JEE Main 2024 — Mathematics Calculus

Given: $f(0)=1,\underset{x\rightarrow -\infty }{\mathrm{lim}}f(x)=1$f(0)=1,x→−∞lim​f(x)=1

And ${f}^{'}(x)-\alpha \cdot f(x)=3$f′(x)−α⋅f(x)=3

$$\Rightarrow \frac{dy}{dx}-\alpha \cdot y=3$$⇒dxdy​−α⋅y=3

Which is a linear differential equation, so

$$\Rightarrow I.F.={e}^{-\alpha x}$$⇒I.F.=e−αx

Solution of the differential equation is given by,

$$y({e}^{-\alpha x})=\int 3\cdot {e}^{-\alpha x}dx$$y(e−αx)=∫3⋅e−αxdx

$$\Rightarrow f(x)\cdot ({e}^{-\alpha x})=\frac{3{e}^{-\alpha x}}{-\alpha }+c$$⇒f(x)⋅(e−αx)=−α3e−αx​+c

$$\Rightarrow f(x)=\frac{-3}{\alpha }+c\cdot {e}^{\alpha x}$$⇒f(x)=α−3​+c⋅eαx

Now, taking case I: when $\alpha >0$α>0 and $\underset{x\rightarrow -\infty }{\mathrm{lim}}f(x)=1$x→−∞lim​f(x)=1 we get,

$$\Rightarrow 1=\frac{-3}{\alpha }+c(0)$$⇒1=α−3​+c(0)

$\alpha =-3$α=−3 (rejected)

Case-II: $\alpha <0$α<0

as $\underset{x\rightarrow -\infty }{\mathrm{lim}}f(x)=1$x→−∞lim​f(x)=1

\Rightarrow c=0&\frac{-3}{\alpha }=1 ${\text{as}{e}^{-(-\infty )}\rightarrow \infty ,\text{so }c=0}$ase−(−∞)→∞,so c=0

$$\Rightarrow \alpha =-3$$⇒α=−3

$\Rightarrow f(x)=1$⇒f(x)=1 and also $f(0)=1$f(0)=1

Hence, $f(x)=1$f(x)=1 is constant function, so $f(-{\mathrm{log}}_{e}2)=1$f(−loge​2)=1