For a differentiable function $\mathrm{f}: I R \rightarrow I R$, suppose $f^{\prime}(\mathrm{x})=3 f(\mathrm{x})+\alpha$, where $\alpha \in \mathrm{IR}, f(0)=1$ and $\lim _{x \rightarrow-\infty} f(x)=7$. Then $9 \mathrm{f}\left(-\log _{e} 3\right)$ is equal to

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Differential Equations

For a differentiable function

f : I R \to I R

, suppose

f^{'} (x) = 3 f (x) + α

, where

α \in IR, f (0) = 1

and

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

. Then

9 f (- lo g_{e} 3)

is equal to

Answer: 61

f^{'} (x) - 3 f (x) = α

. Using integrating factor

e^{- 3 x}

\frac{d}{d x} (f (x) e^{- 3 x}) = α e^{- 3 x} ⟹ f (x) e^{- 3 x} = - \frac{α}{3} e^{- 3 x} + C ⟹ f (x) = C e^{3 x} - \frac{α}{3}

x \to - \infty

C e^{3 x} \to 0

, so

x \to - \infty lim f (x) = - \frac{α}{3} = 7 ⟹ α = - 21

f (0) = 1 ⟹ C - \frac{α}{3} = 1 ⟹ C + 7 = 1 ⟹ C = - 6

, so

f (x) = - 6 e^{3 x} + 7

.

At

x = - lo g_{e} 3

e^{3 x} = 3^{- 3} = \frac{1}{27}

f (- lo g_{e} 3) = - 6 \times \frac{1}{27} + 7 = - \frac{2}{9} + 7 = \frac{61}{9} ⟹ 9 f (- lo g_{e} 3) = 61

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