← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Differentiability · Calculus · asked 3× in 13 yrs · Read the full method →

Question

For the function $f:(0,\infty)\to\mathbb R$ given by $f(x)=x^2\sin\dfrac1x,\ 0<x<\infty$, show that there is a differentiable function $g:\mathbb R\to\mathbb R$ that extends $f$.

Technique

Extend by the same formula and set $g(0)=0$; differentiate by rules off $0$, and use the first-principles difference quotient with the squeeze theorem at $0$.

Solution

We must produce a single $g$ defined on all of $\mathbb R$, differentiable everywhere, agreeing with $f$ on $(0,\infty)$. The natural candidate extends the formula across $0$ and sets $g(0)=0$; the only delicate point is differentiability at $0$, where we use the $x^2$ damping.

Step 1 — Define the extension

Set

On $(0,\infty)$, $g(x)=x^2\sin(1/x)=f(x)$, so $g$ extends $f$. (We have extended the same formula to $x<0$ as well and patched the value at $0$; any choice agreeing with $f$ on $(0,\infty)$ and differentiable on $\mathbb R$ works, and this one does.)

Step 2 — Differentiability at every $x\ne0$

For $x\ne0$, $g$ is a composition/product of differentiable functions ($x\mapsto x^2$, $x\mapsto 1/x$, $\sin$), all differentiable on $\mathbb R\setminus\{0\}$. By the product and chain rules,

Step 3 — Differentiability at $0$ (the crux)

Use the definition of the derivative at $0$:

Since $\left|h\sin(1/h)\right|\le|h|$ and $|h|\to0$, the squeeze theorem gives

Hence the limit exists and $g'(0)=0$. Therefore $g$ is differentiable at $0$.

Step 4 — Conclusion

$g$ is differentiable at every $x\ne0$ (Step 2) and at $x=0$ (Step 3), so $g$ is differentiable on all of $\mathbb R$, and it extends $f$.

Answer