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Question

(D) Let the two numbers be $n$ and $n+2$. The greater number is $n+2$.Given that the greater number is smaller than the denominator $D$,we have $n+2 <

Vedclass · Accepted Answer

$-\frac{1}{5}$

Vedclass · Answer

(D) Let the two numbers be $n$ and $n+2$. The greater number is $n+2$.Given that the greater number is smaller than the denominator $D$,we have $n+2 Let the fraction be $f(n) = \frac{n(n+2)}{D}$.To find the minimum value,we analyze the quadratic expression in the numerator.For a fixed denominator $D$,the expression $n^2 + 2n$ represents a parabola opening upwards.The vertex of this parabola is at $n = -b/2a = -2/2 = -1$.However,since $n$ must be an integer in the context of such problems,we test values near the vertex.If we assume the denominator $D=5$,the fraction is $f(n) = \frac{n^2+2n}{5}$.For $n=0$,$f(0) = 0$.For $n=-1$,$f(-1) = \frac{1-2}{5} = -\frac{1}{5}$.For $n=-2$,$f(-2) = 0$.Thus,the minimum value is $-\frac{1}{5}$.

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