Find the exact value of ((2)).

Write down the expression to be evaluated: ((2)) Quote the exponential definition: (x)=e^x-e^-x2 State the exact value: ((2))=(3)/(4) Answer: (3)/(4)

Which of the following is the exact value of ((3))?

Write down the expression to be evaluated: ((3)) Quote the exponential definition: (x)=e^x-e^-x2 State the exact value: ((3))=(4)/(3) Answer: (4)/(3)

A function is defined by y=(4x). Which of the following is dydx?

Write down the function: y=(4x) Use the derivative of with NO sign change: ddx(kx)=k(kx) Confirm the result from the exponential definition: ddx((4x))=4(4x) Differentiate with respect to x: dydx=4(4x) Evaluate the derivative at x=1: .dydx|_x=1109.1597 Select the correct derivative: dydx=4(4x) Answer: 4(4x)

Which of the following expressions is equal to arcosh(x) for x≥1?

Set y equal to the inverse function: y=arcosh(x) (y)=x Write the forward function in exponential form: e^y+e^-y2=x Multiply through and set w=e^y: w^2-2xw+1=0 Solve for w: w=x±x^2-1 Choose the admissible root: w=x+x^2-1≥1 Take logarithms: y=(x+x^2-1) Recall the exponential definition of: (x)=e^x-e^-x2 Recall the exponential definition of: (x)=e^x+e^-x2 Recall the exponential definition of: (x)=e^x-e^-xe^x+e^-x Quote the fundamental hyperbolic identity: ^2(x)-^2(x)=1 Select the correct logarithmic form: arcosh(x)=(x+x^2-1) Answer: (x+x^2-1)

How many real solutions does the equation (2x)-5(x)+4=0 have?

Write down the equation: (2x)-5(x)+4=0 Substitute u=e^x: u=e^x>0 Remember that is even: (-x)=(x) Count the admissible values of u: n_u=3 List the solutions: x=0, x=±(√(5)2+(3)/(2)) Recall the exponential definition of: (x)=e^x-e^-x2 Recall the exponential definition of: (x)=e^x+e^-x2 Recall the exponential definition of: (x)=e^x-e^-xe^x+e^-x Quote the fundamental hyperbolic identity: ^2(x)-^2(x)=1 Quote the double-angle formula for: (2x)=^2(x)+^2(x) Quote the alternative double-angle form: (2x)=2^2(x)-1 Quote the second alternative double-angle form: (2x)=1+2^2(x) Quote the double-angle formula for: (2x)=2(x)(x) Recall the derivative of: ddx(x)=(x) Recall the derivative of: ddx(x)=(x) State the number of real solutions: n=3 Answer: 3

Further Maths Hyperbolic functions Practice Questions

Free Further Maths Hyperbolic functions practice questions with full step-by-step worked solutions. Covers hyperbolic-functions, exponential-definitions, identities, osborns-rule. Practise exam-style problems and check your method.

hyperbolic-functionsexponential-definitionsidentitiesosborns-ruledifferentiationintegration

Further Maths70 questionsStep-by-step solutions

Question 1

2 markseasy

Find the exact value of

\sinh{\left(\ln{\left(2\right)}\right)}

Show worked solution

Worked solution

Write down the expression to be evaluated
$\sinh{\left(\ln{\left(2\right)}\right)}$
Start from the expression given in the question.
Quote the exponential definition
$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$
Every hyperbolic function is defined directly in terms of $e^{x}$ .
State the exact value
$\sinh{\left(\ln{\left(2\right)}\right)}=\frac{3}{4}$
This is the exact value of the expression.

Answer

\frac{3}{4}

Question 2

2 markseasy

Which of the following is the exact value of

\sinh{\left(\ln{\left(3\right)}\right)}

Show worked solution

Worked solution

Write down the expression to be evaluated
$\sinh{\left(\ln{\left(3\right)}\right)}$
Start from the expression given in the question.
Quote the exponential definition
$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$
Every hyperbolic function is defined directly in terms of $e^{x}$ .
State the exact value
$\sinh{\left(\ln{\left(3\right)}\right)}=\frac{4}{3}$
This is the exact value of the expression.

Answer

\frac{4}{3}

Question 3

4 marksintermediate

A function is defined by

y=\cosh{\left(4x\right)}

. Which of the following is

\frac{\mathrm{d}y}{\mathrm{d}x}

Show worked solution

Worked solution

Write down the function
$y=\cosh{\left(4x\right)}$
Identify the structure before differentiating.
Use the derivative of $\cosh$ with NO sign change
$\frac{\mathrm{d}}{\mathrm{d}x}\cosh(kx)=k\sinh(kx)$
Unlike $\cos$ , differentiating $\cosh$ does not introduce a minus sign.
Confirm the result from the exponential definition
$\frac{\mathrm{d}}{\mathrm{d}x}\left(\cosh{\left(4x\right)}\right)=4\sinh{\left(4x\right)}$
Differentiating the exponential form gives the same answer, which checks every sign.
Differentiate with respect to $x$
$\frac{\mathrm{d}y}{\mathrm{d}x}=4\sinh{\left(4x\right)}$
Apply the standard derivatives together with the chain, product or quotient rule.
Evaluate the derivative at $x=1$
$\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=1}\approx109.1597$
A numerical value gives a quick check on the algebra.
Select the correct derivative
$\frac{\mathrm{d}y}{\mathrm{d}x}=4\sinh{\left(4x\right)}$
This is the required derivative.

Answer

4\sinh{\left(4x\right)}

Question 4

6 markshard

Which of the following expressions is equal to

\operatorname{arcosh}(x)

for

x\ge1

Show worked solution

Worked solution

Set $y$ equal to the inverse function
$y=\operatorname{arcosh}(x)\iff \cosh(y)=x$
The inverse is defined by reversing the forward function.
Write the forward function in exponential form
$\frac{e^{y}+e^{-y}}{2}=x$
Now the equation can be solved algebraically.
Multiply through and set $w=e^{y}$
$w^{2}-2xw+1=0$
A quadratic (or a simple equation) in $w=e^{y}$ results.
Solve for $w$
$w=x\pm\sqrt{x^{2}-1}$
Both roots of the quadratic must be considered before one is chosen.
Choose the admissible root
$w=x+\sqrt{x^{2}-1}\ge1$
Only a positive $w$ is possible. For $\operatorname{arcosh}$ the PRINCIPAL branch additionally requires $y\ge0$ , i.e. $w\ge1$ , which rules out $x-\sqrt{x^{2}-1}$ .
Take logarithms
$y=\ln{\left(x+\sqrt{x^{2}-1}\right)}$
This is the logarithmic form of the inverse function.
Recall the exponential definition of $\sinh$
$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$
Every hyperbolic function is shorthand for a combination of $e^{x}$ and $e^{-x}$ .
Recall the exponential definition of $\cosh$
$\cosh(x)=\frac{e^{x}+e^{-x}}{2}$
This definition is the source of every hyperbolic identity.
Recall the exponential definition of $\tanh$
$\tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
The quotient of the previous two definitions gives $\tanh$ .
Quote the fundamental hyperbolic identity
$\cosh^{2}(x)-\sinh^{2}(x)=1$
Osborn's rule turns $\cos^{2}+\sin^{2}=1$ into a MINUS sign here, because $\sinh^{2}$ is a product of two sines.
Select the correct logarithmic form
$\operatorname{arcosh}(x)=\ln{\left(x+\sqrt{x^{2}-1}\right)}$
This is the standard logarithmic form quoted in the formula book.

Answer

\ln{\left(x+\sqrt{x^{2}-1}\right)}

Question 5

9 markschallenging

How many real solutions does the equation

\cosh{\left(2x\right)}-5\cosh{\left(x\right)}+4=0

have?

Show worked solution

Worked solution

Write down the equation
$\cosh{\left(2x\right)}-5\cosh{\left(x\right)}+4=0$
The NUMBER of real solutions is required, not the solutions themselves.
Substitute $u=e^{x}$
$u=e^{x}>0$
Every real $x$ gives exactly one positive $u$ , and every positive $u$ gives exactly one real $x$ .
Remember that $\cosh$ is even
$\cosh(-x)=\cosh(x)$
Solutions therefore come in $\pm$ pairs, and the negative member is easy to lose.
Count the admissible values of $u$
$n_{u}=3$
Each positive real root of the polynomial in $u$ gives exactly one real value of $x$ .
List the solutions
$x=0,\quad x=\pm\ln{\left(\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}$
Writing them out confirms the count.
Recall the exponential definition of $\sinh$
$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$
Every hyperbolic function is shorthand for a combination of $e^{x}$ and $e^{-x}$ .
Recall the exponential definition of $\cosh$
$\cosh(x)=\frac{e^{x}+e^{-x}}{2}$
This definition is the source of every hyperbolic identity.
Recall the exponential definition of $\tanh$
$\tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
The quotient of the previous two definitions gives $\tanh$ .
Quote the fundamental hyperbolic identity
$\cosh^{2}(x)-\sinh^{2}(x)=1$
Osborn's rule turns $\cos^{2}+\sin^{2}=1$ into a MINUS sign here, because $\sinh^{2}$ is a product of two sines.
Quote the double-angle formula for $\cosh$
$\cosh(2x)=\cosh^{2}(x)+\sinh^{2}(x)$
Osborn's rule flips the sign of the product of two sines, so this is a PLUS, unlike $\cos(2x)$ .
Quote the alternative double-angle form
$\cosh(2x)=2\cosh^{2}(x)-1$
Useful when an equation is to be written entirely in terms of $\cosh(x)$ .
Quote the second alternative double-angle form
$\cosh(2x)=1+2\sinh^{2}(x)$
Useful when an equation is to be written entirely in terms of $\sinh(x)$ .
Quote the double-angle formula for $\sinh$
$\sinh(2x)=2\sinh(x)\cosh(x)$
This one has the same shape as the circular version.
Recall the derivative of $\sinh$
$\frac{\mathrm{d}}{\mathrm{d}x}\sinh(x)=\cosh(x)$
Differentiating the exponential definition returns $\cosh(x)$ .
Recall the derivative of $\cosh$
$\frac{\mathrm{d}}{\mathrm{d}x}\cosh(x)=\sinh(x)$
There is NO minus sign here: unlike $\cos$ , the derivative of $\cosh$ is $+\sinh$ .
State the number of real solutions
$n=3$
This is the complete count of real solutions.

Answer

3

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