For natural numbers \(a\) and \(b\), \((a+b)^2-(a-b)^2\gt 29\) and the smallest value of \(ab\) is \(m\). The value of \(m^2+m+1\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 73
The two squares differ by \((a+b)^2-(a-b)^2=4ab\), because the \(a^2\) and \(b^2\) terms cancel and the \(2ab\) terms add. So the condition is really a condition on the product \(ab\), and \(m\) is the smallest whole number that satisfies it.
The condition becomes \(4ab\gt 29\), that is, \(ab\gt 7.25\)
\(a\) and \(b\) are natural numbers, so \(ab\) is a whole number; the smallest one above 7.25 is 8 (for example \(a=2\), \(b=4\)), so \(m=8\)
Required value: \(m^2+m+1=64+8+1=73\)