Let a function $g:[0,4] \rightarrow R$ be defined as $g(x)=\left\{\begin{array}{ll}\max _{0 \leq t \leq x}\left\{t^{3}-6 t^{2}+9 t-3\right\} & , 0 \leq x \leq 3 \\ 4-x & , 3<x \leq 4\end{array}\right.$, then the number of points in the interval $(0,4)$ where $\mathrm{g}(\mathrm{x})$ is NOT differentiable, is

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Continuity and Differentiability

Let a function

g : [0, 4] \to R

be defined as

g (x) = {max_{0 \leq t \leq x} {t^{3} - 6 t^{2} + 9 t - 3} 4 - x, 0 \leq x \leq 3, 3 < x \leq 4

, then the number of points in the interval

(0, 4)

where

g (x)

is NOT differentiable, is

Answer: 1

Write

h (t) = t^{3} - 6 t^{2} + 9 t - 3

, so

h^{'} (t) = 3 (t - 1) (t - 3)

. On

[0, 1]

the function

h

increases, so the maximum over

[0, x]

h (x)

there; for

1 \leq x \leq 3

the maximum is held at the local maximum value

h (1) = 1

. Hence

g (x) = ⎩ ⎨ ⎧ x^{3} - 6 x^{2} + 9 x - 3, 1, 4 - x, 0 \leq x \leq 1 1 < x \leq 3 3 < x \leq 4

x = 1

the left derivative is

h^{'} (1) = 0

and the right derivative is

0

, so

g

is differentiable there. At

x = 3

the two one-sided values agree at

g (3) = 1

, but the left derivative is

0

while the right derivative is

- 1

, so

g

is not differentiable at

x = 3

and at no other interior point. The number of such points in

(0, 4)

1

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer