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Home » Mr. Freeze Shocked the World—Check Out His Untold Secrets Inside! - jntua results

Mr. Freeze Shocked the World—Check Out His Untold Secrets Inside! - jntua results

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Question: Find the number of functions $ f : \mathbb{R} o \mathbb{R} $ such that $ f(a + b) = f(a) + f(b) + ab $ for all real $ a, b $.

📌 Solution: Assume $ f $ is quadratic. Let $ f(x) = px^2 + qx + r $. Substitute into the equation:

📌 Left-hand side: $ f(a + b) = p(a + b)^2 + q(a + b) + r = p(a^2 + 2ab + b^2) + q(a + b) + r $.

📌 Right-hand side: $ f(a) + f(b) + ab = p(a^2 + b^2) + q(a + b) + 2r + ab $.

📌 Equating coefficients:

📌 $ 2pab = ab \Rightarrow 2p = 1 \Rightarrow p = rac{1}{2} $.

📌 $ r = 2r \Rightarrow r = 0 $.

📌 Thus, $ f(x) = rac{1}{2}x^2 + qx $, where $ q $ is arbitrary. There are infinitely many such functions. However, the original question specifies "number of functions," but the condition allows $ q \in \mathbb{R} $, leading to infinitely many solutions. If additional constraints (e.g., continuity) are implied, the solution is still infinite. But based on the structure, the answer is infinite. However, the original fragment likely intended a finite count. Revisiting, suppose the equation holds for all $ a, b $, but $ f $ is linear: $ f(x) = qx $. Substituting: $ q(a + b) = qa + qb + ab \Rightarrow 0 = ab $, which fails unless $ ab = 0 $. Thus, no linear solutions. The correct approach shows $ f(x) = rac{1}{2}x^2 + qx $, so infinitely many functions exist. But the original question may have intended a specific form. Given the context, the answer is oxed{\infty} (infinite).

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The semiperimeter \(s\) is:
s = \frac{a + b + c}{2}
The inradius is related to area by:
A = r \cdot s \quad \Rightarrow \quad r = \frac{A}{s}
We want to compute \(\frac{A}{r^2}\):
\frac{A}{r^2} = \frac{A}{\left(\frac{A}{s}\right)^2} = \frac{A \cdot s...
Now express \(A\) and \(s\) in terms of \(c\). For a right triangle:
A = \frac{1}{2}ab, \quad \text{and} \quad a^2 + b^2 = c^2
Also, the semiperimeter is:
A = r(a + b + c - c) = r(a + b - c) + rc? \quad \text{(Not helpful dir...

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