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
The gradient ofy= 3x\(^2\) + 11x + 7 at P(x.y) is -1. Find the coordinates of P.
(-3, -2)
(-2,-3)
(-2,3)
(2,2)
Correct answer is B
y= 3x\(^2\) + 11x +7;
dy/dx = 6x + 11= -1;
6x = -12;
x=-2
y= 3x\(^2\) + 11x + 7
At x-2,
y= 3(-2)\(^2\) +11(-2) + 7
= 12 - 22 + 7= -3
p(x, y) = (-2,-3)
12/91
16/91
30/91
32/91
Correct answer is D
Total balls = 8+4+2= 14
n(R) = 8, n (B) =4, n(G) = 2
Without replacement, it is p(RB) or p(BR)
= (8/14 x 4/13) + (4/14 + 8/13) = 16/91 + 16/91
= 32/91