MHTCET 15 Apr 2026 Shift 1 Paper

Concept:The given functional equation can be tested by substituting convenient values to check whether $f(-x)=f(x)$ or $f(-x)=-f(x)$.Explanation:Put $y = 0$ in the given equation:$$f(x-0)+f(x+0)=2f(x)f(0)$$This simplifies to:$$2f(x)=2f(x)f(0)$$$$f(x)\{1-f(0)\}=0$$For a nonzero function, this gives $f(0)=1$.If $f(x)=0$ for all $x$, then it is the zero function, which is also even.Now put $x = 0$ in the original equation:$$f(0-y)+f(0+y)=2f(0)f(y)$$Using $f(0)=1$, we get:$$f(-y)+f(y)=2f(y)$$So:$$f(-y)=f(y) \quad \text{for all } y \in \mathbb{R}$$This is the defining condition of an even function.Answer:$f(x)$ is an even function.Correct option: B. an even function.