MHF4U Marking Scheme Chapter 7 Test PDF

**[MHF4U: Marking Scheme: Chapter 7 Test]** General marking notations: M : marks for method A : marks for application R : marks for reasoning N : marks when no working shown, but answer is correct. (No M,A,R marks will be provided in this case. Only the marks indicated with N will be awarded *...

**[MHF4U: Marking Scheme: Chapter 7 Test]** General marking notations: M : marks for method A : marks for application R : marks for reasoning N : marks when no working shown, but answer is correct. (No M,A,R marks will be provided in this case. Only the marks indicated with N will be awarded **[Q1: \[K=30\]]** a. Isolate logarithmic terms on one side: M1 [log (*x*−1) + log (*x*+2) = 1 ]{.math.inline} M1 Using [log (*a*) + log (*b*) = log (ab)]{.math.inline}: M1R1 [log ((*x*−1)(*x* + 2)) = 1 ]{.math.inline} M1A1 [log (*x*^2^+*x*−2) = 1 ]{.math.inline} A2 [*x*^2^ + *x* − 2 = 10]{.math.inline} A2 [*x*^2^ + *x* − 12 = 0]{.math.inline} A1 Factorize the quadratic to solve: [(*x*−3)(*x*+4) = 0]{.math.inline} M1A2 [*x* = 3]{.math.inline} or [*x* =  − 4]{.math.inline} A1 [*x* =  − 4]{.math.inline} is an extraneous root because both [log (*x*−1)]{.math.inline} and [log (*x*+2) ]{.math.inline}are not defined for this value because logarithm of a negative number is undefined. M1A2 b. [\$\\log\\sqrt\[3\]{x\^{2} + 48x} = \\frac{2}{3}\$]{.math.inline} M1 Using [log (*a*^*b*^) = *b*log *a*]{.math.inline}: M1R1 [\$\\frac{1}{3}\\log{{(x}\^{2} + 48x)} = \\frac{2}{3}\$]{.math.inline} M1A1 [log (*x*^2^ + 48*x*) = 2]{.math.inline} A2 [*x*^2^ + 48*x* = 100]{.math.inline} A1 [*x*^2^ + 48*x* − 100 = 0]{.math.inline} A1 Factorize the quadratic to solve: R1 [(*x*−2)(*x*+50) = 0]{.math.inline} A1 [*x* = 2]{.math.inline} and [*x* =  − 50]{.math.inline} are valid solutions. A1 **Q2: \[A=30\]** a. Decay equation is given by: [\$A(t) = A\_{0}\\left( \\frac{1}{2} \\right)\^{t \\slash h}\$]{.math.inline} M1R1 [*A*~0~ = 50 *mg*]{.math.inline} M1 [*h*=]{.math.inline} Half-life At [*t* = 10 *min*]{.math.inline}, [*A*(*t*) = 41 *mg*]{.math.inline}. So, plugging in the values in the equation: M1R1 [\$41 = 50\\left( \\frac{1}{2} \\right)\^{10 \\slash h}\$]{.math.inline} M1A1 [\$\\frac{41}{50} = \\left( \\frac{1}{2} \\right)\^{10 \\slash h}\$]{.math.inline} A2 Taking log on both sides: M1 [\$\\log{\\left( \\frac{41}{50} \\right) = \\log\\left( \\frac{1}{2} \\right)\^{10 \\slash h}}\$]{.math.inline} A2 [\$\\log{\\left( \\frac{41}{50} \\right) = {\\frac{10}{h}\\log}\\frac{1}{2}}\$]{.math.inline} A2 [\$h = \\frac{10\\log\\left( 0 \\cdot 5 \\right)}{\\log\\left( \\frac{41}{50} \\right)}\$]{.math.inline} A2 [*h* ≈ 35 *min*]{.math.inline} A1 Half-life of radioactive substance is approximately [35 *mins*]{.math.inline} b. 1% of [50 *mg*]{.math.inline} is [0.5 *mg*]{.math.inline}. Let the substance takes [t mins]{.math.inline} to decay to 1% of its initial amount. M1R1 Here, [*A*(*t*) = 0.5 *mg*]{.math.inline} M1 Plugging in the values: M1 [\$0.5 = 50\\left( \\frac{1}{2} \\right)\^{t \\slash 35}\$]{.math.inline} A1 [\$\\frac{1}{100} = \\left( \\frac{1}{2} \\right)\^{t \\slash 35}\$]{.math.inline} A1 Taking log both sides: [\$\\log{\\left( \\frac{1}{100} \\right) = \\log\\left( \\frac{1}{2} \\right)\^{t \\slash 35}}\$]{.math.inline} A2 [\$\\log{0.01 = \\frac{t}{35}\\log{0.5}}\$]{.math.inline} A2 [\$t = \\frac{35\\log\\left( 0 \\cdot 01 \\right)}{\\log{0.5}}\$]{.math.inline} A2 [*t* = 232.5 *mins*]{.math.inline} A1 **Q3: \[ T=20\]** f\) [4^*x*^ = 3 + 18(4^ − *x*^)]{.math.inline} Rewriting the given equation: [4^*x*^ − 3 − 18(4^ − *x*^) = 0]{.math.inline} Take [4^*x*^]{.math.inline} common: M1 [4^2*x*^ − 3(4^*x*^) − 18 = 0]{.math.inline} A1 Let [4^*x*^]{.math.inline} be [*t*:]{.math.inline} M1 [*t*^2^ − 3*t* − 18 = 0]{.math.inline} A1 [(*t*−6)(*t*+3) = 0]{.math.inline} A2 [*t* = 6]{.math.inline} or [*t* =  − 3]{.math.inline} A1 [4^*x*^ = 6]{.math.inline} or [4^*x*^ =  − 3]{.math.inline} A1 [4^*x*^ =  − 3]{.math.inline}: A1 This isn't possible as [4^*x*^]{.math.inline} cannot be negative. M1 [4^*x*^ = 6]{.math.inline}: Take log on both sides: [log 4^*x*^ = log 6 ]{.math.inline} A1 [*x*log 4 = log 6 ]{.math.inline} A1 [\$x = \\frac{\\log 6}{\\log 4}\$]{.math.inline} A1 [*x* = 1.29]{.math.inline} M1 **Q4 \[T=20\]** a. [log *a*^2^bc]{.math.inline} M1 Using product rule: M1 [log *a*^2^bc = log *a*^2^ + log *b* + log *c*]{.math.inline} M1A2 Using exponent rule: M1 [log *a*^2^bc = 2*loga* + log *b* + log *c*]{.math.inline} M1A2 b. [\$\\log\\frac{k}{\\sqrt{m}}\$]{.math.inline} M1 Using product rule: M1 [\$\\log\\frac{A}{\\sqrt{m}} = \\log A + \\log\\sqrt{m}\\ \$]{.math.inline} M1A3 Using exponent rule: M1 [\$\\log\\frac{A}{\\sqrt{m}} = \\log A + \\frac{1}{2}\\log m\\ \$]{.math.inline} M1A3