PUNE MCA Exam 2015 Question Paper with answers for online practice

Concept:Maximizing the difference of two cube roots using the relation between their cubes.Explanation:Let $u = (x+1)^{1/3}$ and $v = (x-1)^{1/3}$.Then $u^3 - v^3 = (x+1) - (x-1) = 2$.Also, $f(x) = u - v$.We have $u^3 - v^3 = (u-v)(u^2 + uv + v^2) = f \cdot (u^2+uv+v^2) = 2$.So $f = \frac{2}{u^2+uv+v^2}$.To maximize $f$, minimize the denominator $D = u^2+uv+v^2$ subject to $u^3 - v^3 = 2$.Set $d = f$, so $u = v + d$.Then $D = (v+d)^2 + (v+d)v + v^2 = 3v^2 + 3vd + d^2$.From $u^3 - v^3 = 2$, we get $d(3v^2+3vd+d^2) = dD = 2$, so $D = 2/d$.Write quadratic in $v$: $3d v^2 + 3d^2 v + (d^3-2) = 0$.For real $v$, discriminant $\ge 0$: $(3d^2)^2 - 4\cdot 3d \cdot (d^3-2) \ge 0$$\Rightarrow 9d^4 - 12d(d^3-2) \ge 0 \Rightarrow 9d^4 - 12d^4 + 24d \ge 0 \Rightarrow -3d^4 + 24d \ge 0$$\Rightarrow 3d(8 - d^3) \ge 0$. Since $d>0$, $8 - d^3 \ge 0 \Rightarrow d^3 \le 8 \Rightarrow d \le 2$.Maximum $d$ is $2$, attained when discriminant zero: $v = -\frac{3d^2}{2\cdot 3d} = -\frac{d}{2} = -1$, then $u = v+d = 1$.Check: $x=0$ gives $f(0)=1 - (-1)=2$.Answer:The maximum value is $2$, which corresponds to option C.