Question

If $ f\left( x \right) $ satisfies the relation $ f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) $ for all $ x,y \in R $ , and $ f\left( 1 \right) = 5 $ , then:
A. $ f\left( x \right) $ is an odd function
B. $ f\left( x \right) $ is an even function
C. $ \sum\limits_{r = 1}^m {f\left( r \right)} = {5^{m + 1}}{C_2} $
D. $ \sum\limits_{r = 1}^m {f\left( r \right)} = \dfrac{{5m\left( {m + 2} \right)}}{3} $

Answer 1

Hint: Check the properties of the function given in the question whether the given function is odd or even. Also check that the sum $ \sum\limits_{r = 1}^m {f\left( r \right)} $ for the given function and check the correct options.

Complete step-by-step answer:
Assume that $ x = y = 0 $ , then according to the relationship given in question:
  $
  f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) \\
  f\left( {0 + 0} \right) = f\left( 0 \right) + f\left( 0 \right) \\
  f\left( 0 \right) = 0 \\
  $
Now let us consider $ y = - x $ and check the nature of function.
  $
  f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) \\
  f\left( {x + \left( { - x} \right)} \right) = f\left( x \right) + f\left( { - x} \right) \\
  f\left( 0 \right) = f\left( x \right) + f\left( { - x} \right) \\
  0 = f\left( x \right) + f\left( { - x} \right) \\
  f\left( x \right) = - f\left( { - x} \right) \\
  $
As the function satisfies the relation $ f\left( x \right) = - f\left( { - x} \right) $ . So, the function $ f\left( x \right) $ is an odd function.
An odd function is never an even function. So, the function $ f\left( x \right) $ is not an even function.
Now let us check the sum $ \sum\limits_{r = 1}^m {f\left( r \right)} $ .
Let $ x = y = 1 $ in the property given,
 $
  f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) \\
  f\left( {1 + 1} \right) = f\left( 1 \right) + f\left( 1 \right) \\
  f\left( 2 \right) = 2f\left( 1 \right) \\
  f\left( 2 \right) = 2 \times 5 \\
  f\left( 2 \right) = 10 \\
  $
Let $ x = 2,y = 1 $ in the property given,
 $
  f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) \\
  f\left( {2 + 1} \right) = f\left( 2 \right) + f\left( 1 \right) \\
  f\left( 3 \right) = 10 + 5 \\
  f\left( 3 \right) = 15 \\
  $
Similarly move up to the term when $ x = k - 1,y = 1 $ in the property given,
 $
  f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) \\
  f\left( {\left( {k - 1} \right) + 1} \right) = f\left( {k - 1} \right) + f\left( 1 \right) \\
  f\left( k \right) = k \times 5 \\
  f\left( k \right) = 5k \\
  $
Now simplify the sum $ \sum\limits_{r = 1}^m {f\left( r \right)} $ .
 $
  \sum\limits_{r = 1}^m {f\left( r \right)} = f\left( 1 \right) + f\left( 2 \right) + \ldots + f\left( m \right) \\
   = 5 + 10 + \ldots + 5m \\
   = 5\left( {1 + 2 + \ldots + m} \right) \\
   = \dfrac{{5m\left( {m + 1} \right)}}{2} \\
   = {5^{m + 1}}{C_2} \\
  $
So, the sum $ \sum\limits_{r = 1}^m {f\left( r \right)} $ is equal to $ {5^{m + 1}}{C_2} $ .
Hence, the correct options are A and C.
So, the correct answer is “Option A and C”.

Note: Check the formula of permutations to check the value of $ {5^{m + 1}}{C_2} $ is equal to $ \dfrac{{5m\left( {m + 1} \right)}}{2} $ . Use the mathematical induction concept to get the value of each functional value in general. Also revise the properties about the nature of a function as is even or odd.