Question

Let f (x) is a non-zero differentiable function $\forall$ x $\in$ R & satisfying f (x + y) = f (x - y) $\forall$ x, y $\in$ R, then -

• A. f (1) + f (0) = f '(0)
• B. f (1) - f (0) = f '(0)
• C. f (0) + f (1) = f (2)
• D. 3f (2) = 2f (3)

Answer 1

Answer: (B)

(B)

Answer 2

The given functional equation is $f(x+y) = f(x-y)$ for all $x, y \in \mathbb{R}$.

The function $f(x)$ is a non-zero differentiable function for all $x \in \mathbb{R}$.

Differentiate the given equation with respect to $y$: $\frac{\partial}{\partial y} f(x+y) = \frac{\partial}{\partial y} f(x-y)$

Using the chain rule, this gives $f'(x+y) \cdot \frac{\partial(x+y)}{\partial y} = f'(x-y) \cdot \frac{\partial(x-y)}{\partial y}$.

$f'(x+y) \cdot 1 = f'(x-y) \cdot (-1)$

$f'(x+y) = -f'(x-y)$ (Equation 1)

Differentiate the given equation with respect to $x$: $\frac{\partial}{\partial x} f(x+y) = \frac{\partial}{\partial x} f(x-y)$

Using the chain rule, this gives $f'(x+y) \cdot \frac{\partial(x+y)}{\partial x} = f'(x-y) \cdot \frac{\partial(x-y)}{\partial x}$.

$f'(x+y) \cdot 1 = f'(x-y) \cdot 1$

$f'(x+y) = f'(x-y)$ (Equation 2)

From Equation 1 and Equation 2, we have:

$-f'(x-y) = f'(x-y)$

$2f'(x-y) = 0$

$f'(x-y) = 0$ for all $x, y \in \mathbb{R}$.

Let $z = x-y$. Since $x$ and $y$ can be any real numbers, $z$ can also be any real number.

Thus, $f'(z) = 0$ for all $z \in \mathbb{R}$.

Since the derivative of $f(x)$ is zero for all $x$, $f(x)$ must be a constant function.

Let $f(x) = c$ for some constant $c$.

The problem states that $f(x)$ is a non-zero function, so $c \neq 0$.

Now we check which of the given options is true for $f(x) = c$ where $c \neq 0$, and $f'(x) = 0$.

(A) $f(1) + f(0) = f'(0)$

Substituting the values, we get $c + c = 0$, which simplifies to $2c = 0$. This implies $c = 0$, which contradicts the condition $c \neq 0$. So, option (A) is incorrect.

(B) $f(1) - f(0) = f'(0)$

Substituting the values, we get $c - c = 0$, which simplifies to $0 = 0$. This statement is true for any value of $c$, including any non-zero constant $c$. So, option (B) is correct.

(C) $f(0) + f(1) = f(2)$

Substituting the values, we get $c + c = c$, which simplifies to $2c = c$, or $c = 0$. This contradicts the condition $c \neq 0$. So, option (C) is incorrect.

(D) $3f(2) = 2f(3)$

Substituting the values, we get $3c = 2c$, which simplifies to $c = 0$. This contradicts the condition $c \neq 0$. So, option (D) is incorrect.

The only option that is true for any non-zero differentiable function satisfying the given functional equation is (B).