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Home » Türk ISA Secrets Revealed: The Shocking Truth Behind This Hidden Gem! - Abu Waleed Tea

Türk ISA Secrets Revealed: The Shocking Truth Behind This Hidden Gem! - Abu Waleed Tea

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 Question: A polynomial $ f(x) $ satisfies $ f(x + 1) - f(x) = 6x + 4 $. Find the leading coefficient of $ f(x) $.

📌 Solution: Since the difference $ f(x+1) - f(x) $ is linear, $ f(x) $ must be a quadratic polynomial. Let $ f(x) = ax^2 + bx + c $. Then:

📌 f(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 + 2x + 1) + b(x + 1) + c = ax^2 + 2ax + a + bx + b + c

📌 f(x+1) - f(x) = (ax^2 + 2ax + a + bx + b + c) - (ax^2 + bx + c) = 2ax + a + b

📌 f(x+1) - f(x) = 6x + 4

📌 2a = 6 \Rightarrow a = 3, \quad a + b = 4 \Rightarrow 3 + b = 4 \Rightarrow b = 1

📌 Thus, the leading coefficient of $ f(x) $ is $ \boxed{3} $.

📌 Question: A function $ f: \mathbb{R} \to \mathbb{R} $ satisfies $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $ for all real $ a, b $. If $ f(1) = 1 $, find $ f(2) $.

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\[ \lambda = rac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}...
\[ \lambda = rac{7 \pm \sqrt{49 - 40}}{2} \]
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The eigenvalues are:
\( \lambda = rac{10}{2} = 5 \)
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