Solve the equation (x - 2) (x - 3) = 12
2, 3
3, 6
-1, 6
1, -6
Correct answer is C
(x - 2) (x - 3) = 12
x2 - 3x - 2x + 6 = 12
x2 - 5x - 6 = 0
(x +1)(x - 6) = 0
x = -1 or 6
What is the nth term of the progression 27, 9, 3,......?
27\(\frac{1}{3}\) n - 1
3n + 2
27 + 18(n - 1)
27 + 6(n - 1)
Correct answer is A
Given 27, 9, 3,......this is a G.P
r = \(\frac{9}{27}\)
= \(\frac{1}{3}\)
T = arn - 1
= 27\(\frac{1}{3}\) n - 1
Find the positive number n, such that thrice its square is equal to twelve times the number
1
2
3
4
Correct answer is D
3n2 = 12n
= 3n2 - 12n = 0
= 3n(n - 4) = 0
∴ n = 4
x < \(\frac{12}{13}\)
x < 13
x < 9
\(\frac{13}{12}\)
Correct answer is A
\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1
= \(\frac{6x + 4x + 3x < 12}{12}\)
i.e. 13 x < 12 = x < \(\frac{12}{13}\)
Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x\(^2\) + xy = 6.
-1 and 5
-5 and 1
1 and 5
1 and 1
Correct answer is A
\(3x + y = 8 ... (i)\)
\(x^{2} + xy = 6 ... (ii)\)
From (i), \(y = 8 - 3x\)
From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\)
Equating the two values of y, we have
\(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\)
\(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\)
\(2x^{2} - 8x + 6 = 0\)
\(x^{2} - 4x + 3 = 0\)
\(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\)
\((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\)
\(y = 8 - 3x \)
When x = 1, \(y = 8 - 3(1) = 5\)
When x = 3, \(y = 8 - 3(3) = -1\)
\(\therefore \text{y = -1 or 5}\)