JEE Main 2021 — Mathematics Calculus

$$h=r\mathrm{sin}\theta +r$$h=rsinθ+r

base $=BC=2r\mathrm{cos}\theta$=BC=2rcosθ

$$\theta \in [0,\frac{\pi }{2})$$θ∈[0,2π​)

Area of $\Delta ABC=\frac{1}{2}(BC).h$ΔABC=21​(BC).h

$$\Delta =\frac{1}{2}(2r\mathrm{cos}\theta ).(r\mathrm{sin}\theta +r)$$Δ=21​(2rcosθ).(rsinθ+r)

$$={r}^{2}(\mathrm{cos}\theta ).(1+\mathrm{sin}\theta )$$=r2(cosθ).(1+sinθ)

$$\frac{d\Delta }{d\theta }={r}^{2}[\mathrm{cos}\theta \frac{d}{d\theta }(1+\mathrm{sin}\theta )+(1+\mathrm{sin}\theta )\frac{d}{d\theta }(\mathrm{cos}\theta )]$$dθdΔ​=r2[cosθdθd​(1+sinθ)+(1+sinθ)dθd​(cosθ)]

$$\frac{d\Delta }{d\theta }={r}^{2}[\mathrm{cos}\theta (0+\mathrm{cos}\theta )+(1+\mathrm{sin}\theta )(-\mathrm{sin}\theta )]$$dθdΔ​=r2[cosθ(0+cosθ)+(1+sinθ)(−sinθ)]

$$\frac{d\Delta }{d\theta }={r}^{2}[{\mathrm{cos}}^{2}\theta -\mathrm{sin}\theta -{\mathrm{sin}}^{2}\theta ]$$dθdΔ​=r2[cos2θ−sinθ−sin2θ]

$$\frac{d\Delta }{d\theta }={r}^{2}[1-{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta -{\mathrm{sin}}^{2}\theta ]$$dθdΔ​=r2[1−sin2θ−sinθ−sin2θ]

$$={r}^{2}[1-\mathrm{sin}\theta -2{\mathrm{sin}}^{2}\theta ]$$=r2[1−sinθ−2sin2θ]

$$={r}^{2}[-2{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta +1]$$=r2[−2sin2θ−sinθ+1]

$$={r}^{2}[-2{\mathrm{sin}}^{2}\theta -2\mathrm{sin}\theta +\mathrm{sin}\theta +1]$$=r2[−2sin2θ−2sinθ+sinθ+1]

$$={r}^{2}[-2\mathrm{sin}\theta (\mathrm{sin}\theta +1)+1(\mathrm{sin}\theta +1)]$$=r2[−2sinθ(sinθ+1)+1(sinθ+1)]

$$=\underset{\mathrm{positive}}{\underset{⏟}{{r}^{2}[1+sin\theta ]}}[1-2\mathrm{sin}\theta ]=0$$=positive⏟r2[1+sinθ]​​[1−2sinθ]=0

$$\Rightarrow \theta =\frac{\pi }{6}$$⇒θ=6π​

$\Rightarrow$⇒ $\Delta$Δ is maximum where $\theta =\frac{\pi }{6}$θ=6π​

${\Delta }_{\mathrm{max}}=\frac{3\sqrt{3}}{4}{r}^{2}=$Δmax​=433​​r2= area of equilateral $\Delta$Δ with $BC=\sqrt{3}r$BC=3​r.