JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, equation ‌

4

sin‌x

+‌

1

1−sin‌x

=α
Let f(x)=‌

4

sin‌x

+‌

1

1−sin‌x

⇒‌‌y=‌

4−3‌sin‌x

sin‌x(1−sin‌x)

Let sin‌x=t
when t∈(0,1) as x∈(0,‌

π

2

).
[

∵t1=sin‌0=0

t2=sin‌ π 2 =1

]
Now, y=‌

4−3t

t−t2

‌

dy

dt

=‌

−3(t−t2)−(1−2t)(4−3t)

(t−t2)2

=0
⇒3t2−3t−(4−11t+6t2)=0
⇒‌‌3t2−8t+4=0
⇒‌‌3t2−6t−2t+4=0
⇒‌‌3t(t−2)−2(t−2)=0
⇒‌‌(t−2)(3t−2)=0
∴‌‌t−2=0 or 3t−2=0
t=2 or t=2∕3
sin‌x=2 or sin‌x=‌

2

3

x=sin−1(‌

2

3

)
Also, y=‌

4−3t

t−t2

=‌

4−3×2∕3

2 3 −(‌ 2 3 )2

y=9
[∵sin‌θ≤1]
[∵t=2∕3]
Now, required graph isFrom the graph, minimum value of
f(x)≥9
α≥9
∴α has minimum value 9 .