2010 - JAMB Mathematics Past Questions and Answers - page 7

3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = ^-20/₄ = -5
From (1) : a + 2(-5) = -9
a - 10 = -9
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44

\(\int\)(Sin x + 2)dx = -cos x + 2x + k

Given that y = -3 - 2x + X²

then, \(\frac{dy}{dx}\) = -2 + 2x

At maximum value, \(\frac{dy}{dx}\) = O

therefore, -2 + 2x

2x = 2

x = ²/₂ = 1

By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m₁ = -4

let the gradient of the line parallel to the given line be m₂,

then, m₂ = m₁ = -4

(condition for parallelism)

using, y - y₁ = m₂(x - x₁)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0

p = \(\frac{M}{5}\)(X + Q) + 1

P - 1 = \(\frac{M}{5}\)(X + Q)

\(\frac{5}{M}\)(p - 1) = X + Q

\(\frac{5}{M}\)(p - 1)- x = Q

Q = \(\frac{5(p -1) - Mx}{M}\)

= \(\frac{5p - 5 - Mx}{M}\)

= \(\frac{5p - Mx - 5}{M}\)