JEE Mathematics 2023 Question with Solution

Given:

$$f'(x) + f(x) = \int_{0}^{2} f(t) \, dt$$

and $f(0) = e^{-2}$.

Find: $2f(0) - f(2)$.

Let

$$k = \int_{0}^{2} f(t) \, dt$$

which is a constant. Then the differential equation becomes

$$\frac{dy}{dx} + y = k$$

Multiplying by the integrating factor $e^x$,

$$y e^x = k e^x + c$$

So,

$$y = k + c e^{-x}$$

Using $f(0) = e^{-2}$,

$$e^{-2} = k + c$$

Hence,

$$c = e^{-2} - k$$

Therefore,

$$f(x) = k + (e^{-2} - k)e^{-x}$$

Now use the definition of $k$:

$$k = \int_{0}^{2} \left(k + (e^{-2} - k)e^{-x}\right) \, dx$$

Solving this gives

$$k = e^{-2} - 1$$

Thus,

$$f(x) = (e^{-2} - 1) + e^{-x}$$

Evaluating at $x=2$,

$$f(2) = 2e^{-2} - 1$$

and

$$f(0) = e^{-2}$$

Therefore,

$$2f(0) - f(2) = 1$$

So the required answer is $1$.

Given:

$$f'(x) + f(x) = \int_{0}^{2} f(t) \, dt$$

with $f(0)=e^{-2}$.

Find: $2f(0)-f(2)$.

Since the right-hand side does not depend on $x$, denote it by

$$k = \int_{0}^{2} f(t) \, dt$$

Then

$$f'(x) + f(x) = k$$

This is a linear first-order differential equation. Its solution is

$$f(x) = k + c e^{-x}$$

Using the initial condition,

$$f(0)=k+c=e^{-2}$$

so

$$c=e^{-2}-k$$

Hence,

$$f(x)=k+(e^{-2}-k)e^{-x}$$

Now integrate from $0$ to $2$:

$$k=\int_{0}^{2} f(x) \, dx = \int_{0}^{2} \left[k+(e^{-2}-k)e^{-x}\right] \, dx$$

That is,

$$k=2k+(e^{-2}-k)\int_{0}^{2} e^{-x} \, dx$$

Also,

$$\int_{0}^{2} e^{-x} \, dx = 1-e^{-2}$$

Therefore,

$$k=2k+(e^{-2}-k)(1-e^{-2})$$

which simplifies to the value stated in the solution:

$$k=e^{-2}-1$$

So,

$$f(x)=(e^{-2}-1)+e^{-x}$$

Now,

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

and

$$2f(0)=2e^{-2}$$

Thus,

$$2f(0)-f(2)=2e^{-2}-(2e^{-2}-1)=1$$

Therefore, the required value is $1$.