NDA 2026 — Mathematics PYQ

We are given the functional equation:

$$2f(x)+f(1-x)=x\text{— (Equation 1)}$$

1. Substitute $x$ with $(1-x)$:

To create a second equation, replace every instance of $x$ in Equation 1 with $(1-x)$:

$$2f(1-x)+f(1-(1-x))=1-x$$

$$2f(1-x)+f(x)=1-x\text{— (Equation 2)}$$

2. Solve the system of equations:

We now have a system of two equations:

1. $2f(x)+f(1-x)=x$

2. $f(x)+2f(1-x)=1-x$

Multiply Equation 2 by 2 to align the $f(1-x)$ terms:

$$2f(x)+4f(1-x)=2(1-x)$$

$$2f(x)+4f(1-x)=2-2x\text{— (Equation 3)}$$

Now, subtract Equation 1 from Equation 3:

$$(2f(x)+4f(1-x))-(2f(x)+f(1-x))=(2-2x)-x$$

$$3f(1-x)=2-3x$$

To isolate $f(x)$, it is easier to multiply Equation 1 by 2:

$$4f(x)+2f(1-x)=2x\text{— (Equation 4)}$$

Subtract Equation 2 from Equation 4:

$$(4f(x)+2f(1-x))-(f(x)+2f(1-x))=2x-(1-x)$$

$$3f(x)=2x-1+x$$

$$3f(x)=3x-1$$

$$f(x)=\frac{3x-1}{3}$$

$$f(x)=x-\frac{1}{3}$$

Final Answer:

The correct option is (b) $x-\frac{1}{3}$.