CUET · MATHS · PYQ PAPER 2025
The area bounded by \(y=3 x+1, x=0, y=0\) and \(x=a\) is 8 Sq.units. Then value of \(a\) (where \(a>0\) ) is
- A 1
- B 2
- C 3
- D \(8 / 3\)
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Area \(A = \int_{0}^{a} (3x+1) dx\) \(8 = \left[ \frac{3x^2}{2} + x \right]_{0}^{a}\) \(8 = \frac{3a^2}{2} + a\) \(16 = 3a^2 + 2a\) \(3a^2 + 2a - 16 = 0\) \((3a+8)(a-2) = 0\) \(a = 2\) (since \(a > 0\))
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