Let $[t]$ denote the greatest integer less than or equal to $t$. Let $\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}]$, $\mathrm{g}(\mathrm{x})=1-\mathrm{x}+[\mathrm{x}]$, and $\mathrm{h}(\mathrm{x})=\min\{\mathrm{f}(\mathrm{x}), \mathrm{g}(\mathrm{x})\}$, $\mathrm{x} \in[-2,2]$ Then h is:

(A) not continuous at exactly three points in $[-2,2]$

(B) not continuous at exactly four points in $[-2,2]$

(C) Continuous in $[-2,2]$ but not differentiable at exactly three points in $(-2,2)$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Continuity and Differentiability

Let

[t]

denote the greatest integer less than or equal to

t

.
Let

f (x) = x - [x]

g (x) = 1 - x + [x]

, and

h (x) = min {f (x), g (x)}

x \in [- 2, 2]

Then h is:

(A)not continuous at exactly three points in $[- 2, 2]$
(B)not continuous at exactly four points in $[- 2, 2]$
(C)Continuous in $[- 2, 2]$ but not differentiable at exactly three points in $(- 2, 2)$
(D)Continuous in $[- 2, 2]$ but not differentiable at more than four points in $(- 2, 2)$ ✓ Correct

Answer: (D)

Both functions depend only on the fractional part

t = x - [x] \in [0, 1)

, with

f (x) = t

and

g (x) = 1 - t

. On each interval

[n, n + 1)

h (x) = min {t, 1 - t} = ⎩ ⎨ ⎧ t, 1 - t, 0 \leq t \leq \frac{1}{2} \frac{1}{2} \leq t < 1

h

rises from

0

\frac{1}{2}

and falls back towards

0

t \to 1^{-}

, while

h = 0

at each integer. The one-sided limits therefore agree everywhere and

h

is continuous on

[- 2, 2]

.
The graph is a chain of triangles, with a corner at each integer (slope

- 1

on the left,

+ 1

on the right) and at each half-integer (slope

+ 1

on the left,

- 1

on the right). Inside

(- 2, 2)

these corners occur at

x = - \frac{3}{2}, - 1, - \frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}

, that is at seven points. So

h

is continuous in

[- 2, 2]

but not differentiable at more than four points in

(- 2, 2)

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option