JEE Main 2024 — Mathematics Calculus

Plotting the diagram of the given data we get,

Area of $\triangle OPQ$△OPQ is given by,

$${S}_{2}=\frac{1}{2}|\begin{matrix}0 & 0 & 1 \\ a & {a}^{2} & 1 \\ -b & {b}^{2} & 1\end{matrix}|$$S2​=21​∣0a−b​0a2b2​111​∣

$$\Rightarrow {S}_{2}=\frac{1}{2}(a{b}^{2}+{a}^{2}b)...(i)$$⇒S2​=21​(ab2+a2b)...(i)

Now, equation of line $PQ:y-{a}^{2}=\frac{{a}^{2}-{b}^{2}}{a+b}(x-a)$PQ:y−a2=a+ba2−b2​(x−a)

$$\Rightarrow y-{a}^{2}=(a-b)x-(a-b)a$$⇒y−a2=(a−b)x−(a−b)a

$$\Rightarrow y=(a-b)x+ab...(ii)$$⇒y=(a−b)x+ab...(ii)

Area bounded by the line $PQ$PQ and the parabola is given by,

$${S}_{1}=\int _{-b}^{a}((a-b)x+ab-{x}^{2})dx$$S1​=∫−ba​((a−b)x+ab−x2)dx

$$\Rightarrow {S}_{1}={[(a-b)\frac{{x}^{2}}{2}+(ab)x-\frac{{x}^{3}}{3}]}_{-b}^{a}$$⇒S1​=[(a−b)2x2​+(ab)x−3x3​]−ba​

$$\Rightarrow {S}_{1}=\frac{{(a-b)}^{2}(a+b)}{2}+ab(a+b)-\frac{({a}^{3}+{b}^{3})}{3}...(iii)$$⇒S1​=2(a−b)2(a+b)​+ab(a+b)−3(a3+b3)​...(iii)

$$\Rightarrow \frac{{S}_{1}}{{S}_{2}}=\frac{\frac{{(a-b)}^{2}}{2}+ab-\frac{({a}^{2}+{b}^{2}-ab)}{3}}{\frac{ab}{2}}$$⇒S2​S1​​=2ab​2(a−b)2​+ab−3(a2+b2−ab)​​

$$\Rightarrow \frac{{S}_{1}}{{S}_{2}}=\frac{3{(a-b)}^{2}+6ab-2({a}^{2}+{b}^{2}-ab)}{3ab}$$⇒S2​S1​​=3ab3(a−b)2+6ab−2(a2+b2−ab)​

$$\Rightarrow \frac{{S}_{1}}{{S}_{2}}=\frac{3({a}^{2}+{b}^{2}-2ab)+6ab-2({a}^{2}+{b}^{2}-ab)}{3ab}$$⇒S2​S1​​=3ab3(a2+b2−2ab)+6ab−2(a2+b2−ab)​

$$\Rightarrow \frac{{S}_{1}}{{S}_{2}}=\frac{3{a}^{2}+3{b}^{2}-6ab+6ab-2{a}^{2}-2{b}^{2}+2ab}{3ab}$$⇒S2​S1​​=3ab3a2+3b2−6ab+6ab−2a2−2b2+2ab​

$$\Rightarrow \frac{{S}_{1}}{{S}_{2}}=\frac{{a}^{2}+{b}^{2}+2ab}{3ab}$$⇒S2​S1​​=3aba2+b2+2ab​

$$\Rightarrow \frac{{S}_{1}}{{S}_{2}}=\frac{1}{3}[\frac{a}{b}+\frac{b}{a}+2]$$⇒S2​S1​​=31​[ba​+ab​+2]

$\Rightarrow {(\frac{{S}_{1}}{{S}_{2}})}_{\mathrm{min}}=\frac{1}{3}[1+1+2]$⇒(S2​S1​​)min​=31​[1+1+2] ${\text{as}\frac{a}{b}+\frac{b}{a}\geq 2}$asba​+ab​≥2

$$\Rightarrow {(\frac{{S}_{1}}{{S}_{2}})}_{\mathrm{min}}=\frac{4}{3}$$⇒(S2​S1​​)min​=34​

$$\Rightarrow m+n=7$$⇒m+n=7