Let be a non-zero real number. Suppose f: R R is a differentiable… - Vidyadip

MCQ

Let $\alpha$ be a non$-$zero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $f(0)=2$ and $\lim _{x \rightarrow-\infty} f ( x )=1$. If $f^{\prime}( x )=\alpha f(x)+3$, for all $x \in R$, then $f\left(-\log _e 2\right)$ is equal to $..........$

A
3
B
5
C
9
D
7

✓

Answer

$f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1$
$f^{\prime}(x)-x \cdot f(x)=3$
$I \cdot F=e^{-\alpha x}$
$y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x$
$f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c$
$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 $
$($But a should be greater than $0$ for finite value of $c)$

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