← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Functions of two/three variables: limits, continuity · Calculus · asked 4× in 14 yrs · Read the full method →

Question

Show that the function

is continuous and differentiable at $(1,-1)$.

Technique

Factor $x^2-y^2=(x-y)(x+y)$ to reveal a removable form; on a neighbourhood off the diagonal $f$ is the polynomial $x+y$; verify the differentiability definition (zero remainder).

Solution

Step 1 — Simplify the defining expression

For $(x,y)\ne(1,1),(1,-1)$ (in particular wherever $x\ne y$),

So away from the diagonal $x=y$ the function equals the polynomial $x+y$. The point of interest is $(1,-1)$, where $x\ne y$, so on a full neighbourhood of $(1,-1)$ (which avoids the line $x=y$ since $1\ne-1$) we have $f(x,y)=x+y$ except possibly at the single prescribed point $(1,-1)$ itself, where $f=0$.

Step 2 — Continuity at $(1,-1)$

As $(x,y)\to(1,-1)$ through points with $x\ne y$,

The defined value is $f(1,-1)=0$. Limit equals value, so $f$ is continuous at $(1,-1)$.

(The point $(1,1)$ lies on the diagonal and far from $(1,-1)$, so it does not interfere; a punctured neighbourhood of $(1,-1)$ contains no point of the line $x=y$.)

Step 3 — Differentiability at $(1,-1)$

A function of two variables is differentiable at $\mathbf a=(1,-1)$ if there is a linear map (the gradient) $L=(A,B)$ with

On the punctured neighbourhood, $f(1+h_1,\,-1+h_2)=(1+h_1)+(-1+h_2)=h_1+h_2$, and $f(1,-1)=0$. Take $A=B=1$ (so $L(\mathbf h)=h_1+h_2$):

The remainder is identically $0$, hence trivially $o(\|\mathbf h\|)$. Therefore $f$ is differentiable at $(1,-1)$ with

Answer