← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

Fundamental theorems of integral calculus · Real Analysis · asked 2× in 13 yrs · Read the full method →

Question

Let $f(t)=\displaystyle\int_0^t [x]\,dx$, where $[x]$ denotes the largest integer less than or equal to $x$.

• (i) Determine all the real numbers $t$ at which $f$ is differentiable.
• (ii) Determine all the real numbers $t$ at which $f$ is continuous but not differentiable.

Technique

FTC at points of continuity of the integrand; one-sided derivatives at the jump points (integers) to detect the slope mismatch; Lipschitz bound for global continuity.

Solution

The integrand $[x]$ (floor) is bounded and piecewise constant, with jump discontinuities at the integers. Key tool: the Fundamental Theorem of Calculus for an integral of a function that is continuous except at isolated points.

Step 1 — $f$ is continuous everywhere

$[x]$ is bounded on every finite interval (Riemann integrable), and the integral of a bounded function with respect to its upper limit is continuous (indeed Lipschitz on bounded sets): for $|t-t_0|$ small,

So $f$ is continuous on all of $\mathbb R$.

Step 2 — Differentiability at non-integer $t$

Fix $t\notin\mathbb Z$. Then $[x]$ is constant, equal to $[t]=n$, on a neighbourhood $(n,n+1)$ containing $t$. By the Fundamental Theorem of Calculus, where the integrand is continuous the derivative of the integral equals the integrand:

Explicitly, for $t\in(n,n+1)$, $f(t)=f(n)+\int_n^t n\,dx=f(n)+n(t-n)$, a line of slope $n$, so $f'(t)=n=[t]$.

Step 3 — One-sided derivatives at an integer $t=n$

At an integer $n$, $[x]=n-1$ just to the left and $[x]=n$ just to the right. So $f$ is piecewise linear with a slope change. Compute one-sided derivatives:

• Left derivative: for $t\uparrow n$, the slope is $[t]=n-1$, so $f'_-(n)=n-1$.
• Right derivative: for $t\downarrow n$, the slope is $[t]=n$, so $f'_+(n)=n$.

Then

The one-sided derivatives differ (by $1$), so $f$ is not differentiable at any integer $n$ — including $n=0$: there $f'_-(0)=-1$ and $f'_+(0)=0$, which differ. (Note for $t<0$, e.g. $t\in(-1,0)$, $[t]=-1$, consistent with $f'_-(0)=-1$.)

Step 4 — Answers

Answer