Question

Mr. X is a teacher of mathematics. His students want to know the ages of his son’s ${S_1}$ and ${S_2}$ . He told that their ages are $'a'$ and $'b'$ respectively such that $f\left( {x + y} \right) - axy = f\left( x \right) + b{y^2}\forall x,y \in R$. After some time students said that information is insufficient, please give more information. Teacher says that $f\left( 1 \right) = 8$ and $f\left( 2 \right) = 32$.

Answer 1

We have given an equation that establishes a relation between the ages of two sons of a mathematics teacher and unknown variables $x$ and $y$. We have to find the ages of two sons. Additional values of the functions are also given. To find the ages of the sum, first we simplify the given equation and then by integrating, first we find the function $f\left( x \right)$.
After that, using the additional information, we will determine the values of the unknown ages of the son’s.

Complete step-by-step answer:
Step1: Separate the alike terms in the given equation
We have given an equation $f\left( {x + y} \right) - axy = f\left( x \right) + b{y^2}\forall x,y \in R$. First, we take out all the functions on the LHS of the equation and all the other terms on the RHS of the equation, we get
$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = axy + b{y^2}$
Now we take $y$ common on LHS, we get
$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = y\left( {ax + by} \right)$
Now we divide by $y$ on both side of the equation and simplify
$\Rightarrow \dfrac{{f\left( {x + y} \right) - f\left( x \right)}}{y} = \dfrac{{y\left( {ax + by} \right)}}{y} \\\ \Rightarrow \dfrac{{f\left( {x + y} \right) - f\left( x \right)}}{y} = ax + by \\\$
Step2: Take limit on both side of equation
Now we take limit on both side of the above equation, we get
$\Rightarrow \mathop {\lim }\limits_{y \to 0} \dfrac{{f\left( {x + y} \right) - f\left( x \right)}}{y} = \mathop {\lim }\limits_{y \to 0} \left( {ax + by} \right)$
The LHS of the above equation is the standard formula of differentiation of $x$ that is $f'\left( x \right)$ and on RHS, we substitute and simplify the limits, we get

$$\Rightarrow f'\left( x \right) = ax + b\mathop {\lim }\limits_{y \to 0} \left( y \right) \\\ \Rightarrow f'\left( x \right) = ax + b\left( 0 \right) \\\ \Rightarrow f'\left( x \right) = ax \\\$$

Step3: Integrate the above equation
Now integrating the above equation on both side with respect to $x$ , we get
$\Rightarrow \int {f'\left( x \right)} dx = \int {axdx}$
On LHS of the above equation, integration and differentiation are cancel out each other and on RHS we use the power rule of integration, we get
$\Rightarrow f\left( x \right) = a\left( {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right) + C$
On simplification, we get
$\Rightarrow f\left( x \right) = \dfrac{{a{x^2}}}{2} + C$
Step 4: Substitute $x = 1$ and $x = 2$ and formulate equation
Now we substitute the value of $x = 1$ as it is given that $f\left( 1 \right) = 8$ so
$\Rightarrow f\left( 1 \right) = \dfrac{{a{{\left( 1 \right)}^2}}}{2} + C$
$\Rightarrow 8 = \dfrac{a}{2} + C$
$\Rightarrow 8 - \dfrac{a}{2} = C$ …..(1)
Now we substitute the value of $x = 2$ as it is given that $f\left( 1 \right) = 32$ so
$\Rightarrow f\left( 2 \right) = \dfrac{{a{{\left( 2 \right)}^2}}}{2} + C$
$\Rightarrow 32 = 2a + C$ …..(2)
Step 5: Substitute the value of $C$ from equation (1) into equation (2)
Now, substituting the value of $C$ from equation (1) into equation (2), we get
$\Rightarrow 32 = 2a + \left( {8 - \dfrac{a}{2}} \right)$
Now simplifying the above equation , we get

\Rightarrow 24 = \dfrac{{3a}}{2} \\\ \Rightarrow 24 \times \dfrac{2}{3} = a \\\ \Rightarrow 16 = a \\\ $$ Step 6: Substitute $x = 1,y = 1$ in given equation Now to determine the value of $b$ , we consider $x = 1,y = 1$ and substitute in the given equation, we get $ f\left( {x + y} \right) - f\left( x \right) = axy + b{y^2} \\\ \Rightarrow f\left( {1 + 1} \right) - f\left( 1 \right) = a \times 1 \times 1 + b \times {\left( 1 \right)^2} \\\ $ Simplifying above equation, we get $ \Rightarrow f\left( 2 \right) - f\left( 1 \right) = a + b$ Now we substitute the known values of $f\left( 1 \right) = 8x$, $f\left( 2 \right) = 32$ and $a = 16$ , we get $ \Rightarrow 32 - 8 = 16 + b$ Simplifying above equation, we get $ \Rightarrow 32 - 8 - 16 = b \\\ \Rightarrow 8 = b \\\ $ **So the value of the age of the first son is $16$ years and age of second son is $8$ years.** **Note:** In this type of questions, students did not get an approach how to solve such questions. In many questions the function is directly not given, so first, we need to determine the function. After correctly choosing the values of $x$ and $y$ such that the additional information gets utilized. Commit to memory: $\int {f'\left( x \right)dx} = f\left( x \right)$ $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$