Exams Joint Entrance Examination Mathematics Differential Equations Question 44 Question 44 of 490 Medium MCQ
Let α \alpha α be a non-zero real number. Suppose f : R → R f:R\to R f : R → R is a differentiable function such that f ( 0 ) = 2 f(0)=2 f ( 0 ) = 2 and l i m x → − ∞ f ( x ) = 1 \underset{x\to -\infty}{lim}f(x)=1 x → − ∞ l im f ( x ) = 1 . If f ′ ( x ) = α f ( x ) + 3 {f}^{'}(x)=\alpha f(x)+3 f ′ ( x ) = α f ( x ) + 3 , for all x ∈ R x\in R x ∈ R , then f ( − log e 2 ) f(-{\log }_{e}2) f ( − log e 2 ) is equal to :
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f ( 0 ) = 2 , lim x → − ∞ f ( x ) = 1 f ′ ( x ) − x ⋅ f ( x ) = 3 I.F = e − α x y ( e − α x ) = ∫ 3 ⋅ e − α x d x f ( x ) ⋅ ( e − α x ) = 3 e − α x − α + c x = 0 ⇒ 2 = − 3 α + c ⇒ 3 α = c − 2 f ( x ) = − 3 α + c ⋅ e α x x → − ∞ ⇒ 1 = − 3 α + c ( 0 ) α = − 3 ∴ c = 1 f ( − ln 2 ) = − 3 α + c ⋅ e α x = 1 + e 3 ln 2 = 9 \begin{aligned}
& f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 \\\\
& f^{\prime}(x)-x \cdot f(x)=3 \\\\
& \text { I.F }=e^{-\alpha x} \\\\
& y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x \\\\
& f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\\\
& x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\\\
& f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\
& x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\\\
& \alpha=-3 \therefore c=1 \\\\
& f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\
& =1+e^{3 \ln 2}=9
\end{aligned} f ( 0 ) = 2 , x → − ∞ lim f ( x ) = 1 f ′ ( x ) − x ⋅ f ( x ) = 3 I.F = e − α x y ( e − α x ) = ∫ 3 ⋅ e − α x d x f ( x ) ⋅ ( e − α x ) = − α 3 e − α x + c x = 0 ⇒ 2 = α − 3 + c ⇒ α 3 = c − 2 f ( x ) = α − 3 + c ⋅ e α x x → − ∞ ⇒ 1 = α − 3 + c ( 0 ) α = − 3 ∴ c = 1 f ( − ln 2 ) = α − 3 + c ⋅ e α x = 1 + e 3 l n 2 = 9
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