Logarithmic Inequality

Algebra Level 5

Find all values of the parameter a R a \in \mathbb{R} for which the following inequality is valid for all x R : x \in \mathbb{R}:

1 + log 5 ( x 2 + 1 ) log 5 ( a x 2 + 4 x + a ) . 1+\log _{ 5 }{ { (x }^{ 2 }+1) } \ge \log _{ 5 }{ (a{ x }^{ 2 }+4x+a) }.

If the range of values of a a can be expressed in the form of ( A , B ] , (A,B], then find the value of A + B . A + B.


The answer is 5.

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1 solution

Tanishq Varshney
Apr 2, 2015

( a 5 ) x 2 + 4 x + ( a 5 ) 0 (a-5)x^2+4x+(a-5) \leq 0 \quad and \quad a x 2 + 4 x + a > 0 ax^2+4x+a >0

a 5 < 0 a-5<0 Its discriminant D 0 D \leq 0 \quad \quad \quad \quad \quad its D < 0 D <0

16 4 ( a 5 ) 2 0 16-4(a-5)^{2} \leq 0 \quad \quad \quad \quad \quad a 2 4 > 0 a^2-4 >0

( a 7 ) ( a 3 ) 0 (a-7)(a-3) \geq 0 \quad \quad \quad \quad \quad ( a 2 ) ( a + 2 ) > 0 (a-2)(a+2)>0

The solution is ( 2 , 3 ] (2,3]

The above solution is incomplete. You didn't mention the conditions, a 5 < 0 a-5<0 for first inequality and a > 0 a>0 for the second one.

Abhishek Sharma - 6 years, 2 months ago

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Sorry i forgot to do that. edited thanx

Tanishq Varshney - 6 years, 2 months ago

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The solution is perfect now. +1

Abhishek Sharma - 6 years, 2 months ago

Do u have any suggestion about this

Tanishq Varshney - 6 years, 2 months ago

Can you tell me how you know a-5 < 0 and D <= 0?

JJ Chai - 5 years, 2 months ago

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Just basic quadratic equations. Refer a book.

Abhishek Sharma - 5 years, 2 months ago

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