Question

If $ f(x)=\dfrac{3x+5}{2x-3} $, then $ f\left( 15 \right)+f\left( -10 \right)= $
A. $ \dfrac{1855}{629} $
B. $ \dfrac{1875}{621} $
C. $ \dfrac{1825}{621} $
D. $ \dfrac{1875}{629} $

Answer 1

Hint: We have given a function $ f(x)=\dfrac{3x+5}{2x-3} $ and we have to find the value of $ f\left( 15 \right)+f\left( -10 \right)= $ we put the value of $ x=15 $ and $ x=-10 $ in the given function one by one and calculate the value of $ f\left( 15 \right) $ and $ f\left( -10 \right) $ . Then, add both to obtain a desired result.

Complete step-by-step answer:
We have given that $ f(x)=\dfrac{3x+5}{2x-3} $
We have to find the value of $ f\left( 15 \right)+f\left( -10 \right) $ .
Now, first we calculate the value of $ f(15) $ , for this we put the value $ x=15 $ in the given equation.
When we put value, we get
 $ \begin{align}
  & f(x)=\dfrac{3x+5}{2x-3} \\
 & f(15)=\dfrac{3\times 15+5}{2\times 15-3} \\
\end{align} $
Now, simplify the equation to solve further
\[\begin{align}
  & f(15)=\dfrac{45+5}{30-3} \\
 & f(15)=\dfrac{50}{27}.......(i) \\
\end{align}\]
Similarly, we calculate the value of $ f(-10) $ .
When we put the value $ x=-10 $ in the given equation, we get
 $ \begin{align}
  & f(x)=\dfrac{3x+5}{2x-3} \\
 & f(-10)=\dfrac{3\times \left( -10 \right)+5}{2\times \left( -10 \right)-3} \\
 & f(-10)=\dfrac{-30+5}{-20-3} \\
 & f(-10)=\dfrac{-25}{-23} \\
 & f(-10)=\dfrac{25}{23}.................(ii) \\
\end{align} $
Now, to find the value of $ f\left( 15 \right)+f\left( -10 \right) $ , substitute the value from equation (i) and equation (ii)
 $ f\left( 15 \right)+f\left( -10 \right)=\dfrac{50}{27}+\dfrac{25}{23} $
Now, take LCM to solve further, as $ 27 $ and $ 23 $ don’t have common factor, we directly cross multiply to solve
 $ \begin{align}
  & f\left( 15 \right)+f\left( -10 \right)=\dfrac{50\times 23+27\times 25}{27\times 23} \\
 & f\left( 15 \right)+f\left( -10 \right)=\dfrac{1150+675}{621} \\
 & f\left( 15 \right)+f\left( -10 \right)=\dfrac{1825}{621} \\
\end{align} $
Hence, the value of $ f\left( 15 \right)+f\left( -10 \right)=\dfrac{1825}{621} $.
So, the correct answer is “Option C”.

Note: One may relate this question with the differentiation as the given equation is of the form $ f(x)=\dfrac{3x+5}{2x-3} $. But we have to find the value of $ f\left( 15 \right)+f\left( -10 \right) $, which are not derivatives so it is not a question of differentiation. If it is asked to find the value of $ f'\left( 15 \right)+f'\left( -10 \right) $ instead of $ f\left( 15 \right)+f\left( -10 \right) $, then we first differentiate the given equation and find the values.