closure
identity
positive
inverse
Correct answer is D
a \(\oplus\) b = ab
The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab
The number 1 is the identity element under multiplication
Find the nth term of the sequence 3, 6, 10, 15, 21.....
\(\frac{n(n - 1)}{2}\)
\(\frac{n(n + 1)}{2}\)
\(\frac{(n + 1)(n + 2)}{2}\)
n(2n + 1)
Correct answer is C
\(\frac{(n + 1)(n + 2)}{2}\)
If n = 1, the expression becomes 3
n = 2, the expression becomes 6
n = 4, the expression becomes 15
n = 5, the expression becomes 21
Find the value of log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63
10-1
10o
10
102
Correct answer is C
log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63
log10r63 = 63
63 = 1063
∴ r = 10
\(\frac{12}{5}\)
\(\frac{12}{5}\)
-2
2
Correct answer is D
In an AP, Tn = a + (n - 1)d
T6 = a + 5d = 11
The first term = a = 1
∴ T6 = 1 + 5d = 11
5d = 11 - 1
5d = 10
∴ d = 2
Find the range of values of x for which \(\frac{1}{x}\) > 2 is true
x < \(\frac{1}{2}\)
x < 0 or x < \(\frac{1}{2}\)
0 < x < \(\frac{1}{2}\)
1 < x < 2
Correct answer is C
\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2
x > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < \(\frac{1}{2}\) (solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , \(\frac{1}{2}\) = 0