JEE Main 2024 — Mathematics Calculus

Given: $(2x+3y-2)dx+(4x+6y-7)dy=0$(2x+3y−2)dx+(4x+6y−7)dy=0

$$\Rightarrow \frac{dy}{dx}=\frac{-(2x+3y-2)}{(4x+6y-7)}...(i)$$⇒dxdy​=(4x+6y−7)−(2x+3y−2)​...(i)

Let, $2x+3y-2=t$2x+3y−2=t

$$\Rightarrow 2+3\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}...(ii)$$⇒2+3dxdy​=dxdt​...(ii)

Also, $4x+6y-4=2t$4x+6y−4=2t

$$\Rightarrow 4x+6y-7=2t-3...(iii)$$⇒4x+6y−7=2t−3...(iii)

Using $(i),(ii)\text{and}(iii)$(i),(ii)and(iii),

$$\Rightarrow \frac{1}{3}(\frac{dt}{dx}-2)=\frac{-t}{2t-3}$$⇒31​(dxdt​−2)=2t−3−t​

$$\Rightarrow \frac{dt}{dx}-2=\frac{-3t}{2t-3}$$⇒dxdt​−2=2t−3−3t​

$$\Rightarrow \frac{dt}{dx}=\frac{-3t}{2t-3}+2$$⇒dxdt​=2t−3−3t​+2

$$\Rightarrow \frac{dt}{dx}=\frac{-3t+4t-6}{2t-3}$$⇒dxdt​=2t−3−3t+4t−6​

$$\Rightarrow \frac{dt}{dx}=\frac{t-6}{2t-3}$$⇒dxdt​=2t−3t−6​

$$\Rightarrow \int \frac{2t-3}{t-6}dt=\int dx$$⇒∫t−62t−3​dt=∫dx

$$\Rightarrow \int (\frac{2t-12}{t-6}+\frac{9}{t-6})\cdot dt=x+c$$⇒∫(t−62t−12​+t−69​)⋅dt=x+c

$$\Rightarrow 2t+9\mathrm{log}(t-6)=x+c$$⇒2t+9log(t−6)=x+c

$$\Rightarrow 2(2x+3y-2)+9\mathrm{log}(2x+3y-8)=x+c...(iv)$$⇒2(2x+3y−2)+9log(2x+3y−8)=x+c...(iv)

It is given that $y(0)=3$y(0)=3.

$$\Rightarrow 2(0+9-2)+9\mathrm{log}(0+9-8)=0+c$$⇒2(0+9−2)+9log(0+9−8)=0+c

$$\Rightarrow 14+9\mathrm{log}(1)=c$$⇒14+9log(1)=c

$$\Rightarrow c=14$$⇒c=14

Using $(iv)$(iv),

$$\Rightarrow 4x+6y-4+9\mathrm{log}(2x+3y-8)=x+14$$⇒4x+6y−4+9log(2x+3y−8)=x+14

$$\Rightarrow 3x+6y+9\mathrm{log}(2x+3y-8)=18$$⇒3x+6y+9log(2x+3y−8)=18

$$\Rightarrow x+2y+3\mathrm{log}(2x+3y-8)=6$$⇒x+2y+3log(2x+3y−8)=6

Comparing with $\alpha x+\beta y+3{\mathrm{log}}_{e}|2x+3y-\gamma |=6$αx+βy+3loge​∣2x+3y−γ∣=6.

$$\Rightarrow \alpha =1,\beta =2,\gamma =8$$⇒α=1,β=2,γ=8

$$\Rightarrow \alpha +2\beta +3\gamma =1+4+24$$⇒α+2β+3γ=1+4+24

$$\Rightarrow \alpha +2\beta +3\gamma =29$$⇒α+2β+3γ=29