Mathematics Questions and Answers

Given the sequence 3, 8, 13, 18, ... which is an arithmetic sequence

a = 3

d = T$_2$ - T$_1$ = 8 - 3 = 5

The general term of an A.P is:

T$_n$ = a + (n - 1)d

⇒ T$_n$ = 3 + (n - 1)5

= T$_n$ = 3 + 5n - 5

∴ T$_n$ = 5n - 2

$x^2 - x - 4 ≤ 2$
Subtract two from both sides to rewrite it in the quadratic standard form:
= $x^2 - x - 4 - 2 ≤ 2 - 2$
= $x^2 - x - 6 ≤ 0$
Now set it = 0 and factor and solve like normal.
= $x^2 - x$ - 6=0
= $(x - 3)(x + 2)$=0
$x$ + 2 = 0 or $x$ - 3 = 0
$x$ = -2 or $x$ = 3
So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, $x$ = 0) and evaluate the original inequality.
= $x2 - x - 4 ≤ 2$
= $(0)^2 - (0) - 4 ≤ 2$
= $0 - 0 - 4 ≤ 2$
$−4 ≤ 2$
Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.
∴ $-2 ≤ x ≤ 3.$

A binary operation '*' defined on a set A is said to be commutative only if a*b=b*a, ∀a, b∈A. If (a*b)*c=a*(b*c), then the operation is said to associative ∀ a, b∈ A. If (b ο c)*a=(b*a) ο (c*a), then the operation is said to be distributive ∀ a, b, c ∈ A.

Let $p(x) = -2x^3 + 6x^2 + 17x - 21$

Using the remainder theorem

Let $x + 1 = 0$

∴ $x = -1$

Since, $(x + 1)$ divides $p(x)$, then, remainder will be p(-1)

⇒ p(-1) = -2(-1)$^3 + 6(-1)^2$ + 17(-1) - 21

∴ p(-1) = -30

Number of students who scored atleast 25% = 5 + 3 + 8 = 16