Question

# If f(x + y) = f(x) + f(y), then f(x) may be(a) x(b) x + 1 (c) ${{\text{x}}^{2}}$ + 1(d) log x

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Hint: We will solve this question by option verification method. We will take some arbitrary value for x and y and verify that for which of the given options, the left hand side is equal to the right hand side. The option in which the left hand side is equal to the right hand side, that will be the answer.

Let us assume some arbitrary value for x and y, say x = 2 and y = 3.
According to option (d), f(x) = log x.
Thus, f(2) = log 2 and f(3) = log 3.
Now, left hand side = f(x + y)
$\Rightarrow$ LHS = f(2 + 3)
$\Rightarrow$ LHS = f(5)
As we know, according to option (d), f(5) = log 5
$\Rightarrow$ LHS = log 5
And, right hand side = f(x) + f(y)
$\Rightarrow$ RHS = log 2 + log 3
$\Rightarrow$ LHS $\ne$ RHS
So, option (d) does not satisfy our conditions.
According to option (c), f(x) = ${{\text{x}}^{2}}$ + 1
Thus, f(2) = 5 and f(3) = 10.
Now, left hand side = f(x + y)
$\Rightarrow$ LHS = f(2 + 3)
$\Rightarrow$ LHS = f(5)
As we know, according to option (d), f(5) = 26
$\Rightarrow$ LHS = 26
And, right hand side = f(x) + f(y)
$\Rightarrow$ RHS = 5 + 10
$\Rightarrow$ RHS = 15
$\Rightarrow$ LHS $\ne$ RHS
So, option (c) also does not satisfy our conditions.
Now, we will verify option (b).
According to option (b), f(x) = x + 1
Thus, f(2) = 3 and f(3) = 4.
Now, left hand side = f(x + y)
$\Rightarrow$ LHS = f(2 + 3)
$\Rightarrow$ LHS = f(5)
As we know, according to option (d), f(5) = 6
$\Rightarrow$ LHS = 6
And, right hand side = f(x) + f(y)
$\Rightarrow$ RHS = 3 + 4
$\Rightarrow$ RHS = 7
$\Rightarrow$ LHS $\ne$ RHS
So, option (b) also does not satisfy our conditions.
The option left with us is option (a).
According to option (b), f(x) = x
Thus, f(2) = 2 and f(3) = 3.
Now, left hand side = f(x + y)
$\Rightarrow$ LHS = f(2 + 3)
$\Rightarrow$ LHS = f(5)
As we know, according to option (d), f(5) = 5
$\Rightarrow$ LHS = 5
And, right hand side = f(x) + f(y)
$\Rightarrow$ RHS = 2 + 3
$\Rightarrow$ RHS = 5
$\Rightarrow$ LHS = RHS
So, option (a) satisfies the conditions.
So, the correct answer is “Option A”.

Note: Whenever, option verification is possible, students should opt for the method as it is quicker. While choosing the values of x and y, students are advised to be careful that all the options yield different values. If two of the given values satisfy with the value of x and y, we shall take some other values.