WBJEE · Maths · Quadratic Equation
For real \(x\), the greatest value of \(\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}\) is
- A 1
- B -1
- C \(\frac{1}{2}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Let \(y=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}\) \(\Rightarrow 2 x^{2} y+4 x y+9 y=x^{2}+2 x+4\) \(\Rightarrow(2 y-1) x^{2}+(4 y-2) x+9 y-4=0\) \(\Rightarrow x=\frac{-(4 y-2) \pm \sqrt{(4 y-2)^{2}-4(2 y-1)(9 y-4)}}{2(2 y-1)}\) Since, \(x\) is real number.…
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