← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q4d — Step-by-Step Solution

12 marks · Section A

Functions of two/three variables: limits, continuity · Calculus · asked 4× in 13 yrs · Read the full method →

Question

For the function

examine the continuity and differentiability.

Technique

Path test ($y=tx^2$) to expose direction-dependent limit; conclude discontinuity; non-continuity ⇒ non-differentiability; partial-derivative check for completeness.

Solution

Note on domain. The expression $\sqrt y$ requires $y\ge 0$. We assume the function is defined on $\{(x,y):y\ge 0\}$, i.e. the upper half-plane (including the $x$-axis). Continuity/differentiability questions are at the origin $(0,0)$.

Step 1 — Continuity at $(0,0)$

Approach along the path $y=tx^2$ for $t\ge 0$, $x\to 0^+$ (taking $x>0$ for clarity; $x<0$ similar):

(Using $\sqrt{tx^2}=|x|\sqrt t$, and assuming $x>0$ so $|x|=x$.)

This limit depends on $t$:

• $t=0$: $1$.
• $t=1$: $0$.
• $t=4$: $(1-2)/5=-1/5$.

So $f$ has different limits along different parabolic paths — the limit at $(0,0)$ does not exist. Therefore $f$ is not continuous at $(0,0)$.

(For $x<0$: $\sqrt{tx^2}=|x|\sqrt t=-x\sqrt t$, so $f=\dfrac{x^2+x^2\sqrt t}{x^2(1+t)}=\dfrac{1+\sqrt t}{1+t}$, also depending on $t$. Different paths from the left also yield different limits.)

Step 2 — Differentiability at $(0,0)$ — direct conclusion

Differentiability at a point requires continuity at that point. Since $f$ is not continuous at $(0,0)$, $f$ is not differentiable at $(0,0)$.

Step 3 — Partial derivatives at $(0,0)$ (do they exist?)

For completeness, check whether $f_x(0,0)$ and $f_y(0,0)$ exist (they can exist even if $f$ is discontinuous).

$f_x(0,0)=\displaystyle\lim_{h\to 0}\dfrac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\dfrac{h^2-0}{h\cdot(h^2+0)}=\lim_{h\to 0}\dfrac{1}{h}$, which does not exist.

Wait: $f(h,0)=\dfrac{h^2-h\cdot 0}{h^2+0}=\dfrac{h^2}{h^2}=1$ for $h\ne 0$.

So $f_x(0,0)=\lim_{h\to 0}\dfrac{1-0}{h}=\lim_{h\to 0}\dfrac{1}{h}=\pm\infty$ — does not exist.

$f_y(0,0)=\displaystyle\lim_{k\to 0^+}\dfrac{f(0,k)-0}{k}$ (one-sided since $y\ge 0$). $f(0,k)=\dfrac{0-0}{0+k}=0$ for $k>0$.

So $f_y(0,0)=\lim_{k\to 0^+}\dfrac{0}{k}=0$. The partial $f_y$ exists (and equals 0).

But $f_x$ does not exist, confirming non-differentiability.

Conclusion

Answer