MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Use substitution $t = 2 - x$ to simplify the integral and then apply standard power rules for integration.Explanation:Given integral is $\int_{0}^{2} x(2-x)^b \, dx = \frac{32}{7}$, where $b \in \mathbb{N}$.Let $t = 2 - x$.Then $x = 2 - t$ and $dx = -dt$.When $x = 0$, $t = 2$; when $x = 2$, $t = 0$.Substituting, the integral becomes $\int_{2}^{0} (2-t) t^b \, (-dt)$.Swapping the limits changes the sign, giving $\int_{0}^{2} (2-t) t^b \, dt$.Now expand the integrand: $\int_{0}^{2} 2t^b \, dt - \int_{0}^{2} t^{b+1} \, dt$.Integrate term by term: $2\left[\frac{t^{b+1}}{b+1}\right]_{0}^{2} - \left[\frac{t^{b+2}}{b+2}\right]_{0}^{2}$.Evaluating the limits: $\frac{2 \cdot 2^{b+1}}{b+1} - \frac{2^{b+2}}{b+2}$.Simplify the first term: $\frac{2^{b+2}}{b+1} - \frac{2^{b+2}}{b+2}$.Factor out $2^{b+2}$: $2^{b+2}\left(\frac{1}{b+1} - \frac{1}{b+2}\right)$.Combine the fractions: $\frac{2^{b+2}}{(b+1)(b+2)}$.This equals $\frac{32}{7}$. Checking the options, for $b = 6$, we get $\frac{2^{8}}{7 \cdot 8} = \frac{256}{56} = \frac{32}{7}$.Hence, the value of $b$ that satisfies the equation is $6$.Answer:$b = 6$, which is Option B.