If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R
\(\frac{-32}{7}\)
\(\frac{-23}{7}\)
\(\frac{23}{7}\)
\(\frac{32}{7}\)
Correct answer is C
\(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) + \(\frac{9}{7(x-3)}\)
15-2x = R(x - 3) +9(x +4)/7
Put x =-4, we have 15 -2(-4) = -7R
23 -7R;
R = 23/7
{21, 91, 221}
{21, 91, 221, 381}
{1,21, 91, 221}
{1,21, 91, 221,381}
Correct answer is A
multiples of 5 less than 20 = 5, 10 and 15
= [ 5, 10 and 15 ]x\(^2\) - x + 1
x\(^2\) - x + 1
when x = 5
5\(^2\) - 5 + 1; 25 - 5 + 1 --> 21
when x = 10
10\(^2\) - 10 + 1
100 - 9 = 91
when x = 15
15\(^2\) - 15 + 1
225 - 15 + 1 = 211
Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0.
17/4
17/2
17/√2
√17/2
Correct answer is C
2x\(^3\) - 4x + 2y\(^2\) - 6y - 2 = 0
Divide through by 2: x\(^2\) - 2x + y\(^2\) -3y -1 = 0
x\(^2\) -2x + y\(^2\) - 3y = 1
x\(^2\) -2x + 1 + y\(^2\) - 3y + 9/4
= 1+ 1 + 9/4
= (x- 1)\(^2\) (y - 3/2)\(^2\)
= √[17/4]
r = √17/2
Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2.
72
84
96
120
Correct answer is B
y 3x\(^2\) (2x +1) = 6x\(^3\) + 3x\(^2\)
dy/dx = 18x\(^2\) + 6x
At x = 2, 18(2)\(^2\) + 6(2)
= 72 + 12 = 84
Find the equation of the normal to the curve y= 2x\(^2\) - 5x + 10 at P(1, 7)
y+x-3 =0
y-x+6=0
y - x - 6=0
y -x+ 3 =0
Correct answer is C
From y= 2x- 5x 10 ; dy/dx= 4x-5
But at (1, 7); m\(_1\) = (dy/dx)\(_{1,7}\)
= 4(1) - 5 = -1
Using m\(_1\)m\(_2\),= -1; m\(_2\) = 1.
Gradient m\(_2\), of normal at (1,7) is 1.
Using y - y\(_1\) = m(x - x\(_1\))
y - 7 = 1(x - 1) ;
y - 7 = x - 1
y - x - 6 = 0
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