JEE Mathematics 2023 Question with Solution

Given: $f(x) = \int_{0}^{2} e^{|k-t|} \, dt$

Find: The minimum value of $f(x)$.

For $0 < k < 2$, split the integral at $t = k$ where the absolute value changes:

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

Now evaluate each part:

$$\int_{0}^{k} e^{k-t} \, dt = e^k \int_{0}^{k} e^{-t} \, dt = e^k(1-e^{-k}) = e^k - 1$$

and

$$\int_{k}^{2} e^{t-k} \, dt = e^{-k} \int_{k}^{2} e^t \, dt = e^{-k}(e^2-e^k) = e^{2-k} - 1$$

Hence,

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

Let $z = e^k$. Then

$$f(x) = z + \frac{e^2}{z} - 2$$

Using differentiation,

$$\frac{df}{dz} = 1 - \frac{e^2}{z^2}$$

Set this equal to zero:

$$1 - \frac{e^2}{z^2} = 0 \implies z^2 = e^2 \implies z = e$$

So $e^k = e$, giving $k = 1$. Therefore,

$$f_{\min} = e + e - 2 = 2(e-1)$$

Thus, the minimum value is $2(e-1)$, so the correct option is A.

The solution labels option B, but the working clearly gives $2(e-1)$, which matches option A.

Given: $f(x) = \int_{0}^{2} e^{|k-t|} \, dt$

Find: The minimum value of $f(x)$.

From the extracted working:

• For $k \le 0$,

$$f(x) = \int_{0}^{2} e^{t-k} \, dt = e^{-k}(e^2-1)$$

This decreases as $k$ increases.

• For $0 < k < 2$,

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

• For $k \ge 2$,

$$f(x) = \int_{0}^{2} e^{k-t} \, dt = e^{k-2}(e^2-1)$$

This increases as $k$ increases.

Therefore, the minimum must occur for $k \in (0,2)$. In that interval,

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

Apply A.M. $\ge$ G.M. to $e^k$ and $e^{2-k}$:

$$e^k + e^{2-k} \ge 2\sqrt{e^k \cdot e^{2-k}} = 2e$$

So,

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

Equality holds when

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

Hence the minimum value is $2(e-1)$. Therefore, the correct option is A.