Question

Let \(a\) be an integer selected at random from the set \(\{0,1,2,3,\ldots ,9\}\). The probability that the equation \(ax^2-ax+1 = 0\) has real roots is ...

• A. \(\frac{3}{5}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{2}{5}\)
• D. \(\frac{5}{9}\)

Hint:

A quadratic needs a nonzero a and discriminant at least zero; count the integers from 0 to 9 that qualify.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The equation \(ax^2-ax+1=0\) is a quadratic only when \(a\neq 0\). For \(a=0\) it becomes \(1=0\), which has no solution at all.
Real roots need a non-negative discriminant.

Step 2: Discriminant Condition:
\[ D = (-a)^2 - 4(a)(1) = a^2 - 4a = a(a-4) \geq 0 \]
With \(a>0\) this gives \(a-4\geq 0\), so \(a\geq 4\).

Step 3: Count Favourable Values:
From \(\{0,1,2,\ldots,9\}\) the values \(a=4,5,6,7,8,9\) qualify. That is 6 values.
\(a=1,2,3\) give a negative discriminant, and \(a=0\) gives no equation.

Step 4: Probability:
Total outcomes \(=10\), so
\[ P = \frac{6}{10} = \frac{3}{5} \]
Options (B), (C), (D) correspond to 5, 4 and 5.55 favourable values out of 10, which do not match the count.

Final Answer:
The probability is \(\tfrac35\), option (A). \[ \boxed{\text{(A) } \frac{3}{5}} \]