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Published on: 13/05/2022
QB365 provides detailed and simple solution for every Creative Questions in class 12 Maths Subject. It will helps to get more idea about question pattern in every Creative questions with solution.
latest Creative QuestionsDownload Tamil Nadu 12th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Find the area enclosed by the parabolas 5x2-y=0 and 2x2-y+9=0.
2.
Find the area bounded by the curve y2(2a-x)=x2 and the line x=2a.
3.
Find the area bounded by the curves y=|x|-1 and y=-|x|+1
4.
Find the volume of the solid generated by the revolution of the loop of the curve x = t2 y = t - \(\frac { { t }^{ 3 } }{ 3 } \) about x-axis.
5.
Show that the area under the curve y = sin x and y = sin 2x between x = 0 and x = \(\frac { \pi }{ 3 } \) and x axis are as 2:3
1.
\(12\sqrt { 3 } \)
2.
3πa2
3.
2 sq.units
4.
Given x = t2, y = t - \(\frac { { t }^{ 3 } }{ 3 } \)
Point of intersection of the curve with x-axis is obtained by putting y = 0.
∴ y = 0
⇒ \(t-\frac { { t }^{ 3 } }{ 3 } =0\)
⇒ t = 0 or ± \(\sqrt3\)
∴ The limit is from t = 0 to t = \(\sqrt { 3 } \)
∴ Volume = \(\pi \int _{ 0 }^{ \sqrt { 3 } }{ { y }^{ 2 }dx } =\pi \int _{ 0 }^{ \sqrt { 3 } }{ { \left( t-\frac { { t }^{ 3 } }{ 3 } \right) }^{ 2 }}(2t \ dt)\)
[∵ x = t2 ⇒ dx = 2t dt]
\(=2\pi \int _{ 0 }^{ \sqrt { 3 } }{ { \left( \frac { 3t-{ t }^{ 3 } }{ 3 } \right) }^{ 2 }t \ dt } \)
\(=\frac { 2\pi }{ 9 } \int _{ 0 }^{ \sqrt { 3 } }{ (9{ t }^{ 2 }-6{ t }^{ 4 }+{ t }^{ 6 })t \ dt } \)
\(=\frac { 2\pi }{ 9 } \int _{ 0 }^{ \sqrt { 3 } }{ (9{ t }^{ 3 }-6{ t }^{ 5 }+{ t }^{ 7 })dt } \)
\(=\frac { 2\pi }{ 9 } { \left[ \frac { { 9 }t^{ 4 } }{ 4 } -\frac { 6{ t }^{ 6 } }{ 6 } +\frac { { t }^{ 8 } }{ 8 } \right] }_{ 0 }^{ \sqrt { 3 } }\)
\(=\frac { 2\pi }{ 9 } \left[ \frac { 81 }{ 4 } -27+\frac { 81 }{ 8 } -0 \right] \)
Volume \(=\frac { 2\pi }{ 9 } \left( \frac { 27 }{ 8 } \right) =\frac { 3\pi }{ 4 } \) cubic units
5.
Area under the curve y = sin x between x = 0 and x = \(\frac { \pi }{ 3 } \) is
\({ A }_{ 1 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ ydx } =\int _{ 0 }^{ \frac { \pi }{ 3 } }{ sinxdx } =-{ \left[ cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)
\(=-(cos\frac { \pi }{ 3 } -cos0)=-\left( \frac { 1 }{ 2 } -1 \right) \)
\(=-\left( -\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } \)
Area under the curve y = sin 2x between x = 0 and \(\frac { \pi }{ 3 } \) is
\({ A }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ { sin2 \ x \ dx=-\left[ \frac { cos2 \ x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 3 } } } \)
\(=-\frac { 1 }{ 2 } [cos2\frac { \pi }{ 3 } -cos0]-\frac { 1 }{ 2 } \left[ -\frac { 1 }{ 2 } -1 \right] =-\frac { 1 }{ 2 } \left( -\frac { 3 }{ 2 } \right) =\frac { 3 }{ 4 } \)
\(\therefore \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 2 } \times \frac { 4 }{ 3 } =\frac { 2 }{ 3 } \)
∴ A1:A2 =2 : 3
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