JEE Main 16 March 2021 Shift 1 Solved Paper

f(x)={

x+2	‌	x<0
x2	‌	x≥0.

g(x)={

x3	‌	x<1
3x−2	‌	x≥1.

f[g(x)]={

g(x)+2	‌	g(x)<0
g2(x)	‌	g(x)≥0.

When g(x)<0⇒g(x)=x3,x<0
When g(x)≥0⇒g(x)={

x3	‌	0≤x<1
3x−2	‌	x≥1.

f[g(x)]={

x3+2	‌	x<0
x6	‌	0≤x<1
(3x−2)2	‌	x≥1.

As, polynomial function is continuous everywhere in its domain. So, f[g(x)] will be continuous everywhere at x<0,0<x<1 and x>1.
We will check the behaviour of fog(x) only at boundary points which is x=0 and x=1.
At x=0,lim−((x3+2)=2

lim

x→0+

x6=0
Clearly, LHL≠ RHL at x=0
So, f∘g(x) is discontinuous at x=0.
⇒fog(x) is not differentiable at x=0
At x=1,

lim

x→1−

x6=1

lim

x→1+

(3x−2)2=1. Also f(1)=1
fog (x) is continuous at x=1
Derivative test at x=1,
LHD=

lim

h→0

‌

f(1)−f(1−h)

h

=

lim

h→0

‌

1−(1−h)6

h

=

lim

h→0

6(1−h)5=6
RHD=

lim

h→0

‌

f(1+h)−f(1)

h

∴f∘g(x) is continuous and differentiable at x=1.
∴fog(x) is discontinuous and non-differentiable at x=0.
So, number of points of non-differentiability of f∘g(x) is 1 .