JEE Mains · Maths · STD 12 - 8. Application and integration
If the line \(y=m x\) bisects the area enclosed by the lines \(x=0, y=0, x=\frac{3}{2}\) and the curve \(\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^{2}\), then \(12 \mathrm{~m}\) is equal to ..... .
- A \(4\)
- B \(15\)
- C \(28\)
- D \(26\)
Answer & Solution
Correct Answer
(D) \(26\)
Step-by-step Solution
Detailed explanation
Total area \(=\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) \,d x\) \(=x+2 x^{2}-\left.\frac{x^{3}}{3}\right|_{0} ^{3 / 2}=\frac{39}{8}\) \(\ \,\frac{39}{16}=\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \,m\) \(\Rightarrow 3\, m=\frac{13}{2} \Rightarrow 12 \,m=26\)
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