TS EAMCET · Maths · Quadratic Equation
The smallest value of the constant \(m>0\) for which \(f(x)=9 m x-1+\frac{1}{x} \geq 0\) for all \(x>0\), is
- A \(\frac{1}{9}\)
- B \(\frac{1}{16}\)
- C \(\frac{1}{36}\)
- D \(\frac{1}{81}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{36}\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=9 m x-1+\frac{1}{x} \geq 0\) Now, \(\quad 9 m x^2-x+1 \geq 0\) For \(f(x) \geq 0, D\) should be less than 0 . \[ D=1-36 m \] \[ \begin{array}{rlrl} & & 1-36 m & \leq 0 \\ \therefore & m & \geq \frac{1}{36} \end{array} \]
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