Question
If $u = {x^2} + {y^2}$ and $x = s + 3t,$$y = 2s - t,$ then ${{{d^2}u} \over {d{s^2}}} = $
Now $\frac{{dx}}{{ds}} = 1,$ $\frac{{dy}}{{ds}} = 2$ .....$(i)$
$\frac{{{d^2}x}}{{d{s^2}}} = 0,$ $\frac{{{d^2}y}}{{d{s^2}}} = 0$ ......$(ii)$
Now $u = {x^2} + {y^2}$, $\frac{{du}}{{ds}} = 2x\,.\,\frac{{dx}}{{ds}} + 2y\,.\,\frac{{dy}}{{ds}}$
$\frac{{{d^2}u}}{{d{s^2}}} = 2{\left( {\frac{{dx}}{{ds}}} \right)^2} + 2x\frac{{{d^2}x}}{{d{s^2}}} + 2{\left( {\frac{{dy}}{{ds}}} \right)^2} + 2y\left( {\frac{{{d^2}y}}{{d{s^2}}}} \right)$
From $(i)$ and $(ii),$
$\frac{{{d^2}u}}{{d{s^2}}} = 2 \times 1 + 0 + 2 \times 4 + 0 = 10$.
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$f(x)=\min \{x-[x], 1-x+[x]\}$
$g(x)=\max \{x-[x], 1-x+[x]\}$
where $[x]$ denotes the largest integer not exceeding $x$ :
The positive integer $n$ for which
$\int_0^n(g(x)-f(x)) d x=100$ is