If x^2-4x +〖log〗_(1/2) a^2 =0 does not have 2 distinct real roots and maximum value of a = m1 /m2 then m1 +m2 ( m1 and m2 are co-primes) equals?
Answers
x2 - 4x + log1/2 a = 0 ⇒ 16 - 4log1/2 a ⇒ log1/2 a ≥ log1/2 (1/2)4 ⇒ a ≤ 1/16 and a > 0 ⇒ a ∈ (0,1/16) Therefore, maximum value of a = 1/16.
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Answer:
Step-by-step explanation:
Query:
If x^2-4x +〖log〗_(1/2) a^2 =0 does not have 2 distinct real roots and maximum value of a = m_1/m_2 then m_1 +m_2 (m_1 and m_2 are co-primes) equals?
Solution:
If the given equation x^2-4x+log_(1/2)" " a^2=0 does not have two distinct real roots, then discriminant (D) will be less than 0.
So, D=b^2-4ac<0
(-4)^2-4×1×log_(1/2)" " a^2<0
16<4log_(1/2)" " a^2
4<log_(1/2)" " a^2
(1/2)^4 >a^2
1/16>a^2
±1/4>a
So, the maximum value of a can be 1/4
Maximum value of a = m_1/m_2 (Given)
1/4 = m_1/m_2
∴m_2=4m_1
m_1 +m_2= 4m_1+m_1=5 m_1