Q93 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If $\int_0^2 x(2 - x)^b\,dx = \dfrac{32}{7}$, where $b \in N$ then $b = $
- A$5$
- Correct answerB$6$
- C$7$
- D$8$
Solution
Let $I = \int_0^2 x(2 - x)^b \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$, we get:
$I = \int_0^2 (2 - x)(2 - (2 - x))^b \, dx$
$I = \int_0^2 (2 - x)x^b \, dx$
$I = \int_0^2 (2x^b - x^{b+1}) \, dx$
Integrating the terms, we get:
$I = \left[ \dfrac{2x^{b+1}}{b+1} - \dfrac{x^{b+2}}{b+2} \right]_0^2$
$I = \dfrac{2 \cdot 2^{b+1}}{b+1} - \dfrac{2^{b+2}}{b+2}$
$I = 2^{b+2} \left( \dfrac{1}{b+1} - \dfrac{1}{b+2} \right)$
$I = \dfrac{2^{b+2}}{(b+1)(b+2)}$
It is given that $I = \dfrac{32}{7}$.
$\dfrac{2^{b+2}}{(b+1)(b+2)} = \dfrac{32}{7}$
Substituting $b = 6$ from the options:
$\dfrac{2^{6+2}}{(6+1)(6+2)} = \dfrac{2^8}{7 \times 8} = \dfrac{256}{56} = \dfrac{32}{7}$
This satisfies the given equation. Therefore, $b = 6$.
Answer: $6$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$, we get:
$I = \int_0^2 (2 - x)(2 - (2 - x))^b \, dx$
$I = \int_0^2 (2 - x)x^b \, dx$
$I = \int_0^2 (2x^b - x^{b+1}) \, dx$
Integrating the terms, we get:
$I = \left[ \dfrac{2x^{b+1}}{b+1} - \dfrac{x^{b+2}}{b+2} \right]_0^2$
$I = \dfrac{2 \cdot 2^{b+1}}{b+1} - \dfrac{2^{b+2}}{b+2}$
$I = 2^{b+2} \left( \dfrac{1}{b+1} - \dfrac{1}{b+2} \right)$
$I = \dfrac{2^{b+2}}{(b+1)(b+2)}$
It is given that $I = \dfrac{32}{7}$.
$\dfrac{2^{b+2}}{(b+1)(b+2)} = \dfrac{32}{7}$
Substituting $b = 6$ from the options:
$\dfrac{2^{6+2}}{(6+1)(6+2)} = \dfrac{2^8}{7 \times 8} = \dfrac{256}{56} = \dfrac{32}{7}$
This satisfies the given equation. Therefore, $b = 6$.
Answer: $6$
Hence, the correct answer is option: 'B'.