The solution set of the inequality ( 3 1 ) l o g ( 9 1 ) ( x 2 − 3 10 x + 1 ) ≤ 1 is written as x ∈ [ 0 , a 1 ) ∪ ( a , a 10 ] then find a .

0 < x 2 − 3 10 x + 1 ≤ 1 ⇒ x ∈ [ 0 , 3 1 ) ∪ ( 3 , 3 10 ]

Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationLocation of rootsHard2 minQB

MathematicsHardinteger

The solution set of the inequality $(\frac{1}{3})^{l o g_{(\frac{1}{9})} (x^{2} - \frac{10}{3} x + 1)} \leq 1$ is written as $x \in [0, \frac{1}{a}) \cup (a, \frac{10}{a}]$ then find a .

Answer:

Solution:

$0 < x^{2} - \frac{10 x}{3} + 1 \leq 1$

$\Rightarrow x \in [0, \frac{1}{3}) \cup (3, \frac{10}{3}]$

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Location of roots

⏱ 2mℹ️ Source: QB

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