The fourth term of a geometric sequence is 2 and the sixth term is 8. Find the common ratio.
±1
±2
±3
±4
Correct answer is B
\(T_{n} = ar^{n - 1}\) (Exponential sequence)
\(T_{4} = ar^{3} = 2 .... (1)\)
\(T_{6} = ar^{5} = 8 ......(2)\)
Divide (2) by (1),
\(\frac{ar^{5}}{ar^{3}} = \frac{8}{2} \)
\(r^{2} = 4\)
\(r = \pm 2\)
5
6
7
8
Correct answer is A
\(\begin{pmatrix} 3 & 2 \\ 7 & x \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\)
\(\begin{pmatrix} 3 \times 2 + 2 \times 3 \\ 7 \times 2 + x \times 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\)
\(\implies 14 + 3x = 29 \)
\(3x = 29 - 14 = 15\)
\(x = 5\)
The inverse of a function is given by \(f^{-1} : x \to \frac{x + 1}{4}\).
\(f : x \to 4x - 1\)
\(f : x \to 4x + 1\)
\(f : x \to \frac{4x - 1}{4}\)
\(f : x \to \frac{x - 1}{2}\)
Correct answer is A
The inverse of the inverse of a function gives the function
i.e \(f^{-1}(f^{-1}(x)) = f(x)\)
\(f^{-1}(x) = \frac{x + 1}{4}\)
Take y = x, so
\(f^{-1}(y) = \frac{y + 1}{4}\)
Let \(x = f^{-1}(y)\),
\(x = \frac{y + 1}{4} \implies 4x = y + 1\)
\(y = f(x) = 4x - 1\)
Solve \(9^{2x + 1} = 81^{3x + 2}\)
\(\frac{-3}{4}\)
\(\frac{-2}{3}\)
\(\frac{4}{5}\)
\(\frac{3}{2}\)
Correct answer is A
\(9^{2x + 1} = 81^{3x + 2}\)
\(9^{2x + 1} = (9^{2})^{3x + 2}\)
\(9^{2x + 1} = 9^{6x + 4}\)
Equating powers,
\(2x + 1 = 6x + 4 \implies -3 = 4x\)
\(\therefore x = \frac{-3}{4}\)
3y - x -14 = 0
3x + y + 1 = 0
3y + x + 1 = 0
3y + x + 14 = 0
Correct answer is D
\(3x - y + 11 = 0 \implies y = 3x + 11\)
\(Gradient = 3\)
Gradient of perpendicular line = \(\frac{-1}{3}\)
\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)
\(3(y + 5) = 1 - x\)
\(3y + x + 15 - 1 = 3y + x + 14 = 0\)