MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Use the fundamental theorem of calculus and differentiate the given integral equation to find $f'(x)$, then integrate.Explanation:Differentiate both sides of the given equation with respect to $x$.$$\frac{d}{dx}\left(\int f'(x)e^{x^{2}}\,dx\right)=\frac{d}{dx}\left((x-1)e^{x^{2}}+k\right)$$The left side becomes $f'(x)e^{x^{2}}$ because differentiation reverses integration.The derivative of $k$ is zero, so use the product rule on $(x-1)e^{x^{2}}$.$$\frac{d}{dx}\left((x-1)e^{x^{2}}\right)=(x-1)(2xe^{x^{2}})+e^{x^{2}}(1)$$Thus,$$f'(x)e^{x^{2}}=\left(2x(x-1)+1\right)e^{x^{2}}$$Cancel $e^{x^{2}}$ from both sides.$$f'(x)=2x(x-1)+1=2x^{2}-2x+1$$Integrate $f'(x)$ to obtain $f(x)$.$$f(x)=\int \left(2x^{2}-2x+1\right)\,dx=\frac{2x^{3}}{3}-x^{2}+x+c$$Answer:$$f(x)=\frac{2x^{3}}{3}-x^{2}+x+c$$This matches Option D.