JEE Mathematics 2024 Question with Solution

Given: $f'(x) = \alpha f(x) + 3$, $f(0) = 2$, and $\lim_{x \to \infty} f(x) = 1$.

Find: $f(-\log 2)$.

Rewrite the differential equation in linear form:

$$f'(x) - \alpha f(x) = 3$$

Using the integrating factor method, take

$$\mu(x) = e^{\int -\alpha \, dx} = e^{-\alpha x}$$

Multiplying throughout by the integrating factor:

$$e^{-\alpha x} f'(x) - \alpha e^{-\alpha x} f(x) = 3e^{-\alpha x}$$

So,

$$\frac{d}{dx}\left(e^{-\alpha x} f(x)\right) = 3e^{-\alpha x}$$

Integrating both sides:

$$e^{-\alpha x} f(x) = \int 3e^{-\alpha x} \, dx = -\frac{3}{\alpha} e^{-\alpha x} + C$$

Hence,

$$f(x) = -\frac{3}{\alpha} + Ce^{\alpha x}$$

Using $f(0) = 2$,

$$2 = -\frac{3}{\alpha} + C$$

so

$$C = 2 + \frac{3}{\alpha}$$

Therefore,

$$f(x) = -\frac{3}{\alpha} + \left(2 + \frac{3}{\alpha}\right)e^{\alpha x}$$

Now use $\lim_{x \to \infty} f(x) = 1$. From the solution, this gives

$$-\frac{3}{\alpha} = 1$$

so

$$\alpha = -3$$

Substituting $\alpha = -3$,

$$f(x) = 1 + e^{-3x}$$

Now evaluate at $x = -\log 2$:

$$f(-\log 2) = 1 + e^{-3(-\log 2)} = 1 + e^{3\log 2}$$

Since

$$e^{3\log 2} = 2^3 = 8$$

we get

$$f(-\log 2) = 1 + 8 = 9$$

Therefore, the correct option is C.

Given: $f'(x) = \alpha f(x) + 3$, $f(0) = 2$, and $\lim_{x \to \infty} f(x) = 1$.

Find: $f(-\log 2)$.

A solved form shown in the solution is

$$f(x) = -\frac{3}{\alpha} + \left(2 + \frac{3}{\alpha}\right)e^{\alpha x}$$

For the limit at infinity to exist and equal $1$, the exponential term must vanish as $x \to \infty$, so the constant term must be the limiting value. Hence,

$$-\frac{3}{\alpha} = 1$$

which gives

$$\alpha = -3$$

Then

$$f(x) = 1 + e^{-3x}$$

Now substitute $x = -\log 2$:

$$\begin{aligned} f(-\log 2) &= 1 + e^{-3(-\log 2)} \\ &= 1 + e^{3\log 2} \\ &= 1 + 2^3 \\ &= 9 \end{aligned}$$

Therefore, the value of $f(-\log 2)$ is $9$, so the correct option is C.