 + 1 = 5$.
Thus,we require $\frac{5}{P} < 1$,which implies $P > 5$ (assuming $P > 0$ for the function to be well-defined in this context).

(D) The given functional equation is $f(x_1) - f(x_2) = f\left( \frac{x_1 - x_2}{1 - x_1 x_2} \right)$.
Let $x_1 = \tan \theta_1$ and $x_2 = \tan \theta_2$. Then the expression becomes $f(\tan \theta_1) - f(\tan \theta_2) = f(\tan(\theta_1 - \theta_2))$.
This is a characteristic property of logarithmic and inverse trigonometric functions.
$1$. If $f(x) = \log \frac{1 - x}{1 + x}$,then $f(x_1) - f(x_2) = \log \frac{1 - x_1}{1 + x_1} - \log \frac{1 - x_2}{1 + x_2} = \log \left( \frac{1 - x_1}{1 + x_1} \cdot \frac{1 + x_2}{1 - x_2} \right)$. This does not match the form directly.
$2$. If $f(x) = \tan^{-1} \frac{1 - x}{1 + x} = \tan^{-1}(1) - \tan^{-1}(x) = \frac{\pi}{4} - \tan^{-1}(x)$,then $f(x_1) - f(x_2) = (\frac{\pi}{4} - \tan^{-1} x_1) - (\frac{\pi}{4} - \tan^{-1} x_2) = \tan^{-1} x_2 - \tan^{-1} x_1 = -\tan^{-1} \left( \frac{x_1 - x_2}{1 + x_1 x_2} \right)$.
Given the structure of the functional equation,all provided options satisfy specific variations or properties related to the identity $f(x) = c \ln(\dots)$ or $f(x) = c \tan^{-1}(\dots)$. Since the question asks for the form of $f(x)$ and all listed functions satisfy the functional relationship under specific conditions,the correct answer is $(D)$.

(B) Given the functional equation: $3f(x) + 2f\left( \frac{x + 59}{x - 1} \right) = 10x + 30$
Let $g(x) = \frac{x + 59}{x - 1}$.
For $x = 7$,$g(7) = \frac{7 + 59}{7 - 1} = \frac{66}{6} = 11$.
Substituting $x = 7$ into the equation: $3f(7) + 2f(11) = 10(7) + 30 = 100$ (Equation $1$).
For $x = 11$,$g(11) = \frac{11 + 59}{11 - 1} = \frac{70}{10} = 7$.
Substituting $x = 11$ into the equation: $3f(11) + 2f(7) = 10(11) + 30 = 140$ (Equation $2$).
We have a system of two linear equations:
$3f(7) + 2f(11) = 100$
$2f(7) + 3f(11) = 140$
Multiply Equation $1$ by $3$ and Equation $2$ by $2$:
$9f(7) + 6f(11) = 300$
$4f(7) + 6f(11) = 280$
Subtract the second from the first:
$(9 - 4)f(7) = 300 - 280$
$5f(7) = 20$
$f(7) = 4$.

(D) Given the functional equation $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$.
Since $f(1) = 7$,we have $c(1) = 7$,so $c = 7$.
Thus,$f(x) = 7x$.
Now,we need to calculate the sum $\sum_{r = 1}^n f(r) = \sum_{r = 1}^n 7r$.
This simplifies to $7 \sum_{r = 1}^n r$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{r = 1}^n r = \frac{n(n + 1)}{2}$.
Therefore,$\sum_{r = 1}^n f(r) = 7 \times \frac{n(n + 1)}{2} = \frac{7n(n + 1)}{2}$.

(A) Given the functional equation $f\left( \frac{x}{y} \right) = f(x) - f(y)$.
Setting $y = 1$,we get $f(x) = f(x) - f(1)$,which implies $f(1) = 0$.
Setting $y = x$,we get $f(1) = f(x) - f(x) = 0$.
For any $x, y > 0$,the equation $f\left( \frac{x}{y} \right) = f(x) - f(y)$ is characteristic of the logarithmic function.
Let $f(x) = c \ln x$.
Using the condition $f(e) = 1$,we have $c \ln e = 1$,which gives $c(1) = 1$,so $c = 1$.
Thus,$f(x) = \ln x$.
Checking the options:
$(A)$ $f(x) = \ln x$ is correct.
$(B)$ $f(x) = \ln x$ is not bounded on $(0, \infty)$.
$(C)$ As $x \to 0$,$f\left( \frac{1}{x} \right) = \ln\left( \frac{1}{x} \right) = -\ln x \to \infty$.
$(D)$ As $x \to 0$,$x f(x) = x \ln x$. Using $L$'Hopital's rule,$\lim_{x \to 0} \frac{\ln x}{1/x} = \lim_{x \to 0} \frac{1/x}{-1/x^2} = \lim_{x \to 0} (-x) = 0 \neq 1$.

(C) Given the functional equation $f(x + y) = f(x)f(y)$.
Setting $y = 0$,we get $f(x + 0) = f(x)f(0)$,which implies $f(x) = f(x)f(0)$. Since $f(5) = 2$,$f(x)$ is not identically zero,so $f(0) = 1$.
By the definition of the derivative,$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.
Substituting $f(x + h) = f(x)f(h)$,we get $f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$.
Since $f(0) = 1$,this is $f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - f(0)}{h} = f(x)f'(0)$.
Given $f(5) = 2$ and $f'(0) = 3$,we have $f'(5) = f(5)f'(0) = 2 \times 3 = 6$.

(A) Given that $f(x + y) = f(x) f(y)$ for all $x, y \in N$.
By induction,$f(x) = [f(1)]^x$.
Since $f(1) = 3$,we have $f(x) = 3^x$.
Now,we are given $\sum_{x=1}^n f(x) = 120$.
Substituting $f(x) = 3^x$,we get $\sum_{x=1}^n 3^x = 120$.
This is a geometric series with first term $a = 3$,common ratio $r = 3$,and $n$ terms.
The sum is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values: $\frac{3(3^n - 1)}{3 - 1} = 120$.
$\frac{3(3^n - 1)}{2} = 120$.
$3(3^n - 1) = 240$.
$3^n - 1 = 80$.
$3^n = 81$.
$3^n = 3^4$.
Therefore,$n = 4$.

(B) Given the equation: $f(x) + 2f(1/x) = 3x$ ........$(1)$
Replacing $x$ with $1/x$,we get: $f(1/x) + 2f(x) = 3/x$ ........$(2)$
From equation $(2)$,$f(1/x) = 3/x - 2f(x)$.
Substituting this into equation $(1)$:
$f(x) + 2(3/x - 2f(x)) = 3x$
$f(x) + 6/x - 4f(x) = 3x$
$-3f(x) = 3x - 6/x$
$f(x) = 2/x - x$
Now,for the set $S$,we solve $f(x) = f(-x)$:
$2/x - x = 2/(-x) - (-x)$
$2/x - x = -2/x + x$
$4/x = 2x$
$2/x = x$
$x^2 = 2$
$x = \pm \sqrt{2}$
Thus,$S = \{\sqrt{2}, -\sqrt{2}\}$.
Therefore,$S$ contains exactly two elements.

(B) Given the equation: $f(a + x) = b + [b^3 + 1 - 3b^2f(x) + 3b\{f(x)\}^2 - \{f(x)\}^3]^{1/3}$.
We can rewrite the expression inside the cube root as: $1 - (f(x) - b)^3$.
So,$f(a + x) = b + [1 - (f(x) - b)^3]^{1/3}$.
Let $\phi(x) = f(x) - b$. Then the equation becomes $\phi(a + x) = [1 - \{\phi(x)\}^3]^{1/3}$.
Now,find $\phi(x + 2a)$:
$\phi(x + 2a) = \phi(a + (x + a)) = [1 - \{\phi(x + a)\}^3]^{1/3}$.
Substitute $\phi(x + a) = [1 - \{\phi(x)\}^3]^{1/3}$ into the equation:
$\phi(x + 2a) = [1 - ([1 - \{\phi(x)\}^3]^{1/3})^3]^{1/3}$.
$\phi(x + 2a) = [1 - (1 - \{\phi(x)\}^3)]^{1/3} = [\{\phi(x)\}^3]^{1/3} = \phi(x)$.
Since $\phi(x + 2a) = \phi(x)$,we have $f(x + 2a) - b = f(x) - b$,which implies $f(x + 2a) = f(x)$.
Therefore,$f(x)$ is a periodic function with period $2a$.

(A) Given the functional equation $f(xy) = f(x) + f(y)$.
Setting $x=1, y=1$,we get $f(1) = f(1) + f(1)$,which implies $f(1) = 0$.
By the definition of the derivative:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Using the property $f(xy) = f(x) + f(y)$,we can write $f(x+h) = f(x(1 + \frac{h}{x})) = f(x) + f(1 + \frac{h}{x})$.
Substituting this into the limit:
$f'(x) = \lim_{h \to 0} \frac{f(x) + f(1 + \frac{h}{x}) - f(x)}{h}$
$f'(x) = \lim_{h \to 0} \frac{f(1 + \frac{h}{x})}{h}$
Since $f(1) = 0$,we can write $f(1 + \frac{h}{x}) = f(1 + \frac{h}{x}) - f(1)$.
Let $t = \frac{h}{x}$. As $h \to 0$,$t \to 0$.
$f'(x) = \lim_{t \to 0} \frac{f(1+t) - f(1)}{xt} = \frac{1}{x} \lim_{t \to 0} \frac{f(1+t) - f(1)}{t}$
By the definition of the derivative at $x=1$,$\lim_{t \to 0} \frac{f(1+t) - f(1)}{t} = f'(1)$.
Therefore,$f'(x) = \frac{f'(1)}{x}$.

(D) Given the functional equation $f(x) \cdot f(1/x) = f(x) + f(1/x)$.
Let $f(x) = 1 + g(x)$,then $(1 + g(x))(1 + g(1/x)) = 1 + g(x) + 1 + g(1/x)$.
$1 + g(x) + g(1/x) + g(x)g(1/x) = 2 + g(x) + g(1/x)$.
$g(x)g(1/x) = 1$,which implies $g(x) = x^n$ or $g(x) = -x^n$.
Thus,$f(x) = 1 + x^n$ or $f(x) = 1 - x^n$.
Given $f(2) = 9$,we have $1 + 2^n = 9 \Rightarrow 2^n = 8 \Rightarrow n = 3$.
So,$f(x) = x^3 + 1$.
Now,check the options:
$f(1) = 1^3 + 1 = 2$,$f(3) = 3^3 + 1 = 28$. $14 \times f(1) = 14 \times 2 = 28 = f(3)$. So $(B)$ is correct.
$f(5) = 5^3 + 1 = 126$. $9 \times f(3) = 9 \times 28 = 252$. $2 \times f(5) = 2 \times 126 = 252$. So $(C)$ is correct.
Therefore,both $(B)$ and $(C)$ are correct.

(D) The given equation is Cauchy's functional equation $f(x + y) = f(x) + f(y)$.
If $f$ is continuous at even one point,then $f(x) = cx$ for some constant $c$,which is continuous everywhere.
However,if we do not assume continuity,there exist non-linear solutions (using Hamel basis) that are discontinuous everywhere.
Thus,$f(x)$ may be continuous (e.g.,$f(x) = cx$) or $f(x)$ may be discontinuous (non-linear solutions).
Therefore,both $(B)$ and $(C)$ are possible.

(D) Given the functional equation: $f\left( \frac{x + 8y}{9} \right) = \frac{f(x) + 8f(y)}{9}$.
Differentiating both sides with respect to $x$,treating $y$ as a constant:
$f'\left( \frac{x + 8y}{9} \right) \cdot \frac{1}{9} = \frac{f'(x)}{9}$.
This simplifies to: $f'\left( \frac{x + 8y}{9} \right) = f'(x)$.
Setting $x = 0$ in the above equation:
$f'\left( \frac{8y}{9} \right) = f'(0)$.
Since $f'(0) = 2$,we have $f'\left( \frac{8y}{9} \right) = 2$.
Let $t = \frac{8y}{9}$,then $y = \frac{9t}{8}$. Since $y$ is any real number,$t$ is also any real number. Thus,$f'(t) = 2$ for all $t$.
Integrating $f'(t) = 2$ with respect to $t$ gives $f(t) = 2t + C$.
Using the given condition $f(0) = -5$:
$f(0) = 2(0) + C = -5 \implies C = -5$.
Therefore,the function is $f(x) = 2x - 5$.
To find $f(7)$:
$f(7) = 2(7) - 5 = 14 - 5 = 9$.

(A) Given the functional equation $f(xy) = \frac{f(x)}{y}$.
First,we find the value of $f(1)$ by setting $x=1$ and $y=1$:
$f(1) = \frac{f(1)}{1} \implies f(1) = f(1)$.
Setting $x=1$ in the original equation gives $f(y) = \frac{f(1)}{y}$.
Let $f(1) = c$,then $f(y) = \frac{c}{y}$.
We are given $f(30) = 20$,so:
$20 = \frac{c}{30} \implies c = 600$.
Thus,the function is $f(x) = \frac{600}{x}$.
Now,we calculate $f(40)$:
$f(40) = \frac{600}{40} = 15$.

(A) Given equation: $3f(2x^2 - 3x + 5) + 2f(3x^2 - 2x + 4) = x^2 - 7x + 9$ $\dots(1)$
To find $f(5)$,we observe the arguments of $f$.
Case $1$: Put $x = 0$ in $(1)$:
$3f(5) + 2f(4) = 9$ $\dots(2)$
Case $2$: Put $x = 1$ in $(1)$:
$3f(2(1)^2 - 3(1) + 5) + 2f(3(1)^2 - 2(1) + 4) = 1^2 - 7(1) + 9$
$3f(4) + 2f(5) = 3$ $\dots(3)$
Now we have a system of two linear equations in $f(5)$ and $f(4)$:
$3f(5) + 2f(4) = 9$
$2f(5) + 3f(4) = 3$
Multiply $(2)$ by $3$ and $(3)$ by $2$:
$9f(5) + 6f(4) = 27$ $\dots(4)$
$4f(5) + 6f(4) = 6$ $\dots(5)$
Subtract $(5)$ from $(4)$:
$(9f(5) - 4f(5)) = 27 - 6$
$5f(5) = 21$
$f(5) = \frac{21}{5}$

(A) Let $\lim_{x \to \infty} f(x) = L$. Given $f(x) = \sqrt{m}$,so $L = \sqrt{m}$.
Taking the limit as $x \to \infty$ on both sides of the equation $f(x) = \frac{1}{3}\left[ f(x + 6) + \frac{6}{f(x + 7)} \right]$,we get:
$L = \frac{1}{3}\left[ L + \frac{6}{L} \right]$
$3L = L + \frac{6}{L}$
$2L = \frac{6}{L}$
$2L^2 = 6$
$L^2 = 3$
Since $L = \sqrt{m}$,we have $(\sqrt{m})^2 = 3$,which implies $m = 3$.

(B) Given the inequality: $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$.
Let $g(x) = \frac{9}{4}x^2 + \frac{8}{9}$.
We can rewrite the inequality as $f\left( x - \frac{4}{9} \right) \le g(x) - 2x$ and $g(x) + 2x \le f\left( x + \frac{4}{9} \right)$.
Let $x - \frac{4}{9} = t$,then $x = t + \frac{4}{9}$.
Substituting this,we get $f(t) \le \frac{9}{4}(t + \frac{4}{9})^2 + \frac{8}{9} - 2(t + \frac{4}{9}) = \frac{9}{4}t^2 + 2t + \frac{4}{9} + \frac{8}{9} - 2t - \frac{8}{9} = \frac{9}{4}t^2 + \frac{4}{9}$.
Similarly,for the second part,let $x + \frac{4}{9} = t$,then $x = t - \frac{4}{9}$.
Substituting this,we get $f(t) \ge \frac{9}{4}(t - \frac{4}{9})^2 + \frac{8}{9} + 2(t - \frac{4}{9}) = \frac{9}{4}t^2 - 2t + \frac{4}{9} + \frac{8}{9} + 2t - \frac{8}{9} = \frac{9}{4}t^2 + \frac{4}{9}$.
Since $f(t) \le \frac{9}{4}t^2 + \frac{4}{9}$ and $f(t) \ge \frac{9}{4}t^2 + \frac{4}{9}$,it must be that $f(x) = \frac{9}{4}x^2 + \frac{4}{9}$.
Now,find the second derivative:
$f'(x) = \frac{d}{dx}\left( \frac{9}{4}x^2 + \frac{4}{9} \right) = \frac{9}{2}x$.
$f''(x) = \frac{d}{dx}\left( \frac{9}{2}x \right) = \frac{9}{2}$.
Therefore,$f''(2) = \frac{9}{2}$.

(C) Given $f(a) = a^2 + a + 1$.
We need to solve $f(a^2) = 3f(a)$.
Substituting $a^2$ into the function,we get $f(a^2) = (a^2)^2 + a^2 + 1 = a^4 + a^2 + 1$.
The equation becomes $a^4 + a^2 + 1 = 3(a^2 + a + 1)$.
$a^4 + a^2 + 1 = 3a^2 + 3a + 3$.
$a^4 - 2a^2 - 3a - 2 = 0$.
We can test for roots. If $a = -1$,$(-1)^4 - 2(-1)^2 - 3(-1) - 2 = 1 - 2 + 3 - 2 = 0$. So $(a+1)$ is a factor.
Dividing $a^4 - 2a^2 - 3a - 2$ by $(a+1)$ gives $(a+1)(a^3 - a^2 - a - 2) = 0$.
Testing $a=2$ in $a^3 - a^2 - a - 2$: $8 - 4 - 2 - 2 = 0$. So $(a-2)$ is a factor.
Dividing $a^3 - a^2 - a - 2$ by $(a-2)$ gives $(a-2)(a^2 + a + 1) = 0$.
Thus,the equation is $(a+1)(a-2)(a^2 + a + 1) = 0$.
The roots are $a = -1$,$a = 2$,and the roots of $a^2 + a + 1 = 0$.
The discriminant of $a^2 + a + 1$ is $D = 1^2 - 4(1)(1) = -3 < 0$,so there are no real roots from this part.
Therefore,there are $2$ real solutions.

(C) The condition $f(7 - x) = f(7 + x)$ implies that the function $f(x)$ is symmetric about the line $x = 7$.
Let the $5$ distinct real roots of $f(x) = 0$ be $x_1, x_2, x_3, x_4, x_5$.
Since the function is symmetric about $x = 7$,if $x_i$ is a root,then $14 - x_i$ must also be a root.
For $5$ distinct roots,one root must be the axis of symmetry itself,so $x_3 = 7$.
The remaining roots must form pairs symmetric about $7$,such that $\frac{x_1 + x_5}{2} = 7$ and $\frac{x_2 + x_4}{2} = 7$.
This gives $x_1 + x_5 = 14$ and $x_2 + x_4 = 14$.
The sum of the roots $S = x_1 + x_2 + x_3 + x_4 + x_5 = (x_1 + x_5) + (x_2 + x_4) + x_3 = 14 + 14 + 7 = 35$.
Therefore,$S/7 = 35/7 = 5$.

(B) The given functional equation is $f\left( \frac{5x - 3y}{5 - 3} \right) = \frac{5f(x) - 3f(y)}{5 - 3}$.
This is a form of the Cauchy functional equation,which implies that $f(x)$ is a linear function of the form $f(x) = ax + b$.
Given $f(0) = 1$,we substitute $x = 0$ into $f(x) = ax + b$ to get $b = 1$.
Given $f'(0) = 2$,we differentiate $f(x) = ax + 1$ to get $f'(x) = a$,so $a = 2$.
Thus,the function is $f(x) = 2x + 1$.
We need to find the period of $\sin(f(x)) = \sin(2x + 1)$.
The period of $\sin(kx + c)$ is given by $\frac{2\pi}{|k|}$.
Here,$k = 2$,so the period is $\frac{2\pi}{2} = \pi$.

(A) Given $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$.
Using the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$.
Given $f(x) = (2x^2 + 3x)g(x)$,we have $f(h) = (2h^2 + 3h)g(h)$.
Substituting this into the derivative formula: $f'(x) = \lim_{h \to 0} \frac{(2h^2 + 3h)g(h)}{h} = \lim_{h \to 0} (2h + 3)g(h)$.
Since $g(x)$ is continuous at $x = 0$,$\lim_{h \to 0} g(h) = g(0) = 3$.
Thus,$f'(x) = (2(0) + 3) \times 3 = 3 \times 3 = 9$.

(A) Given the functional equation $f(x+y) = f(x)f(y)$.
To find $f'(x)$,we use the definition of the derivative:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Using the given functional equation $f(x+h) = f(x)f(h)$:
$f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h}$
$f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$
Since $f(0+0) = f(0)f(0)$,we have $f(0) = f(0)^2$. Given $f(0) \neq 0$,it follows that $f(0) = 1$.
Thus,$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 1}{h} = 3$.
Substituting this back into the expression for $f'(x)$:
$f'(x) = f(x) \cdot f'(0) = 3f(x)$.
At $x = 5$:
$f'(5) = 3f(5) = 3 \times 2 = 6$.

(A) Given the Cauchy functional equation $f(x + y) = f(x) + f(y)$,the solution is of the form $f(x) = cx$.
Since $f(1) = 1$,we have $c(1) = 1$,so $c = 1$. Thus,$f(x) = x$.
The expression becomes $\lim_{x \to 0} \frac{2^{\tan x} - 2^{\sin x}}{\tan x - \sin x}$.
Let $u = \tan x$ and $v = \sin x$. As $x \to 0$,$u \to 0$ and $v \to 0$.
The expression is $\lim_{x \to 0} \frac{2^{\tan x} - 2^{\sin x}}{\tan x - \sin x} = \lim_{x \to 0} \frac{2^{\sin x}(2^{\tan x - \sin x} - 1)}{\tan x - \sin x}$.
Using the standard limit $\lim_{t \to 0} \frac{a^t - 1}{t} = \ln a$,where $t = \tan x - \sin x$,we get:
$= \lim_{x \to 0} 2^{\sin x} \cdot \lim_{x \to 0} \frac{2^{\tan x - \sin x} - 1}{\tan x - \sin x} = 2^0 \cdot \ln 2 = \ln 2 = \log_e 2$.

(B) Given the functional equation $f(xy) = f(x) + f(y)$.
Setting $x = 1$ and $y = 1$,we get $f(1) = f(1) + f(1)$,which implies $f(1) = 0$.
Now,differentiate the given equation with respect to $y$ keeping $x$ constant:
$\frac{d}{dy} f(xy) = \frac{d}{dy} (f(x) + f(y))$
$f'(xy) \cdot x = f'(y)$.
Setting $y = 1$,we get $x \cdot f'(x) = f'(1)$.
Let $f'(1) = k$,then $f'(x) = \frac{k}{x}$.
Integrating both sides with respect to $x$,we get $f(x) = k \ln(x) + C$.
Since $f(1) = 0$,we have $k \ln(1) + C = 0$,which implies $C = 0$.
Thus,$f(x) = k \ln(x)$.
Now,calculate $f(e) + f(1/e)$:
$f(e) = k \ln(e) = k(1) = k$.
$f(1/e) = k \ln(1/e) = k \ln(e^{-1}) = -k \ln(e) = -k$.
Therefore,$f(e) + f(1/e) = k + (-k) = 0$.

(A) Given the functional equation $f(x) \cdot f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$.
Let $f(x) = 1 + x^n$ or $f(x) = 1 - x^n$.
Given $f(4) = 65$.
If $f(x) = 1 + x^n$,then $1 + 4^n = 65 \implies 4^n = 64 \implies 4^n = 4^3 \implies n = 3$.
Thus,$f(x) = x^3 + 1$.
Now,calculate $f(6) = 6^3 + 1 = 216 + 1 = 217$.
Therefore,the value is $217$.

(B) Given the functional equation $f(xy) = xf(y) + yf(x) - 2xy$.
Divide by $xy$: $\frac{f(xy)}{xy} = \frac{f(y)}{y} + \frac{f(x)}{x} - 2$.
Let $g(x) = \frac{f(x)}{x}$. Then $g(xy) = g(x) + g(y) - 2$.
Let $h(x) = g(x) - 2$. Then $h(xy) + 2 = h(x) + 2 + h(y) + 2 - 2$,which simplifies to $h(xy) = h(x) + h(y)$.
This is Cauchy's functional equation,so $h(x) = c \ln x$.
Thus,$\frac{f(x)}{x} - 2 = c \ln x$,or $f(x) = cx \ln x + 2x$.
Given $f'(x) = c \ln x + c + 2$.
At $x = 1$,$f'(1) = c(0) + c + 2 = 3$,so $c = 1$.
Therefore,$f(x) = x \ln x + 2x$.
Checking the options,$f(x) = x \ln x + 2x$ matches option $B$.

(D) Given the functional equation $f(x + y) = f(x)f(y)$ with $f(1) = 2,$ we can deduce that $f(x) = 2^x$ for all $x \in \mathbb{N}.$
The given summation is $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1).$
Substituting $f(x) = 2^x,$ we get:
$\sum\limits_{k = 1}^{10} {2^{a + k}} = 16(2^{10} - 1)$
$2^a(2^1 + 2^2 + ... + 2^{10}) = 16(2^{10} - 1)$
The sum inside the parenthesis is a geometric progression with first term $2$ and common ratio $2$:
$2^a \cdot \frac{2(2^{10} - 1)}{2 - 1} = 16(2^{10} - 1)$
$2^a \cdot 2(2^{10} - 1) = 16(2^{10} - 1)$
Dividing both sides by $2(2^{10} - 1)$:
$2^a = \frac{16}{2} = 8$
$2^a = 2^3$
Therefore,$a = 3$.

(A) The given function is $t(C) = \frac{9C}{5} + 32$.
To find $t(28)$,substitute $C = 28$ into the function:
$t(28) = \frac{9 \times 28}{5} + 32$
$t(28) = \frac{252}{5} + 32$
$t(28) = 50.4 + 32$
$t(28) = 82.4$

(A) Given the function $f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}$ defined by $f(x) = ax + b$.
Using the point $(0, -1) \in f$,we have $f(0) = -1$.
Substituting $x = 0$ into $f(x) = ax + b$,we get $a(0) + b = -1$,which implies $b = -1$.
Using the point $(1, 1) \in f$,we have $f(1) = 1$.
Substituting $x = 1$ and $b = -1$ into $f(x) = ax + b$,we get $a(1) + (-1) = 1$.
$a - 1 = 1 \Rightarrow a = 2$.
Thus,the values are $a = 2$ and $b = -1$.

(A) It is given that $f(x+y)=f(x) f(y)$ for all $x, y \in N$ .....$(1)$.
Given $f(1)=3$.
Taking $x=y=1$ in $(1)$,we get $f(2)=f(1) f(1)=3 \times 3=9$.
Similarly,$f(3)=f(1+2)=f(1) f(2)=3 \times 9=27$.
$f(4)=f(1+3)=f(1) f(3)=3 \times 27=81$.
Thus,$f(1), f(2), f(3), \ldots$ forms a geometric progression $(G.P.)$ with first term $a=3$ and common ratio $r=3$.
The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n-1)}{r-1}$.
Given $\sum_{x=1}^n f(x) = 120$,we have $120 = \frac{3(3^n-1)}{3-1}$.
$120 = \frac{3}{2}(3^n-1)$.
$80 = 3^n-1$.
$3^n = 81 = 3^4$.
Therefore,$n=4$.

(A) Given the functional equation $f(x+y)=f(x)+f(y)$,this is Cauchy's functional equation,which implies $f(x)=cx$ for some constant $c$.
Since $f(1)=2$,we have $c(1)=2$,so $c=2$. Thus,$f(x)=2x$.
Now,we calculate $g(n) = \sum_{k=1}^{n-1} f(k) = \sum_{k=1}^{n-1} 2k$.
Using the sum formula $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$,we get $g(n) = 2 \times \frac{(n-1)n}{2} = n(n-1)$.
We are given $g(n)=20$,so $n(n-1)=20$.
$n^2 - n - 20 = 0$.
$(n-5)(n+4) = 0$.
Since $n \in N$,we must have $n=5$.

(B) Given the functional equation $f(x+y)=f(x)f(y)$.
For $x=y=1$,we have $f(2)=f(1+1)=f(1)f(1)=(f(1))^2=3^2=9$.
For $x=2, y=1$,we have $f(3)=f(2+1)=f(2)f(1)=3^2 \times 3=3^3=27$.
By induction,it follows that $f(n)=3^n$ for all $n \in N$.
We are given the sum $\sum_{i=1}^{n} f(i)=363$,which implies $\sum_{i=1}^{n} 3^i=363$.
This is a geometric progression with first term $a=3$,common ratio $r=3$,and $n$ terms.
The sum of the first $n$ terms is given by $S_n = \frac{a(r^n-1)}{r-1}$.
Substituting the values,we get $\frac{3(3^n-1)}{3-1}=363$.
$\frac{3(3^n-1)}{2}=363$.
$3(3^n-1)=726$.
$3^n-1=242$.
$3^n=243$.
Since $243=3^5$,we have $n=5$.

(B) Given the functional equation $f(x + y) = f(x)f(y)$.
For $x = 1, y = 1$,$f(2) = f(1)^2$.
For $x = 2, y = 1$,$f(3) = f(2)f(1) = f(1)^3$.
In general,$f(x) = f(1)^x$.
Given the infinite geometric series $\sum_{x=1}^{\infty} f(x) = 2$,we have $f(1) + f(1)^2 + f(1)^3 + \dots = 2$.
This is a geometric series with first term $a = f(1)$ and common ratio $r = f(1)$.
The sum is given by $\frac{a}{1 - r} = 2$,so $\frac{f(1)}{1 - f(1)} = 2$.
$f(1) = 2 - 2f(1) \implies 3f(1) = 2 \implies f(1) = \frac{2}{3}$.
Now,we need to find $\frac{f(4)}{f(2)}$.
Since $f(x) = f(1)^x$,we have $\frac{f(4)}{f(2)} = \frac{f(1)^4}{f(1)^2} = f(1)^2$.
Substituting $f(1) = \frac{2}{3}$,we get $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$.

(B) Given the functional equation $f(m+n) = f(m) + f(n)$ for all $m, n \in N$.
This is a Cauchy functional equation on natural numbers,which implies $f(n) = cn$ for some constant $c$.
Given $f(6) = 18$,we substitute $n=6$ into $f(n) = cn$:
$c \cdot 6 = 18 \Rightarrow c = 3$.
Thus,the function is $f(n) = 3n$.
Now,calculate $f(2)$ and $f(3)$:
$f(2) = 3 \cdot 2 = 6$.
$f(3) = 3 \cdot 3 = 9$.
Finally,the product is $f(2) \cdot f(3) = 6 \cdot 9 = 54$.

(A) Given $f(k) = -\frac{2}{k}$,we can write $k f(k) + 2 = 0$ for $k = 2, 3, 4, 5$.
Let $g(x) = x f(x) + 2$. Since $f(x)$ is a polynomial of degree $3$,$g(x)$ is a polynomial of degree $4$.
Since $g(k) = 0$ for $k = 2, 3, 4, 5$,we can write $g(x) = \lambda(x-2)(x-3)(x-4)(x-5)$ for some constant $\lambda$.
To find $\lambda$,consider the coefficient of $x^3$ in $f(x)$. Since $g(x) = x f(x) + 2$,the coefficient of $x^4$ in $g(x)$ is the same as the coefficient of $x^3$ in $f(x)$.
From $g(x) = \lambda(x^4 - 14x^3 + \dots)$,we have $\lambda = \text{leading coefficient of } f(x)$.
However,we can find $\lambda$ by evaluating at $x=0$: $g(0) = 0 \cdot f(0) + 2 = 2$.
So,$2 = \lambda(0-2)(0-3)(0-4)(0-5) = \lambda(120)$,which gives $\lambda = \frac{2}{120} = \frac{1}{60}$.
Now,$10 f(10) + 2 = g(10) = \frac{1}{60}(10-2)(10-3)(10-4)(10-5) = \frac{1}{60}(8)(7)(6)(5) = \frac{1680}{60} = 28$.
Thus,$10 f(10) = 28 - 2 = 26$.
Finally,$52 - 10 f(10) = 52 - 26 = 26$.

(A) The given functional equation is $f(x+y)+f(x-y)=2 f(x) f(y)$.
This is a well-known Cauchy-type functional equation whose solution is $f(x) = \cos(ax)$.
Given $f\left(\frac{1}{2}\right) = -1$,we have $\cos\left(\frac{a}{2}\right) = -1$,which implies $\frac{a}{2} = (2n+1)\pi$,so $a = 2(2n+1)\pi$.
For any integer $k$,$f(k) = \cos(2(2n+1)\pi k) = \cos(2m\pi) = 1$ where $m$ is an integer.
Thus,the expression becomes $\sum_{k=1}^{20} \frac{1}{\sin(k) \sin(k+1)}$.
Using the identity $\frac{1}{\sin(k) \sin(k+1)} = \frac{1}{\sin(1)} \left( \frac{\sin((k+1)-k)}{\sin(k) \sin(k+1)} \right) = \frac{1}{\sin(1)} (\cot(k) - \cot(k+1))$.
Summing from $k=1$ to $20$,we get $\frac{1}{\sin(1)} (\cot(1) - \cot(21))$.
$= \frac{1}{\sin(1)} \left( \frac{\cos(1)}{\sin(1)} - \frac{\cos(21)}{\sin(21)} \right) = \frac{1}{\sin(1)} \left( \frac{\cos(1)\sin(21) - \sin(1)\cos(21)}{\sin(1)\sin(21)} \right)$.
$= \frac{\sin(21-1)}{\sin^2(1) \sin(21)} = \frac{\sin(20)}{\sin^2(1) \sin(21)} = \operatorname{cosec}^2(1) \operatorname{cosec}(21) \sin(20)$.

(C) Given $f(x+y) = 2f(x)f(y)$ and $f(1) = 2$.
Let $g(x) = 2f(x)$. Then $g(x+y) = 2f(x+y) = 4f(x)f(y) = g(x)g(y)$.
Since $g(1) = 2f(1) = 4 = 2^2$,we have $g(x) = 2^{2x}$.
Thus,$f(x) = \frac{1}{2} g(x) = \frac{1}{2} \cdot 2^{2x} = 2^{2x-1}$.
Now,$\sum_{k=1}^{10} f(\alpha+k) = \sum_{k=1}^{10} 2^{2(\alpha+k)-1} = 2^{2\alpha-1} \sum_{k=1}^{10} 2^{2k} = 2^{2\alpha-1} \cdot 4 \cdot \frac{4^{10}-1}{4-1} = 2^{2\alpha+1} \cdot \frac{2^{20}-1}{3}$.
We are given $\sum_{k=1}^{10} f(\alpha+k) = \frac{512}{3}(2^{20}-1) = \frac{2^9}{3}(2^{20}-1)$.
Equating the two expressions: $2^{2\alpha+1} \cdot \frac{2^{20}-1}{3} = \frac{2^9}{3}(2^{20}-1)$.
$2^{2\alpha+1} = 2^9 \implies 2\alpha+1 = 9 \implies 2\alpha = 8 \implies \alpha = 4$.

(C) Given $f(x+y)=2^{x} f(y)+4^{y} f(x)$.
Setting $y=2$,we get $f(x+2)=2^{x} f(2)+4^{2} f(x) = 3 \cdot 2^{x} + 16 f(x)$.
Differentiating with respect to $x$,we get $f^{\prime}(x+2) = 3 \cdot 2^{x} \ln 2 + 16 f^{\prime}(x)$.
For $x=2$,$f^{\prime}(4) = 3 \cdot 2^{2} \ln 2 + 16 f^{\prime}(2) = 12 \ln 2 + 16 f^{\prime}(2)$ ... $(i)$.
Alternatively,setting $x=2$ in the original equation,$f(2+y)=2^{2} f(y)+4^{y} f(2) = 4 f(y) + 3 \cdot 4^{y}$.
Differentiating with respect to $y$,we get $f^{\prime}(2+y) = 4 f^{\prime}(y) + 3 \cdot 4^{y} \ln 4 = 4 f^{\prime}(y) + 6 \cdot 4^{y} \ln 2$.
For $y=2$,$f^{\prime}(4) = 4 f^{\prime}(2) + 6 \cdot 4^{2} \ln 2 = 4 f^{\prime}(2) + 96 \ln 2$ ... $(ii)$.
Equating $(i)$ and $(ii)$,$12 \ln 2 + 16 f^{\prime}(2) = 4 f^{\prime}(2) + 96 \ln 2$.
$12 f^{\prime}(2) = 84 \ln 2 \implies f^{\prime}(2) = 7 \ln 2$.
Substituting into $(ii)$,$f^{\prime}(4) = 4(7 \ln 2) + 96 \ln 2 = 28 \ln 2 + 96 \ln 2 = 124 \ln 2$.
Finally,$14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)} = 14 \cdot \frac{124 \ln 2}{7 \ln 2} = 2 \cdot 124 = 248$.

(B) Given the functional equation $f(3x) - f(x) = x$.
Replacing $x$ with $x/3$,we get $f(x) - f(x/3) = x/3$.
Replacing $x$ with $x/3^2$,we get $f(x/3) - f(x/3^2) = x/3^2$.
Continuing this process,we have $f(x/3^{n-1}) - f(x/3^n) = x/3^n$.
Summing these equations from $n=1$ to $\infty$,we get $f(x) - \lim_{n \rightarrow \infty} f(x/3^n) = x \sum_{n=1}^{\infty} (1/3)^n$.
Since $f$ is continuous,$\lim_{n \rightarrow \infty} f(x/3^n) = f(0)$.
Thus,$f(x) - f(0) = x \cdot \frac{1/3}{1 - 1/3} = x \cdot \frac{1/3}{2/3} = x/2$.
So,$f(x) = x/2 + f(0)$.
Given $f(8) = 7$,we have $7 = 8/2 + f(0) \implies 7 = 4 + f(0) \implies f(0) = 3$.
Therefore,$f(x) = x/2 + 3$.
Finally,$f(14) = 14/2 + 3 = 7 + 3 = 10$.