The area of the region enclosed between the parabola \(y = \displaystyle\frac{3x^2}{4}\) and the line \(3x-2y+12=0\) is,
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 27 sq. units
The enclosed region lies between the points where the line cuts the parabola.
From \(3x - 2y + 12 = 0\),
\( \displaystyle y = \frac{3x + 12}{2} \)
At the points of intersection, both curves give the same y:
\( \displaystyle \frac{3x^2}{4} = \frac{3x + 12}{2} \)
\(3x^2 = 6x + 24\)
\(x^2 - 2x - 8 = 0\)
\((x - 4)(x + 2) = 0,\ \ x = -2, 4\)
Between \(x = -2\) and \(x = 4\) the line is above the parabola (at x = 0 the line gives 6 and the parabola gives 0).
\( \displaystyle \text{Area} = \int_{-2}^{4} \left(\frac{3x + 12}{2} - \frac{3x^2}{4}\right) dx \)
\( \displaystyle = \left[\frac{3x^2}{4} + 6x - \frac{x^3}{4}\right]_{-2}^{4} \)
\(= (12 + 24 - 16) - (3 - 12 + 2)\)
\(= 20 - (-7) = 27\) sq. units
Note: NTA dropped this question. For the benefit of students and to give them additional practice, the question has been suitably modified so that it has one clear answer.