{x : 4 < x < 8}
{x : 2 < x ≤ 4}
{x : 2 ≤ x ≤ 4}
{x : 4 ≤ x ≤ 8}
Correct answer is C
P = {x : 2 ≤ x ≤ 8} = {2, 3, 4, 5, 6, 7, 8}
Q = {x : 4 < x ≤ 12} = {5, 6, 7, 8, 9, 10, 11, 12}
Q1 = R - {5, 6, 7, 8, 9, 10, 11, 12} = {..., 2, 3, 4, 13, 14, 15, ...}
∴ P ⋂ Q\(^1\) = {2, 3, 4} = {x : 2 ≤ x ≤ 4}
Find the fifth term in the binomial expansion of \((q + x)^7\).
\(21q^2x^5\)
\(21q^4x^3\)
\(35q^3x^4\)
\(35q^5x^2\)
Correct answer is C
rth term of a binomial expansion = \(^nC_r - _1a^{n - (r - 1)} b^{r - 1}\)
n = 7, r = 5 ∴ r - 1 = 4
5th term = \(^7C_4 q^{7 - 4} x^4\)
= \(^7C_4 q^3x^4\)
\(\therefore 35q^3x^4\)
3 sec
2 sec
4 sec
1 sec
Correct answer is A
\(u = 27 ms^{-1}; a = -9 ms^{-2}; v = 0; t = ?\)
\(v = u + at; t = \frac{v - u}{a}\)
\(t = \frac{0 - 27}{-9} = \frac{-27}{-9}\)
\(\therefore t = 3 sec\)
Given that M is the midpoint of T (2, 4) and Q (-8, 6), find the length of MQ .
\(√26 units\)
\(√28 units\)
\(√24 units\)
\(√30 units\)
Correct answer is A
\(|MQ| = \frac{1}{2} |TQ|\)
\(|TQ| = √((y2 - y1)^2 + (x2 - x1)^2)\)
\(|TQ| = √((6 - 4)^2 + (-8 - 2)^2)\)
\(|TQ| = √(2^2 + (-10)^2)\)
\(|TQ| = √(4 + 100) = √104\)
\(|TQ| = 2√26 units\)
\(|MQ| = \frac{1}{2} |TQ| = 2 \times 2√26\)
∴ \(|MQ| = √26 units\)
Find the radius of the circle \(2x^2 + 2y^2 - 4x + 5y + 1 = 0\)
\(\frac{\sqrt33}{4}\)
\(\frac{\sqrt5}{6}\)
\(\frac{5}{6}\)
\(\frac{33}{4}\)
Correct answer is A
Standard Form equation of a circle (Center-Radius Form): \((x − a)^2 + (y − b)^2 = r^2\)
Where "a" and "b" are the coordinates of the center and "r" is the radius of the circle
\(2x^2+2y^2-4x+5y+1=0\)
Divide through by 2
= \(x^2+y^2-2x+\frac{5}{2}y+\frac{1}{2}= 0\)
=\(x^2-2x+y^2+\frac{5}{ 2}y=-\frac{1}{ 2}\)
=\(x^2-2x+1^2+y^2+\frac{5}{ 2}y+(\frac{5}{ 4})^2-1-\frac{25}{16}=-\frac{1}{2}\)
=\((x-1)^2+(y+\frac{5}{4})^2=-\frac{1}{2}+1+\frac{25}{16}\)
=\((x-1)^2+(y-(-\frac{5}{4}))^2=\frac{33}{16}\)
=\((x-1)^2+(y-(-\frac{5}{4}))^2=(\frac{\sqrt33}{4})^2\)
\(\therefore a = 1, b = - \frac{5}{4} and\) \(r \frac{\sqrt33}{4} (answer)\)