Question

Let the function \(f\) satisfy the relation, \(f\left(x+y^{3}\right)=f(x)+f\left(y^{3}\right)\), $\forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\( and be differentiable for all \)\mathrm{x}$. Assertion \((\mathbf{A}):\) If \(f^{\prime}(2)=a\), then \(f^{\prime}(-2)=a\). Reason \((\mathbf{R}): \mathrm{f}(\mathrm{x})\) is an odd function

Short answer

Answer: Yes, the assertion (A) is true. We proved that f(x) is an odd function and used this property to establish that if f'(2) = a, then f'(-2) = a.

Step-by-step solution

Proving that f(x) is an odd function#

We need to show that \(f(-x) = -f(x)\). Let's consider \(f\left(x + (-x)^3\right)\) using the given relation: \[f(x + (-x)^3) = f(x) + f((-x)^3).\] Now, since \((-x)^3 = -x^3\), we can rewrite this equation as: \[f(x - x^3) = f(x) + f(-x^3).\] Since, \(x - x^3 = x(1-x^2)\), let \(y = x^2 - 1\). Then, \(1 - x^2 = -y\) and \(x(1-x^2) = -xy\). We can substitute this back into our relation: \[f(-xy) = f(x) + f(-y^3).\] Now, let's substitute \(x \rightarrow -x\): \[f(xy) = f(-x) + f(y^3).\] By subtracting the last two equations, we get: \[f(-xy) - f(xy) = f(x) - f(-x).\] We can rewrite this equation as: \[f(-x) - f(x) = f(-xy) - f(xy).\] Since this equation holds for all \(x\) and \(y\), we can conclude that \(f(-x) = -f(x)\) – this means \(f(x)\) is an odd function.

Applying the odd function property to the given assertion#

Now that we know \(f(x)\) is an odd function, we can use this property to establish the assertion \(f^\prime(2) = a \Rightarrow f^\prime(-2) = a\). Since, \(f\) is differentiable for all \(x\), we can differentiate the given relation with respect to \(x\): \[f'\left(x+y^3\right) = f'(x).\] Now, let \(x = 2\) and \(y^3 = -2\) (i.e., \(y=(-2)^{\frac{1}{3}}\)): \[f'(2 + (-2) ^ {\frac{1}{3}}) = f'(2).\] Using the odd function property (\(f'(2 + (-2) ^ {\frac{1}{3}}) = -f'((-2)^\frac{1}{3} + 2)\)): \[-f'((-2)^\frac{1}{3} + 2) = f'(2).\] But we know according to our original assertion that \(f'(2) = a\). Then replacing \(f'(2)\) with \(a\): \[-f'((-2)^\frac{1}{3} + 2) = a.\] Now let's plug in \(x = -2\) and \( y^3=(-2)^{\frac{2}{3}}\): \[f'\left(x+y^3\right) = f'(-2).\] Using again the odd function property: \[-f'(2 - (-2) ^{\frac{2}{3}}) = f'(-2).\] As we derived above earlier: \[-f'((-2) ^ {\frac{1}{3}} + 2) = a.\] Comparing these two equations, we get: \[f'(-2) = a.\] Therefore, if \(f'(2) = a\), then \(f'(-2) = a\), as per our assertion \((A)\).