If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the

1.	If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x2 + px + \(\frac{p}{4}+\frac{1}{2}\) = 0 are real is(a) \(\frac{2}{3}\) (b) \(\frac{2}{5}\)(c) \(\frac{3}{5}\) (d) \(\frac{4}{5}\)
Answer» (a) \(\frac{2}{3}\) The equation x² + px + \(\frac{p}{4}+\frac{1}{2}\) = 0 has real roots if the discriminant D ≥ 0. ⇒ p² – 4 \(\bigg(\frac{p}{4}+\frac{1}{2}\bigg)\)≥ 0 ⇒ p² - p - 2 ≥ 0 ⇒ p² – 2p + p – 2 ≥ 0 ⇒ p(p – 2) + 1 (p – 2) ≥ 0 ⇒ (p – 2) (p + 1) ≥ 0 ⇒ (p – 2) ≥ 0 and (p + 1) ≥ 0 ⇒ p ≥ 2 or p ≤ –1 The condition p ≤ –1 is not admissible as 0 ≤ p ≤ 5. Now p ≥ 2 ⇒ p can take up the value 2 or 3 or 4 or 5 from the given values. {0, 1, 2, 3, 4, 5} ∴ Probability (Roots of given equation are real) = \(\frac{\text{Number of values p can take}}{\text{Given number of values}}\) = \(\frac{4}{6}=\frac{2}{3}.\)

If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x2 + px + \(\frac{p}{4}+\frac{1}{2}\) = 0 are real is(a) \(\frac{2}{3}\) (b) \(\frac{2}{5}\)(c) \(\frac{3}{5}\) (d) \(\frac{4}{5}\)

Answer»

(a) \(\frac{2}{3}\)

The equation x² + px + \(\frac{p}{4}+\frac{1}{2}\) = 0 has real roots if the discriminant D ≥ 0.

⇒ p² – 4 \(\bigg(\frac{p}{4}+\frac{1}{2}\bigg)\)≥ 0 ⇒ p² - p - 2 ≥ 0

⇒ p² – 2p + p – 2 ≥ 0 ⇒ p(p – 2) + 1 (p – 2) ≥ 0

⇒ (p – 2) (p + 1) ≥ 0

⇒ (p – 2) ≥ 0 and (p + 1) ≥ 0

⇒ p ≥ 2 or p ≤ –1

The condition p ≤ –1 is not admissible as 0 ≤ p ≤ 5.

Now p ≥ 2 ⇒ p can take up the value 2 or 3 or 4 or 5 from the given values. {0, 1, 2, 3, 4, 5}

∴ Probability (Roots of given equation are real)

= \(\frac{\text{Number of values p can take}}{\text{Given number of values}}\) = \(\frac{4}{6}=\frac{2}{3}.\)