WBJEE · Maths · Probability
Each of \(a\) and \(b\) can take values 1 or 2 with equal probability. The probability that the equation \(a x^{2}+b x+1=0\) has real roots, is equal to
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{8}\)
- D \(\frac{1}{16}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
The given equation \[ a x^{2}+b x+1=0 \] has real roots. \(\therefore\) Discriminant \((D) \geq 0\) \[ \Rightarrow \quad b^{2}-4 a \geq 0 \] From Eq. (ii), we observe that a has to be 1 and \(b\) has to be 2 . So, the required probability…
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