Q89 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If $\int f'(x) \cdot e^{x^2}\,dx = (x - 1) \cdot e^{x^2} + k$, where $k$ is constant of integration, then $f(x) = \ldots$
- A$2x^3 - \dfrac{x^2}{2} + x + c$, where $c$ is constant of integration.
- B$\dfrac{x^3}{2} + 3x^2 + 4x + c$, where $c$ is constant of integration.
- C$x^3 + 4x^2 + 6x + c$, where $c$ is constant of integration.
- Correct answerD$\dfrac{2x^3}{3} - x^2 + x + c$, where $c$ is constant of integration.
Solution
Differentiating both sides of the given equation with respect to $x$:
$\dfrac{d}{dx} \left[ \int f'(x) \cdot e^{x^2}\,dx \right] = \dfrac{d}{dx} \left[ (x - 1) \cdot e^{x^2} + k \right]$
$f'(x) \cdot e^{x^2} = 1 \cdot e^{x^2} + (x - 1) \cdot e^{x^2} \cdot (2x)$
$f'(x) \cdot e^{x^2} = e^{x^2} (1 + 2x^2 - 2x)$
Dividing both sides by $e^{x^2}$:
$f'(x) = 2x^2 - 2x + 1$
Integrating both sides with respect to $x$ to find $f(x)$:
$f(x) = \int (2x^2 - 2x + 1)\,dx$
$f(x) = \dfrac{2x^3}{3} - x^2 + x + c$
where $c$ is the constant of integration.
Answer: $\dfrac{2x^3}{3} - x^2 + x + c$, where $c$ is constant of integration.
$\dfrac{d}{dx} \left[ \int f'(x) \cdot e^{x^2}\,dx \right] = \dfrac{d}{dx} \left[ (x - 1) \cdot e^{x^2} + k \right]$
$f'(x) \cdot e^{x^2} = 1 \cdot e^{x^2} + (x - 1) \cdot e^{x^2} \cdot (2x)$
$f'(x) \cdot e^{x^2} = e^{x^2} (1 + 2x^2 - 2x)$
Dividing both sides by $e^{x^2}$:
$f'(x) = 2x^2 - 2x + 1$
Integrating both sides with respect to $x$ to find $f(x)$:
$f(x) = \int (2x^2 - 2x + 1)\,dx$
$f(x) = \dfrac{2x^3}{3} - x^2 + x + c$
where $c$ is the constant of integration.
Answer: $\dfrac{2x^3}{3} - x^2 + x + c$, where $c$ is constant of integration.
Hence, the correct answer is option: 'D'.