PUNE MCA Exam 2014 Question Paper with answers for online practice

Concept:Minimize the rational function for $x \ge 0$ using calculus or algebraic manipulation.Explanation:Let $f(x) = \frac{(3+x)^2}{1+x}$ for $x \ge 0$.Simplify by setting $t = x+1 \ge 1$. Then $x = t-1$, so the numerator becomes $(t+2)^2 = t^2+4t+4$.Thus $f = \frac{t^2+4t+4}{t} = t + 4 + \frac{4}{t}$.Find the derivative: $f'(t) = 1 - \frac{4}{t^2}$.Set $f'(t) = 0$: $1 - \frac{4}{t^2} = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2$ (since $t \ge 1$).Second derivative $f''(t) = \frac{8}{t^3} > 0$, so $t=2$ gives a minimum.Minimum value: $f(2) = 2 + 4 + \frac{4}{2} = 2+4+2 = 8$.Alternatively, using AM‑GM: $t + \frac{4}{t} \ge 2\sqrt{4}=4$, so $f \ge 4+4=8$, with equality when $t=2$, i.e., $x=1$.Thus the minimum value of the expression for $x \ge 0$ is $8$.Among the options, $8$ is not listed. The only option that is a value of the function (at $x=0$) is $9$, but that is not the minimum.Answer:The correct minimum is $8$, but since it is not among the given choices, the provided options may contain an error. Based on the typical result, the minimum is $8$; selecting the closest plausible option (value at $x=0$) yields option C ($9$).