JEE Mains · Maths · STD 12 - 8. Application and integration
One of the points of intersection of the curves \(\mathrm{y}=1+3 \mathrm{x}-2 \mathrm{x}^2\) and \(\mathrm{y}=\frac{1}{\mathrm{x}}\) is \(\left(\frac{1}{2}, 2\right)\). Let the area of the region enclosed by these curves be \(\frac{1}{24}(\ell \sqrt{5}+\mathrm{m})-\operatorname{nlog}_{\mathrm{e}}(1+\sqrt{5})\), where \(\ell, \mathrm{m}, \mathrm{n} \in\) \(\mathrm{N}\). Then \(\ell+\mathrm{m}+\mathrm{n}\) is equal to
- A \(32\)
- B \(30\)
- C \(29\)
- D \(31\)
Answer & Solution
Correct Answer
(B) \(30\)
Step-by-step Solution
Detailed explanation
\( A=\int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\left(1+3 x-2 x^2-\frac{1}{x}\right) d x \) \( A=\left[x+\frac{3 x^2}{2}-\frac{2 x^3}{3}-\ln x\right]_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}} \)…
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