MATH101 Calculus Assessed Exercise Solutions 2

Total [10] marks

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not by deducting marks for errors.

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A2.1. Recall that $\cosh x=(e^{x}+e^{-x})/2$ and that $\sinh x=(e^{x}-e^{-x})/2.$

(i) Hence

$$\eqalign{{\hbox{LHS}}&=\cosh^{2}x-\sinh^{2}x\cr&={{1}\over{4}}\bigl(e^{2x}+2+e^{-2x}\bigr)-{{1}\over{4}}\bigl(e^{2x}-2+e^{-2x}\bigr)\cr&=1={\hbox{RHS}},\cr}$$

as required.

[1] for calculation

Solutions with incorrect use of implication signs do not have correct logic.

(ii) Here we have

$$\eqalign{{\hbox{RHS}}&=\sinh x\cosh y-\cosh x\sinh y\cr&={{1}\over{4}}\bigl(e^{x}-e^{-x}\bigr)\bigl(e^{y}+e^{-y}\bigr)-{{1}\over{4}}\bigl(e^{x}+e^{-x}\bigr)\bigl(e^{y}-e^{-y}\bigr)\cr&={{1}\over{4}}\bigl(e^{x-y}+e^{x+y}-e^{-x-y}-e^{-x+y}+e^{x-y}-e^{x+y}+e^{-x-y}-e^{-x+y}\bigr)\cr&={{1}\over{4}}\bigl(2e^{x-y}-2e^{-x+y}\bigr)\cr&=\sinh(x-y)={\hbox{LHS}},\cr}$$

as required.

[2] for correct calculation

(iii) In this case we can use the definitions of the hyperbolic functions and calculate

$$\eqalign{{\hbox{RHS}}&={{\tanh x-\tanh y}\over{1-\tanh x\tanh y}}\cr&={{{{e^{x}-e^{-x}}\over{e^{x}+e^{-x}}}-{{e^{y}-e^{-y}}\over{e^{y}+e^{-y}}}}\over{1-{{e^{x}-e^{-x}}\over{e^{x}+e^{-x}}}{{e^{y}-e^{-y}}\over{e^{y}+e^{-y}}}}}\cr&={{{{e^{x}-e^{-x}}\over{e^{x}+e^{-x}}}-{{e^{y}-e^{-y}}\over{e^{y}+e^{-y}}}}\over{1-{{e^{x}-e^{-x}}\over{e^{x}+e^{-x}}}{{e^{y}-e^{-y}}\over{e^{y}+e^{-y}}}}}{{(e^{x}+e^{-x})(e^{y}+e^{-y})}\over{(e^{x}+e^{-x})(e^{y}+e^{-y})}}\cr\omit\span\omit\@@LTX@noalign{{\hbox{which simplifies when we take common denominators of the fractions to}}}\\ &={{(e^{x}-e^{-x})(e^{y}+e^{-y})-(e^{x}+e^{-x})(e^{y}+e^{-y})}\over{(e^{x}+e^{-x})(e^{y}+e^{-y})-(e^{x}-e^{-x})(e^{y}+e^{-y})}}\cr&={{e^{x-y}+e^{x+y}-e^{-x-y}-e^{-x+y}+e^{x-y}-e^{x+y}+e^{-x-y}-e^{-x+y}}\over{e^{x-y}+e^{x+y}+e^{-x-y}+e^{-x+y}+e^{x-y}-e^{x+y}-e^{-x-y}+e^{-x+y}}}\cr&={{e^{x-y}-e^{-x+y}}\over{e^{x-y}+e^{-x+y}}}\cr&=\tanh(x+y)={\hbox{LHS}}.\cr}$$

An equivalent, but more elegant, argument is to use

$$\eqalign{{\hbox{RHS}}&={{{{\sinh x}\over{\cosh x}}-{{\sinh y}\over{\cosh y}}}\over{1-{{\sinh x}\over{\cosh x}}{{\sinh y}\over{\cosh y}}}},\cr\omit\span\omit\@@LTX@noalign{\hbox{an expression which we take over common denominators to get}}\\ &={{\sinh x\cosh y-\cosh x\sinh y}\over{\cosh x\cosh y-\sinh y\sinh x}}\cr\omit\span\omit\@@LTX@noalign{\hbox{an expression which reduces by part (ii) and a result of lectures to}}\\ &={{\sinh(x-y)}\over{\cosh(x-y)}}\cr&=\tanh(x-y)={\hbox{LHS}}.\cr}$$

You may have seen a similar proof used to prove the corresponding identity for $\tan(x-y)$.

[2] for complete and correct calculation

award partial marks for steps

(iv) The function $\cosh x$ with $x\geq 0$ has range $[1,\infty)$ and is strictly increasing, so we can introduce $f:[0,\infty)\rightarrow[1,\infty)$ with inverse $\cosh^{-1}:[1,\infty)\rightarrow[0,\infty)$; so $\cosh^{-1}$ has domain $[1,\infty)$ and codomain $[0,\infty)$.

[1] for correct statement

[6] total for question

A2.2. (i) Intercepts. When $x=0$ we have $y=3/4$; further the intercept with the $x$-axis is given by the solutions of the quadratic equation $y=0$; that is

$$x^{2}+6x-3=0,$$

so the roots are

$$x={{-6\pm\sqrt{6^{2}+4\times 3}}\over{2}}=-3\pm 2\sqrt{3}.$$

[1] for intercepts

surd better than decimal

(ii) By polynomial long division we have

$$y={{x^{2}+6x-3}\over{x-4}}=x+10+{{37}\over{x-4}}$$

*

since one can divide

$$\matrix{&&&x&+10\cr x-4&|&x^{2}&+6x&-3\cr&&x^{2}&-4x&\cr&&&10x&-3\cr&&&10x&-40\cr&&&&37.\cr}$$

[1] any method acceptable

From $(*)$, we see that $y\rightarrow\infty$ as $x\rightarrow\infty$; likewise, $y\rightarrow-\infty$ as $x\rightarrow-\infty$. (One can say that $f(x)$ has asymptote $y=x+10$ as $x\rightarrow\pm\infty$.)

[1] for a proper statement

(iii) Also from $(*)$ we see that $y\rightarrow-\infty$ as $x\rightarrow(4)-$, and $y\rightarrow\infty$ as $x\rightarrow(4)+$; (One can say that $x=4$ is a vertical asymptote.)

[1] needs justification

[4] total for question