Which of the following is a factor of the polynomial \(6x^{4} + 2x^{3} + 15x + 5\)?
3x + 1
x + 1
2x + 1
x + 2
Correct answer is A
Using the remainder theorem, if (x - a) is a factor of f(x), then f(a) = 0.
Check the options and get the answer.
\(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)
\(\begin{pmatrix} 27 & 12 \\ 16 & -15 \end{pmatrix}\)
\(\begin{pmatrix} -20 & -6 \\ 12 & -8 \end{pmatrix}\)
\(\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}\)
Correct answer is D
\(P = \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix}; Q = \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix}\)
= \(PQ = \begin{pmatrix} -10+2 & 6-1 \\ 15+8 & -9-4 \end{pmatrix}\)
= \(\begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix}\)
\(QP = \begin{pmatrix} -10-9 & 5-12 \\ -4-3 & 2-4 \end{pmatrix}\)
= \(\begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}\)
\(PQ - QP = \begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix} - \begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}\)
= \(\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}\)
Given that \(\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}\), find P and Q.
P = 4 and Q = 2
P = 2 and Q = 4
P = 4 and Q = -2
P = -2 and Q = 4
Correct answer is C
\(\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}\)
\(\frac{2x}{(x + 6)(x + 3)} = \frac{P(x + 3) + Q(x + 6)}{(x + 6)(x + 3)}\)
Comparing equations, we have
\(2x = Px + 3P + Qx + 6Q\)
\(\implies 3P + 6Q = 0 ... (1) ; P + Q = 2 .... (2)\)
From equation (1), \(3P = -6Q \implies P = -2Q\)
\(\therefore -2Q + Q = -Q = 2 \)
\(Q = -2\)
\(P = -2Q = -2(-2) = 4\)
\(P = 4, Q = -2\)
Given that \(f : x \to x^{2}\) and \(g : x \to x + 3\), where \(x \in R\), find \(f o g(2)\).
25
9
7
5
Correct answer is A
\(f : x \to x^{2} ; g : x \to x + 3\)
\(g(2) = 2 + 3 = 5\)
\(f o g(2) = f(5) = 5^{2} = 25\)
Find the 3rd term of \((\frac{x}{2} - 1)^{8}\) in descending order of x.
\(\frac{x^{7}}{8}\)
\(\frac{7x^{6}}{16}\)
\(\frac{7x^{5}}{4}\)
\(\frac{35x^{4}}{8}\)
Correct answer is B
\((\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...\)
\(\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}\)
= \(\frac{8!}{6!2!}(\frac{x^{6}}{64})(1) \)
= \(28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}\)