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

✓
$3$
B
$5$
C
$9$
D
$7$

✓

Answer

Correct option: A.

$3$

a
$ f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 $
$ f^{\prime}(x)-\alpha \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 $ $....(1)$
$ f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x}$
Case-I $\alpha > 0$
$\mathrm{x} \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+\mathrm{c}(0)$
$\alpha=-3 \quad$ (rejected)
Case-II $\alpha < 0$
as $\lim _{x \rightarrow-\infty} \mathrm{f}(\mathrm{x})=1 \Rightarrow \mathrm{c}=0$ and $\frac{-3}{\alpha}=1 \Rightarrow \alpha=-3$
$\Rightarrow \mathrm{f}(\mathrm{x})=1 \quad \text { (rejected) }$
as $f(0)=2$
$\Rightarrow$ data is inconsistent

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