Grade 11 maths problems with answers and solutions are presented.
Problems
An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.
Find the area between two concentric circles defined by
x^{2} + y^{2} -2x + 4y + 1 = 0
x^{2} + y^{2} -2x + 4y - 11 = 0
Find all values of parameter m (a real number) so that the equation 2x^{2} - m x + m = 0 has no real solutions.
The sum an integer N and its reciprocal is equal to 78/15. What is the value of N?
m and n are integers so that 4^{m} / 125 = 5^{n} / 64. Find values for m and n.
P is a polynomial such that P(x^{2} + 1) = - 2 x^{4} + 5 x^{2} + 6. Find P(- x^{2} + 3)
For what values of r would the line x + y = r be tangent to the circle x^{2} + y^{2} = 4?
Solutions to the Above Problems
Let x = speed of airplane in still air, y = speed of wind and D the distance between A and B. Find the ratio x / y
Against the wind: D = 8(x - y), with the wind: D = 7(x + y)
8x - 8y = 7x + 7y, hence x / y = 15
Rewrite equations of circles in standard form. Hence equation x^{2} + y^{2} -2x + 4y + 1 = 0 may be written as
(x - 1)^{2} + (y + 2) ^{2} = 4 = 2^{2}
and equation x^{2} + y^{2} -2x + 4y - 11 = 0 as
(x - 1)^{2} + (y + 2) ^{2} = 16 = 4^{2}
Knowing the radii, the area of the ring is ? (4)^{2} - ? (2)^{2} = 12 ?
The given equation is a quadratic equation and has no solutions if it discriminant D is less than zero.
D = (-m)^{2} - 4(2)(m) = m^{2} - 8 m
We nos solve the inequality m^{2} - 8 m < 0
The solution set of the above inequality is: (0 , 8)
Any value of m in the interval (0 , 8) makes the discriminant D negative and therefore the equation has no real solutions.
Write equation in N as follows
N + 1/N = 78/15
Multiply all terms by N, obtain a quadratic equation and solve to obtain N = 5.
4^{m} / 125 = 5^{n} / 64
Cross multiply: 64 4^{m} = 125 5^{n}
Note that 64 = 4^{3} and 125 = 5^{3}
The above equation may be written as: 4^{m + 3} = 5^{n + 3}
The only values of the exponents that make the two exponential expressions equal are: m + 3 = 0 and n + 3 = 0, which gives m = - 3 and n = - 3.
P(x^{2} + 1) = - 2 x^{4} + 5 x^{2} + 6
Let t = x^{2} + 1 which also gives x^{2} = t - 1
Substitute x^{2} by t - 1 in P to obtain: P(t) = - 2 (t - 1)^{2} + 5 (t - 1) + 6 = -2 t ^{2} + 9t - 1
Now let t = - x^{2} + 3 and substitute in P(t) above to obtain
P(- x^{2} + 3) = -2 (- x^{2} + 3) ^{2} + 9 (- x^{2} + 3) - 1 = -2 x ^{4} + 3 x ^{2} + 8
Solve x + y = r for y: y = r - x
Substitute in the equation of the circle:
x^{2} + (r - x)^{2} = 4
Expand: 2 x^{2} -2 r x + r ^{2} - 4 = 0
If we solve the above quadratic equation (in x) we will obtain the x coordinates of the points of intersection of the line and the circle. The 2 points of intersection "become one" and therefore the line and the circle become tangent if the discriminant D of the quadratic equation is zero. Hence
D = (-2r)^{2} - 4(2)(r^{2} - 4) = 4(8 - r^{2}) = 0
Solve for r to obtain: r = 2 √2 and r = - 2√2