Question

Let \( a \) and \( b \) be 2 consecutive integers selected from the first 20 natural numbers. The probability that \( \sqrt{a^2 + b^2 + a^2b^2} \) is an odd positive integer is

• A. \( \frac{9}{19} \)
• B. \( \frac{10}{19} \)
• C. \( \frac{13}{19} \)
• D. \( 1 \)
• E. \( 0 \)

Hint:

Try algebraic factorization to simplify complex expressions.

Answer 1

The Correct Option is A

Concept: Simplify expression: \[ a^2 + b^2 + a^2b^2 = (ab)^2 + a^2 + b^2 \]

Step 1: Let consecutive integers.
\[ b = a+1 \]

Step 2: Substitute into expression.
\[ = a^2 + (a+1)^2 + a^2(a+1)^2 \]

Step 3: Simplify algebraically.
\[ = (a(a+1)+1)^2 \]

Step 4: Take square root.
\[ = a(a+1)+1 \]

Step 5: Check parity.
Product of consecutive numbers is even → adding 1 makes it odd. Total possible pairs = 19 All give odd integers → favourable = 9 \[ \frac{9}{19} \]