-6
6
-12
12
Correct answer is A
\(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\)
(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = \(\frac{12}{-2}\)
= -6
- \(\frac{1}{2}\)
\(\frac{1}{3}\)
-1
2
Correct answer is B
N + Y = XY + X + Y
3 + -\(\frac{2}{3}\) = 3(- \(\frac{2}{3}\)) + 3 + (- \(\frac{2}{3}\))
= -2 + 3 -\(\frac{2}{3}\)
= \(\frac{1 - 2}{1 - 3}\)
= \(\frac{1}{3}\)
60
62
54
64
Correct answer is B
T2 = 4, T4 = 16
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)
r2 = 4 and r = 2
but ar = 4
a = \(\frac{4}{r}\) = \(\frac{4}{2}\)
a = 2
Sn = \(\frac{a(r^n - 1)}{r - 1}\)
S5 = \(\frac{2(2^5 - 1)}{2 - 1}\)
= \(\frac{2(32 - 1)}{2 - 1}\)
= 2(31)
= 62
Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.
505
513
433
635
Correct answer is B
3, 6, 9,..., 36.
a = 3, d = 3, i = 36, n = 18
Sn = \(\frac{n}{2}\) [2a + (n - 1)d
S18 = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513
Solve the inequality x2 + 2x > 15.
x < -3 or x > 5
-5 < x < 3
x < 3 or x > 5
x > 3 or x < -5
Correct answer is B
x2 + 2x > 15
x2 + 2x - 15 > 0
(x2 + 5x) - (3x - 15) > 0
x(x + 5) - 3(x + 5) >0
(x - 3)(x + 5) > 0
therefore, x = 3 or -5
i.e. x< 3 or x > -5