JEE Main 26 Feb 2021 Shift 2 Solved Paper

Let a △ABC inscribed in a circle with centre O and radius r.Let ∠OBC=θ
Now, area of △ABC=‌

1

2

× Base × Height
A=‌

1

2

×(BC)×(AP) . . . (i)
Now, BC=2BP
Consider ∆OBP, where OB=r
Then, BP=r‌cos‌θ
Hence, BC=2r‌cos‌θ
Again, AP=AO+OP
where, AO=r
Consider ∆OBP, where OB=r
Then, OP=r‌sin‌θ
⇒‌‌AP=r+r‌sin‌θ
From Eq. (i), we get
Area =‌

1

2

×(2r‌cos‌θ)×(r+r‌sin‌θ)
A=r2‌cos‌θ(1+sin‌θ)
Now, ‌

dA

dθ

=r2(−sin‌θ)(1+sin‌θ)+r2cos2θ
‌‌=r2(cos2θ−sin‌θ−sin2θ)
‌‌=r2(1−2sin2θ−sin‌θ)
‌‌=r2(1+sin‌θ)(1−2‌sin‌θ)
Equate ‌

dA

dθ

=0
⇒‌‌r2(1+sin‌θ)(1−2‌sin‌θ)=0
r=r2(1+sin‌θ)(1−2‌sin‌θ)=0
⇒‌‌sin‌θ=‌

1

2

⇒θ=‌

π

6

Now, ‌

d2A

dθ2

<0, when θ=‌

π

6

⇒A is maximum, when θ=‌

π

6

∴ Maximum area =r2‌cos(‌

π

6

)(1+sin‌

π

6

)=‌

3√3

4

r2
Height =AP=‌

3

2

r
Consider △ABP,
(AB)2‌=(AP)2+(BP)2‌
‌=(‌

3

2

r)2+(‌

√3

2

r)2‌[∵BP=r‌cos‌θ]
‌=‌

9

4

r2+‌

3

4

r2=3r2‌
⇒‌‌AB‌=√3r‌
Hence, the △ABC is an equilateral triangle with side √3.