Q94 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If the area bounded by $y = x^3 + ax$ (where $a > 0$), the $x$-axis and the lines $x = -2$ and $x = 1$ is $\dfrac{37}{4}$ square units, then $\ldots$
- A$a = 4$
- Correct answerB$a = 2$
- C$a = 10$
- D$a = 20$
Solution
Given the curve $y = x^3 + ax$ with $a > 0$, the roots of $y = 0$ are given by $x(x^2 + a) = 0$. Since $a > 0$, the only real root is $x = 0$.
For $x \in [-2, 0]$, $y \le 0$, and for $x \in [0, 1]$, $y \ge 0$.
The required area $A$ is given by:
$A = \int_{-2}^{0} -(x^3 + ax) dx + \int_{0}^{1} (x^3 + ax) dx$
$A = -\left[ \dfrac{x^4}{4} + \dfrac{ax^2}{2} \right]_{-2}^{0} + \left[ \dfrac{x^4}{4} + \dfrac{ax^2}{2} \right]_{0}^{1}$
$A = -\left( 0 - \left( \dfrac{16}{4} + \dfrac{4a}{2} \right) \right) + \left( \dfrac{1}{4} + \dfrac{a}{2} - 0 \right)$
$A = (4 + 2a) + \left( \dfrac{1}{4} + \dfrac{a}{2} \right)$
$A = \dfrac{17}{4} + \dfrac{5a}{2}$
Given that the area is $\dfrac{37}{4}$ square units:
$\dfrac{17}{4} + \dfrac{5a}{2} = \dfrac{37}{4}$
$\dfrac{5a}{2} = \dfrac{20}{4}$
$\dfrac{5a}{2} = 5 \Rightarrow a = 2$
Answer: $a = 2$
For $x \in [-2, 0]$, $y \le 0$, and for $x \in [0, 1]$, $y \ge 0$.
The required area $A$ is given by:
$A = \int_{-2}^{0} -(x^3 + ax) dx + \int_{0}^{1} (x^3 + ax) dx$
$A = -\left[ \dfrac{x^4}{4} + \dfrac{ax^2}{2} \right]_{-2}^{0} + \left[ \dfrac{x^4}{4} + \dfrac{ax^2}{2} \right]_{0}^{1}$
$A = -\left( 0 - \left( \dfrac{16}{4} + \dfrac{4a}{2} \right) \right) + \left( \dfrac{1}{4} + \dfrac{a}{2} - 0 \right)$
$A = (4 + 2a) + \left( \dfrac{1}{4} + \dfrac{a}{2} \right)$
$A = \dfrac{17}{4} + \dfrac{5a}{2}$
Given that the area is $\dfrac{37}{4}$ square units:
$\dfrac{17}{4} + \dfrac{5a}{2} = \dfrac{37}{4}$
$\dfrac{5a}{2} = \dfrac{20}{4}$
$\dfrac{5a}{2} = 5 \Rightarrow a = 2$
Answer: $a = 2$
Hence, the correct answer is option: 'B'.