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Published on: 13/05/2022
QB365 provides detailed and simple solution for every Book back Questions in class 12 Business Maths Subject. It will helps to get more idea about question pattern in every book back questions with solution.
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Take MCQ Business Maths and Statistics Test

1.
If MR = 20 − 5x + 3x2, find total revenue function.
2.
Find the area bounded by the lines y − 2x − 4 = 0, y = 1, y = 3 and the y-axis
3.
Find the area of the region bounded by the parabola \(y=4{ - }x^{ 2 }\) , x −axis and the lines x = 0, x = 2.
4.
For the marginal revenue function MR = 6 − 3x2 − x3, Find the revenue function and demand function.
5.
If MR = 14 − 6x + 9x2, find the demand function.
1.
Given MR = 20-5x+3x2
\(\Rightarrow \frac { dR }{ dx } =20-5x+3{ x }^{ 2 }\)
⇒ dR = (20 - 5x + 3x2)dx
⇒ഽdR = ഽ(20-5x+3x2)dx
\(\Rightarrow R=20x-\frac { { 5x }^{ 2 } }{ 2 } +\frac { { 3x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
∴ R = 20x - \(\frac { 5{ x }^{ 2 } }{ x } +{ x }^{ 3 }\)
2.
y - 2x - 4 = 0
| x | 0 | -2 |
| y | 4 | 0 |

Given y - 2.x - 4 = 0
⇒ y-4 = 2x
⇒
Since the area lies to the left of Y-axis, with the limits y = 1 &y = 3.
Area =\(\int _{ 1 }^{ 3 }{ -xdy } \)
\(=\int _{ 1 }^{ 3 }{ -\left( \frac { 1 }{ 2 } \right) (y-4)dy } \)
\(=\frac { 1 }{ 2 } \int _{ 1 }^{ 3 }{ (4-y)dy } =\frac { 1 }{ 2 } { \left[ 4y-\frac { { y }^{ 2 } }{ 2 } \right] }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 2 } \left[ \left( 4(3)-\frac { { 3 }^{ 2 } }{ 2 } \right) -\left( 4(1)-\frac { { 1 }^{ 2 } }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \left( 12-\frac { 9 }{ 2 } \right) -\left( 4-\frac { 1 }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \left( \frac { 24-9 }{ 2 } \right) -\left( \frac { 8-1 }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } -\frac { 7 }{ 2 } \right] =\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } \right] \)

3.
\(y=4{ - }x^{ 2 }\)
Required area = \(\int _{ 0 }^{ 2 }{ ydx } =\int _{ 0 }^{ 2 }{ (4- } { x }^{ 2 })dx\)
= \({ \left[ 4x-\frac { { x }^{ 2 } }{ 3 } \right] }_{ 0 }^{ 2 }=8-\frac { 8 }{ 3 } \)
= \(\frac { 16 }{ 3 } \) sq.units

4.
Given MR = 6 − 3x2 − x3
⇒ ഽMR =ഽ(6 − 3x2 − x3)dx
\(\Rightarrow \mathrm{R}=6 x-\frac{\not{3} x^{3}}{\not3}-\frac{x^{4}}{4}+k\)
\(\Rightarrow 6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } \)
Demand function \(P=\frac { R }{ x } =6-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 4 } \)
5.
Given MR = 14 - x + 9x2
⇒ \(\frac{dR}{dx}\) = 14 - 6x + 9x2
⇒ dR = (14 - 6x + 9x2)dx
⇒ ഽdR = ഽ(14 - 6x + 9x2)dx
\(\Rightarrow \ R=14x-\frac { 6{ x }^{ 2 } }{ 2 } +\frac { { 9x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
R = 14x - 3x2+ 3x3
Demand function \(P=\frac { R }{ x } =\frac { 14x-3{ x }^{ 2 }+{ 3x }^{ 3 } }{ x } \)
⇒ P = 14 − 3x + 3x2
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