JEE Advanced 2021 Paper 2

ψ1(x)=e−x+x,x≥0
ψ1′(x)=1−e−x>0
⇒ψ1(x) is increasing
ψ1(x)≥ψ1(0)‌∀x≥0⇒ψ1(x)≥1
Option (a) is incorrect.
(b) ‌‌ψ2(x)=x2−2x+2−2e−x‌‌x≥0
⇒ψ2(x) is increasing
⇒‌‌ψ2−(x)≥ψ2(0)‌‌⇒‌‌ψ2(x)≥0
So, option (b) is incorrect.
(c) f(x)=2‌

x

∫

0

(t−t2)e−t2dt\;x∈(0,‌

1

2

)
=

x

∫

0

2te−t2dt−

x

∫

0

2t2e−t2dt
=−e−x2|0x−‌‌

x

∫

0

2t2e−t2dt
Let H(0)=0
H′(x)=2(x−x2)e−x2−2xe−x2+2x2−2x4
=−2x2e−x2+2x2−2x4=2x2(1−x2−e−x2)
∵‌‌e−x≥1−x‌∀x≥0
⇒H′(x)≤0
⇒‌‌H(x) is decreasing
⇒‌‌1−1(x)<0‌∀x∈(0,‌

1

2

)
Let P(x)=g(x)−‌

2

3

x3+‌

2

5

x5−‌

1

7

x7x∈(0,‌

1

2

)
P′(x)=2x2e−x2−2x2+2x4−x6
=−‌

x8

3

+‌

x10

12

.........
⇒P′(x)≤0⇒P(x) is decreasing
⇒P(x)≤0
So, option (d) is correct.