Q88 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
If $f'(x) = \dfrac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}}$ and $f(0) = e$ then $f(1) = \ldots\ldots\ldots\ldots$
- A$e$
- B$2e$
- Correct answerC$3e$
- D$4e$
Solution
$f(x) = \int \dfrac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}} dx$
Let $\sqrt{x} = t \Rightarrow \dfrac{1}{2\sqrt{x}} dx = dt \Rightarrow \dfrac{dx}{\sqrt{x}} = 2 dt$
$f(x) = \int (t+1)e^t (2 dt) = 2 \int (t+1)e^t dt$
$f(x) = 2 t e^t + C$
Substituting $t = \sqrt{x}$:
$f(x) = 2 \sqrt{x} e^{\sqrt{x}} + C$
Given $f(0) = e$:
$2(0)e^0 + C = e \Rightarrow C = e$
Thus, $f(x) = 2 \sqrt{x} e^{\sqrt{x}} + e$
For $x = 1$:
$f(1) = 2(1)e^1 + e = 3e$
Answer: $3e$
Let $\sqrt{x} = t \Rightarrow \dfrac{1}{2\sqrt{x}} dx = dt \Rightarrow \dfrac{dx}{\sqrt{x}} = 2 dt$
$f(x) = \int (t+1)e^t (2 dt) = 2 \int (t+1)e^t dt$
$f(x) = 2 t e^t + C$
Substituting $t = \sqrt{x}$:
$f(x) = 2 \sqrt{x} e^{\sqrt{x}} + C$
Given $f(0) = e$:
$2(0)e^0 + C = e \Rightarrow C = e$
Thus, $f(x) = 2 \sqrt{x} e^{\sqrt{x}} + e$
For $x = 1$:
$f(1) = 2(1)e^1 + e = 3e$
Answer: $3e$
Hence, the correct answer is option: 'C'.