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This quiz contains questions on functions, limits, logs, exponential functions, simultaneous equations and quadratic equations.

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You have 5 minutes to complete this quiz.

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Evaluate \$$f(\\var{a3})\$$

\n

\$$f(\\var{a3})\$$ = [[0]]

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Evaluating a function

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\$$f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x-\\var{c1}\$$

\n

\$$x=\\var{a3}\$$

\n

\$$f(\\var{a3})=\\var{a1}*(\\var{a3})^{\\var{a2}}+\\var{b1}*(\\var{a3})-\\var{c1}\$$

\n

\$$f(\\var{a3})=\\simplify{{a1}*{a3}^{{a2}}}+\\simplify{{b1}*{a3}}-\\var{c1}\$$

\n

\$$f(\\var{a3})=\\simplify{{a1}*{a3}^{{a2}}+{b1}*{a3}-{c1}}\$$

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Given the function:

\n

\$$f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x-\\var{c1}\$$

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\$$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{x^2-\\simplify{{a}+{b}}x+\\simplify{{a}*{b}}}\$$

\n

We can factorise the denominator

\n

\$$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{(x-\\var{a})(x-\\var{b})}\$$

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Cancel out the common factor

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\$$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{1}{x-\\var{b}}\$$

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Insert the value \$$\\var{a}\$$ in for \$$x\$$ to evaluate the limit

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\$$-\\frac{1}{\\simplify{{b}-{a}}}\$$

\n

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Evaluate the following limit

\n

\$$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{x^2-\\simplify{{a}+{b}}x+\\simplify{{a}*{b}}}\$$

\n
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\$$\\var{k}=\\var{a}e^{\\var{m}x+{\\var{c}}}\$$

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\$$\\frac{\\var{k}}{\\var{a}}=e^{\\var{m}x+\\var{c}}\$$

\n

\$$ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)=\\var{m}x+\\var{c}\$$

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\$$ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}=\\var{m}x\$$

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\$$\\frac{ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}}{\\var{m}}=x\$$

Solve an exponential equation

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Given the equation \$$f(x)=\\var{a}e^{\\var{m}x+\\var{c}}\$$

\n

Determine the value for \$$x\$$ that satisfies the relation \$$f(x)=\\var{k}\$$

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\n

\$$x = \$$ [[0]]

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Calculate the value of \$$x\$$ that satisfies the equation when  \$$y=\\var{d}\$$.

\n

\n

\$$x = \$$ [[0]]

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Given the following logarithmic equation:

\n

\$$y=\\var{a}log(\\var{b}x+\\var{c}))\$$

\n

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Solve a logarithmic equation

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\$$\\var{a}log(\\var{b}x+\\var{c})=\\var{d}\$$

\n

Divide across by \$$\\var{a}\$$

\n

\$$log(\\var{b}x+\\var{c})=\\var{d}/\\var{a}=\\simplify{{d}/{a}}\$$

\n

\$$\\var{b}x+\\var{c}=10^{\\simplify{{d}/{a}}}\$$

\n

\$$\\var{b}x+\\var{c}=\\simplify{10^{{d}/{a}}}\$$

\n

\$$\\var{b}x=\\simplify{10^{{d}/{a}}}-\\var{c}\$$

\n

\$$\\var{b}x=\\simplify{10^{{d}/{a}}-{c}}\$$

\n

\$$x=\\simplify{(10^{{d}/{a}}-{c})/{b}}\$$

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Manipulation of an exponential function

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\$$x =\$$ [[0]]

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\$$y=\\var{k}(1-\\var{c}e^{\\var{m}x+\\var{d}})\$$

\n

Working from the outside in, we divide across by \$$\\var{k}\$$

\n

\$$\\frac{y}{\\var{k}}=1-\\var{c}e^{\\var{m}x+\\var{d}}\$$

\n

We can bring the \$$x\$$ variable to the left hand side and move the \$$y\$$ variable to the right hand side

\n

\$$\\var{c}e^{\\var{m}x+\\var{d}}=1-\\frac{y}{\\var{k}}\$$

\n

Again working from the outside in we divide across by \$$\\var{c}\$$

\n

\$$e^{\\var{m}x+\\var{d}}=\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\$$

\n

Taking the natural log of both sides eliminates the \$$e\$$ from the left hand side.

\n

\$$\\var{m}x+\\var{d}=ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)\$$

\n

Subtract \$$\\var{d}\$$ from both sides

\n

\$$\\var{m}x=ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)-\\var{d}\$$

\n

and finally divide by \$$\\var{m}\$$ to get

\n

\$$x=\\frac{ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)-\\var{d}}{\\var{m}}\$$

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Rearrange the following expression to make \$$x\$$ the subject:

\n

\$$y=\\var{k}(1-\\var{c}e^{\\var{m}x+\\var{d}})\$$

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Manipulation of algebraic fractions

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\n

\$$V =\$$ [[0]]

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When one fraction equals another fraction we can clear both fractions by cross-multiplying:

\n

\$$(\\var{a}V+1)*(\\var{d}R+7)=(\\var{b}R+3)*(\\var{c}V+5)\$$

\n

\$$\\simplify{{a}*{d}}VR+\\simplify{7*{a}}V+\\var{d}R+7=\\simplify{{b}*{c}}VR+\\simplify{5*{b}}R+\\simplify{3*{c}}V+15\$$

\n

Gathering all the terms involving \$$V\$$ to the left hand side and moving all other terms to the right hand side gives

\n

\$$\\simplify{{a}*{d}-{b}*{c}}VR+\\simplify{7*{a}-3*{c}}V=\\simplify{5*{b}-{d}}R+8\$$

\n

Factoring \$$V\$$ out on the left hand side

\n

\$$V(\\simplify{{a}*{d}-{b}*{c}}R+\\simplify{7*{a}-3*{c}})=\\simplify{5*{b}-{d}}R+8\$$

\n

Thus

\n

\$$V=\\frac{\\simplify{5*{b}-{d}}R+8}{\\simplify{{a}*{d}-{b}*{c}}R+\\simplify{7*{a}-3*{c}}}\$$

", "statement": "

Rearrange the following expression to make V the subject:

\n

\$$\\frac{\\var{a}V+1}{\\var{b}R+3}=\\frac{\\var{c}V+5}{\\var{d}R+7}\$$

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There are two values that satisfy the quadratic equation:

\n

\$$\\var{a1}x^2+\\simplify{{{a1}*{b1}*{c1}}}=\\simplify{{a1}*{b1}+{a1}{c1}}x\$$

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Type in the greater of the two values that satisfies the equation.

\n

Input your answer correct to three decimal places.  \$$x = \$$ [[0]]

\n

Type in the lesser of the two values that satisfies the equation.

\n

Input your answer correct to three decimal places.  \$$x = \$$ [[1]]

Solving quadratic equations using a formula,

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The formula for solving a quadratic equation of the form  \$$ax^2+bx+c=0\$$  is given by

\n

\$$x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\$$

\n

In this example  \$$a=\\var{a1},\\,\\,\\,b=\\simplify{+-{a1}*({b1}+{c1})}\$$  and  \$$c=\\simplify{{a1}*{b1}*{c1}}\$$

\n

\$$x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\$$

\n

\$$x=\\frac{\\var{b}\\pm \\sqrt{\\simplify{{b}^2-4*{a1}*{c}}}}{\\simplify{2*{a1}}}\$$

\n

\$$x=\\simplify{{b}+({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}+({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\$$        or        \$$x=\\simplify{{b}-({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}-({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\$$

\n

\n

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Type in the greater of the two values that satisfies the equation. Input your answer correct to three decimal places.

\n

\$$x\$$ = [[0]]

\n

Type in the lesser of the two values that satisfies the equation. Input your answer correct to three decimal places.

\n

\$$x\$$ = [[1]]

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There are two values that satisfy the quadratic function below when  \$$y=\\var{c1}\$$:

\n

\$$y=\\var{a1}x^2+\\var{b1}x\$$

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Solving quadratic equations using a formula,

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The formula for solving a quadratic equation of the form  \$$ax^2+bx+c=0\$$  is given by

\n

\$$x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\$$

\n

In this example  \$$a=\\var{a1},\\,\\,\\,b=\\var{b1}\$$  and  \$$c=\\var{c1}\$$

\n

\$$x=\\frac{-\\var{b1}\\pm \\sqrt{\\var{b1}^2-4\\times\\var{a1}\\times\\var{c1}}}{2\\times\\var{a1}}\$$

\n

\$$x=\\frac{-\\var{b1}\\pm \\sqrt{\\simplify{{b1}^2-4*{a1}*{c1}}}}{\\simplify{2*{a1}}}\$$

\n

\$$x=\\simplify{(-{b1}+ ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\$$   or   \$$x=\\simplify{(-{b1}- ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\$$

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The following equation can be converted into a quadratic equation:

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\$$\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\$$

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Solving quadratic equations using a formula

\$$\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\$$

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We clear the fraction in the equation by multiplying across by \$$x\$$

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\$$\\var{a1}x^2+\\simplify{{a1}*{b1}*{c1}}=\\simplify{{a1}*({b1}+{c1})}x\$$

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Bringing all the terms to the left hand side and putting them in order of their powers of \$$x\$$ gives

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\$$\\var{a1}x^2-\\simplify{{a1}*({b1}+{c1})}x+\\simplify{{a1}*{b1}*{c1}}=0\$$

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The formula for solving a quadratic equation of the form  \$$ax^2+bx+c=0\$$  is given by

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\$$x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\$$

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In this example  \$$a=\\var{a1},\\,\\,\\,b=\\simplify{+-{a1}*({b1}+{c1})}\$$  and  \$$c=\\simplify{{a1}*{b1}*{c1}}\$$

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\$$x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\$$

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\$$x=\\frac{\\var{b}\\pm \\sqrt{\\simplify{{b}^2-4*{a1}*{c}}}}{\\simplify{2*{a1}}}\$$

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\$$x=\\simplify{{b}+({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}+({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\$$        or        \$$x=\\simplify{{b}-({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}-({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\$$

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Type in the greater of the two values that satisfies the equation.

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\$$x = \$$ [[0]]

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Type in the lesser of the two values that satisfies the equation.

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\$$x = \$$ [[1]]

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