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in JEE by (15 points)
Let \( g : R \rightarrow R \) be a non constant twice differentiable function such that \( g ^{\prime}\left(\frac{1}{2}\right)= g \left(\frac{3}{2}\right) \). If a real valued function \( f \) is defined as \( f(x)=\frac{1}{2}[g(x)+g(2-x)] \), then

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Sure, I can help you with this problem. Let's analyze the given information and solve for the properties of the function f(x).


Given information:


g(x) is a non-constant, twice differentiable function over the real numbers (ℝ).

g'(1) = g(1). This means the slope of g(x) at x = 1 is equal to the function's value at x = 1.

f(x) = [g(x) + g(2 - x)]. This defines f(x) as the sum of g(x) and its reflection across x = 1.

What we need to find:


The problem asks for properties of the function f(x) without explicitly stating what those properties are. We can approach this by analyzing the given information and making logical deductions about f(x)'s characteristics.


Analysis:


Symmetry: Since f(x) is defined as the sum of g(x) and its reflection across x = 1, it follows that f(x) is an even function. This means f(x) = f(-x) for all x in ℝ.


Derivative at x = 1: We can differentiate f(x) to find its derivative f'(x). Using the sum rule and the chain rule, we get:


f'(x) = g'(x) - g'(2 - x)


Substituting g'(1) = g(1) into this expression, we get:


f'(1) = g(1) - g'(2 - 1) = g(1) - g'(1)


We don't have enough information to determine the exact value of f'(1), but we know it depends on the difference between g(1) and its slope at x = 1.


In conclusion:


We can definitively say that f(x) is an even function due to its definition.

We cannot definitively determine the value of f'(1) without more information about g(x). However, we can express it in terms of g(1) and g'(1).

If you have any further information about g(x) or specific properties you're interested in for f(x), please provide them, and I'll be happy to help you analyze them.





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