Question

Let R be the set of all real numbers and let f be a function R to R such that \[f\left( x \right) + \left( {x + \dfrac{1}{2}} \right)f\left( {1 - x} \right) = 1\], for all \[x \in R\]. Then \[2f\left( 0 \right) + 3f\left( 1 \right)\] is equal to.
A.2
B.0
C.-2
D.-4

Answer 1

Hint: Here in this question, we have to find the value of the function \[2f\left( 0 \right) + 3f\left( 1 \right)\] by using \[f\left( x \right) + \left( {x + \dfrac{1}{2}} \right)f\left( {1 - x} \right) = 1\]. To solve this first we have to find the value of \[f\left( 0 \right)\] and \[f\left( 1 \right)\] by giving the value of \[x = 0\] and \[x = 1\] to the function \[f\left( x \right) + \left( {x + \dfrac{1}{2}} \right)f\left( {1 - x} \right) = 1\] and further substitute these values in required function \[2f\left( 0 \right) + 3f\left( 1 \right)\] on simplification we get the required solution.

Complete step by step solution:
Given that let f be a function R to R i.e., \[f:R \to R\] where R be a set of all real numbers.
Consider the function
\[ \Rightarrow f\left( x \right) + \left( {x + \dfrac{1}{2}} \right)f\left( {1 - x} \right) = 1\]--------(1)
Now, find the values of \[f\left( 0 \right)\] and \[f\left( 1 \right)\].
Let, \[x = 0\], then equation (1) becomes
\[ \Rightarrow f\left( 0 \right) + \left( {0 + \dfrac{1}{2}} \right)f\left( {1 - 0} \right) = 1\]
On simplification, we get
\[ \Rightarrow f\left( 0 \right) + \dfrac{1}{2}f\left( 1 \right) = 1\]------(2)
When \[x = 1\], then equation (1) becomes
\[ \Rightarrow f\left( 1 \right) + \left( {1 + \dfrac{1}{2}} \right)f\left( {1 - 1} \right) = 1\]
On simplification, we get
\[ \Rightarrow f\left( 1 \right) + \dfrac{3}{2}f\left( 0 \right) = 1\]--------(3)
Multiplying equation (2) by 2 and subtracting equation (3) from that, we have
\[ \Rightarrow 2\left( {f\left( 0 \right) + \dfrac{1}{2}f\left( 1 \right)} \right) - \left( {f\left( 1 \right) + \dfrac{3}{2}f\left( 0 \right)} \right) = 2\left( 1 \right) - 1\]
\[ \Rightarrow 2f\left( 0 \right) + f\left( 1 \right) - f\left( 1 \right) - \dfrac{3}{2}f\left( 0 \right) = 2 - 1\]
On simplification, we have
\[ \Rightarrow 2f\left( 0 \right) - \dfrac{3}{2}f\left( 0 \right) = 1\]
Take 2 as LCM in LHS, then
\[ \Rightarrow \dfrac{{4f\left( 0 \right) - 3f\left( 0 \right)}}{2} = 1\]
\[ \Rightarrow \dfrac{{f\left( 0 \right)}}{2} = 1\]
Multiply 2 by both side, then
\[ \Rightarrow f\left( 0 \right) = 2\]
Substitute, the value of \[f\left( 0 \right)\] in equation (3), we have
 \[ \Rightarrow f\left( 1 \right) + \dfrac{3}{2}\left( 2 \right) = 1\]
\[ \Rightarrow f\left( 1 \right) + 3 = 1\]
Subtract, both side by 3, then
\[ \Rightarrow f\left( 1 \right) = 1 - 3\]
\[ \Rightarrow f\left( 1 \right) = - 2\]
Now, consider the given equation
 \[ \Rightarrow 2f\left( 0 \right) + 3f\left( 1 \right)\]
Substitute the value of \[f\left( 0 \right)\] and \[f\left( 1 \right)\], then
\[ \Rightarrow 2\left( 2 \right) + 3\left( { - 2} \right)\]
\[ \Rightarrow 4 - 6\]
\[ \Rightarrow - 2\]
Hence, The value of \[2f\left( 0 \right) + 3f\left( 1 \right) = - 2\].
Therefore, option (C) is correct.
So, the correct answer is “Option C”.

Note: The function is an image of the domain value. To solve these kinds of problems the student must know the tables of multiplication and the simple arithmetic operations. in some of the questions the function term will be given directly. Like these kinds of problems we have to try to simplify.