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JAMB Mathematics 2023 Paper

Question : 25

Total: 80

2x^2

4x + 16

Solution:

To find the point of intersection, equate the two equations

⇒ 2x

^2

+ 10 = 4x + 16

⇒ 2x

^2

+ 10 - 4x - 16 = 0

⇒ 2x

^2

- 4x - 6 = 0

Factorize

⇒ 2(x + 1)(x - 3) = 0

∴ x = -1 or 3

The curves will intersect at -1 and 3

Area =

\int\limits_{a}^{b}

} [upper function] - [lower function] dx

= $\int\limits_{1}^{3} 4x + 16 - (2x^2 + 10) dx$

= $\int\limits_{1}^{3} -2x^2 + 4x + 6dx$

= $\left( -\frac{2x^3}{3} + 2x^2 + 6x \right)_{1}^{3}$

= $\left( -\frac{2(3)^3}{3} + 2(3)^2 + 6(3) \right) - \left( -\frac{2(-1)^3}{3} + 2(-1)^2 + 6(-1) \right)$

= 18 - $\left( \frac{10}{3} \right)$

= 18 + $\frac{10}{3}$

= $\frac{64}{3}$

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