For a differentiable function f : R R , suppose f ' ( x ) = 3 f ( x ) , where , f ( 0 ) = 1 and lim x ? - ? f ( x ) = 7 . Then 9 f ( - log e 3 ) is equal to ______ . [2024]

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Solution

Q.
For a differentiable function $f : R \to R,$ suppose $f' (x) = 3 f (x) + α,$ where $α \in R,$ $f (0) = 1$ and $\lim_{x \to - \infty} f (x) = 7 .$ Then $9 f (- \log_{e} 3)$ is equal to ______ . [2024]

Ans.
(61)

We have $f' (x) = 3 f (x) + α$

Let $f (x) = y$ and $\frac{d y}{d x} = f' (x) \Rightarrow \frac{d y}{d x} = 3 y + α \Rightarrow \frac{d y}{3 y + α} = d x$

$\Rightarrow \frac{1}{3} \log (3 y + α) = x + c ... (i) [On integrating]$

Now, $f (0) = 1,$ so $y (0) = 1$

$\Rightarrow \frac{1}{3} \log (3 + α) = c$

$\Rightarrow \frac{1}{3} \log (\frac{3 y + α}{3 + α}) = x$ [Putting value of $c$ in eq. (i)]

$\Rightarrow \frac{3 y + α}{3 + α} = e^{3 x} \Rightarrow y = \frac{e^{3 x} (3 + α) - α}{3} = f (x)$

Now, $\lim_{x \to - \infty} f (x) = 7$

$\Rightarrow \lim_{x \to - \infty} \frac{e^{3 x} (3 + α) - α}{3} = 7 \Rightarrow \frac{- α}{3} = 7$

$\Rightarrow α = - 21$

$∴ f (x) = 7 - 6 e^{3 x}$

So, $9 f (- \log_{e} 3) = 9 (7 - 6 e^{3 (- \log_{e} 3)})$

$= 9 (7 - \frac{6}{27}) = 63 - 2 = 61$

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