Complete worked solutions for the Math Questions (Set 2) .
Given \( y = \log(x) \), rewrite in exponential form:
\[ x = 10^y \]Interchange \(x\) and \(y\):
\[ y = 10^x \]Therefore, the inverse function is:
\[ f^{-1}(x) = 10^x \]Use the change of base formula:
\[ \log_4(65) = \frac{\ln(65)}{\ln(4)} \]Using a calculator:
\[ \log_4(65) \approx 3.01 \]Rewrite \(16\) as a power of \(2\):
\[ 2^{3x-1} = 2^4 \]Since the bases are equal, the exponents must be equal:
\[ 3x - 1 = 4 \] \[ x = \frac{5}{3} \]Given:
\[ \log_x 9 = 2 \]Rewrite in exponential form:
\[ x^2 = 9 = 3^2 \]Thus:
\[ x = 3 \]Expand both sides:
\[ x^2 + kx - 6 = x^2 + x - 6 \]Matching coefficients gives:
\[ k = 1 \]Complete the square:
\[ y = 2x^2 + 8x - 3 \] \[ = 2(x^2 + 4x + 4) - 11 \] \[ = 2(x+2)^2 - 11 \]Vertex:
\[ (-2, -11) \]Substituting the proposed roots into each option shows that Option D has roots \(3\) and \(5\).
Note:
\[ (x+2)(x-4) = x^2 - 2x - 8 \]Multiply the entire equation by \((x+2)(x-4)\):
\[ x(x-4) + 3(x+2) = 4x + 2 \] \[ x^2 - 5x + 4 = 0 \] \[ (x-1)(x-4) = 0 \]Solutions: \(x=1\), \(x=4\). However, \(x=4\) is not allowed (division by zero).
Final solution:
\[ x = 1 \]Zeros at \(x=0\) and \(x=6\). Since the parabola opens downward, the maximum occurs at:
\[ x = \frac{-b}{2a} = 3 \] \[ f(3) = 9 \]Range:
\[ [0, 9] \]Solve:
\[ -x^2 + 3x + 18 = 0 \] \[ (x-6)(x+3)=0 \]x-intercepts:
\[ (6,0),\;(-3,0) \]The function is undefined when:
\[ |x-2| = 0 \]Domain:
\[ (-\infty,2)\cup(2,\infty) \]A coterminal angle:
\[ -1280^\circ + 4(360^\circ) = 160^\circ \]Reference angle:
\[ 180^\circ - 160^\circ = 20^\circ \]False. \[ \tan(-x) = -\tan x \]
Option A is not an identity.
Valid solution:
\[ \cos x = \frac{1}{2} \]