Find the coefficient of x\(^2\)in the binomial expansion of \((x + \frac{2}{x^2})^5\)
10
40
32
80
Correct answer is A
\((x + \frac{2}{x^2})^5\)
n = 5, r = 4, p = x and q = \(\frac{2}{x^2}\)
5C\(_4\)x\(^4\) (\(\frac{2}{x^2}\))1 = 5C\(_4\) \(\frac{2x^4}{x^2}\)
5C\(_4\) 2x\(^2\) = \(\frac{5!}{[5-4]!4!}\) * 2x\(^2\)
\(\frac{5*4!}{4!} * 2x^2\) = 5 * 2x\(^2\) = 10x\(^2\)
The coefficient is 10.
Evaluate \(4p_2 + 4C_2 - 4p_3\)
18
6
-6
-18
Correct answer is C
\(4p_2 + 4C_2 - 4p_3\)
\(np_r = \frac{n!}{[n-r]!} and nC_r = \frac{n!}{[n-r]!r!} \)
= \(\frac{4!}{[4-2]!} + \frac{4!}{[4-2]!2!} - \frac{4!}{[4-3]!} = \frac{4!}{2!} + \frac{4!}{2!2!} - \frac{4!}{1!}\)
= \(\frac{4*3*2!}{2!} + \frac{4*3*2!}{2!2!} - \frac{4*3*2*1}{1!}\)
12 + 6 - 24 = -6
(7, -2)
(5, -2)
(-2, 7)
(-7, 2)
Correct answer is A
Let (x1, y1) be the image of the point (x, y) under the given transformations.
x1 = 3x - y
y1 = x + 4y
\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y = \end{vmatrix} \begin{vmatrix} x_1 \\ y_1 \end{vmatrix}\)
\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} 2 \\ 1 = \end{vmatrix} \begin{vmatrix} 7 \\ -2 \end{vmatrix}\)
-8
-5
-4
-3
Correct answer is B
\(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix} \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\)
\(\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)
\(\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)
-12 - 3k = 3
-3k = 3 + 12
k = \(\frac{15}{-3}\)
k = -5
Evaluate\({1_0^∫} x^2(x^3+2)^3\)
\(\frac{56}{12}\)
\(\frac{65}{12}\)
12
65
Correct answer is B
\({1_0^∫} x^2(x^3+2)^3\)dx
let \( u = x^3 + 2, du = 3x^2dx\)
when x = 1, u = 3
when x = 0, u = 2
dx = \(\frac{du}{3x^2}\)
\({3_2^∫}\) \(\frac{x^2[u]^3}{3x^2}\)
\({3_2^∫}\) \(\frac{u^3}{3}\) du
= \(\frac{u^4}{3*4}\)\(_2\)3
\(\frac{1}{12} [u^4]\)\(_2\)3
\(\frac{1}{12} [3^4 - 2^4]\)
\(\frac{1}{12}[81 - 16]\)
\(\frac{65}{12}\)