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Find the smallest integer value of a for which the inequality 1 + log₅(1 + x²) <= log₅(a*x² + 4x + a) holds for all real x.

1. 5
2. 6
3. 7
4. 8

Correct answer: 7

Solution

Rewriting gives 5(1+x²) <= a*x²+4x+a, i.e., (a-5)*x²+4x+(a-5) >= 0 for all x; requiring a-5 > 0 and discriminant <= 0 yields a >= 7, so the smallest integer is 7.

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