MCQ
If $x - 2y = 4,$ the minimum value of $xy$ is
- ✓$-2$
- B$2$
- C$0$
- D$-3$
Let $P = xy$
From $(i),$ $P = y(2y + 4) = 4y + 2{y^2}$;
And $\frac{{dP}}{{dy}} = 4 + 4y = 0$
$y = - 1 \Rightarrow x = 2$ and $\frac{{{d^2}P}}{{d{y^2}}} = 4$(+ve)
$\therefore$ ${P_{{\rm{min}}{\rm{.}}}} = xy = (2)\,( - 1) = - 2$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $1:$ $f(x)\, \le \,g\,(x)$ for $x$ in $(0,\infty )$
Statement $2:$ $f(x)\, \le \,1$ for $(x)$ in $(0,\infty )$ but $g(x)\,\to \infty$ as $x\,\to \infty$