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

Which of the following is equal to (2x) for all real x?

Write down the expression: (2x) Replace each hyperbolic function by its exponential definition: (2x)=e^2x2-e^-2x2 Expand and collect the exponential terms: e^2x2-e^-2x2=e^2x2-e^-2x2 Write the result back in hyperbolic form: (2x)=2(x)(x) 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) Select the equivalent expression: (2x)=2(x)(x) Answer: 2(x)(x)

Given that (x)=-(3)/(4), find the exact value of (2x).

Write down what is given: (x)=-(3)/(4) Quote the identity linking the hyperbolic functions: ^2(x)-^2(x)=1 Solve for x: x=-(2) Substitute into the required expression: (2x)=(17)/(8) Check the result numerically: (2x)2.12500 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 exact value: (2x)=(17)/(8) Answer: (17)/(8)

Evaluate _0^(3)(x)(x) dx.

Write down the integral: _0^(3)(x)(x) dx Write down the antiderivative: ^2(x)2 Differentiate the antiderivative as a check: ddx(^2(x)2)=(x)(x) Substitute the limits: [^2(x)2]_0^(3)=((25)/(18))-((1)/(2)) Simplify the difference: (8)/(9) Give a decimal check: 0.88889 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) State the exact value of the definite integral: _0^(3)(x)(x) dx=(8)/(9) Answer: (8)/(9)

Hard Further Maths Hyperbolic functions Questions

Challenging, exam-style Further Maths Hyperbolic functions questions with worked solutions. Stretch yourself on the hardest hyperbolic-functions, solving-equations, exponential-substitution, two-roots problems.

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Further Maths34 questionsStep-by-step solutions

Question 1

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

Question 2

9 markschallenging

Which of the following is equal to

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

for all real

x

Show worked solution

Worked solution

Write down the expression
$\sinh{\left(2x\right)}$
Identify which identity or definition will simplify it.
Replace each hyperbolic function by its exponential definition
$\sinh{\left(2x\right)}=\frac{e^{2x}}{2}-\frac{e^{-2x}}{2}$
The exponential definitions turn the expression into ordinary algebra.
Expand and collect the exponential terms
$\frac{e^{2x}}{2}-\frac{e^{-2x}}{2}=\frac{e^{2x}}{2}-\frac{e^{-2x}}{2}$
Multiplying out lets the exponentials cancel.
Write the result back in hyperbolic form
$\sinh{\left(2x\right)}=2\sinh{\left(x\right)}\cosh{\left(x\right)}$
This is the required simplified form.
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 $+\sinh$ .
Select the equivalent expression
$\sinh{\left(2x\right)}=2\sinh{\left(x\right)}\cosh{\left(x\right)}$
The two expressions are identically equal for every real $x$ .

Answer

2\sinh{\left(x\right)}\cosh{\left(x\right)}

Question 3

9 markschallenging

Which of the following gives the complete solution set of the equation

\cosh{\left(x\right)}=3

Show worked solution

Worked solution

Write down the equation
$\cosh{\left(x\right)}=3$
All real solutions are required.
Replace every hyperbolic function by its exponential definition
$\cosh{\left(x\right)}-3=0$
This turns the equation into an equation in $e^{x}$ alone.
Substitute $u=e^{x}$ and clear the denominator
$u^{2}-6u+1=0$
Multiplying through by a power of $u$ leaves a polynomial equation in $u$ .
Solve for $u$
$u=3-2\sqrt{2},\quad u=2\sqrt{2}+3$
Each root of the polynomial is a candidate value of $e^{x}$ .
Take natural logarithms of every admissible root
$x=\ln\left(3-2\sqrt{2}\right),\quad x=\ln\left(2\sqrt{2}+3\right)$
Each admissible value of $u$ gives exactly one real value of $x$ .
Simplify each logarithm
$x=\pm\ln{\left(2\sqrt{2}+3\right)}$
Reciprocal roots $u$ and $\frac{1}{u}$ produce $x$ and $-x$ .
Keep BOTH members of the $\pm$ pair
$\cosh(-x)=\cosh(x)$
Because $\cosh$ is even, the roots come in $\pm$ pairs; dropping the negative one is the classic error.
Check the solution $x=-\ln{\left(2\sqrt{2}+3\right)}$
$\cosh{\left(\ln{\left(2\sqrt{2}+3\right)}\right)}=3$
Substituting the root back reproduces the right-hand side.
Count the solutions
$n=2$
The polynomial in $u$ has exactly 2 admissible positive roots, so the equation has exactly 2 real solutions.
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 $\sinh(x)$ .
Select the option showing the complete solution set
$x=\pm\ln{\left(2\sqrt{2}+3\right)}$
Every real solution is listed; none has been dropped.

Answer

x=\pm\ln{\left(2\sqrt{2}+3\right)}

Question 4

9 markschallenging

Given that

\sinh{\left(x\right)}=-\frac{3}{4}

, find the exact value of

\cosh{\left(2x\right)}

Show worked solution

Worked solution

Write down what is given
$\sinh{\left(x\right)}=-\frac{3}{4}$
This fixes the value of one hyperbolic function.
Quote the identity linking the hyperbolic functions
$\cosh^{2}(x)-\sinh^{2}(x)=1$
The MINUS sign here is Osborn's rule at work.
Solve for $x$
$x=-\ln{\left(2\right)}$
The given information leaves exactly one admissible value of $x$ .
Substitute into the required expression
$\cosh{\left(2x\right)}=\frac{17}{8}$
This is the exact value asked for.
Check the result numerically
$\cosh{\left(2x\right)}\approx2.12500$
A decimal check confirms the exact value.
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 exact value
$\cosh{\left(2x\right)}=\frac{17}{8}$
This is the required exact value.

Answer

\frac{17}{8}

Question 5

9 markschallenging

Evaluate

\int_{0}^{\ln{\left(3\right)}}\sinh{\left(x\right)}\cosh{\left(x\right)}\,\mathrm{d}x

Show worked solution

Worked solution

Write down the integral
$\int_{0}^{\ln{\left(3\right)}}\sinh{\left(x\right)}\cosh{\left(x\right)}\,\mathrm{d}x$
Identify the standard form that the integrand matches.
Write down the antiderivative
$\frac{\cosh^{2}{\left(x\right)}}{2}$
Reverse the corresponding standard derivative.
Differentiate the antiderivative as a check
$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\cosh^{2}{\left(x\right)}}{2}\right)=\sinh{\left(x\right)}\cosh{\left(x\right)}$
Differentiating back reproduces the integrand exactly, so the antiderivative is right.
Substitute the limits
$\left[\frac{\cosh^{2}{\left(x\right)}}{2}\right]_{0}^{\ln{\left(3\right)}}=\left(\frac{25}{18}\right)-\left(\frac{1}{2}\right)$
Evaluate the antiderivative at the upper limit and subtract its value at the lower limit.
Simplify the difference
$\frac{8}{9}$
Exponentials, surds and logarithms combine to the exact value.
Give a decimal check
$\approx0.88889$
The decimal value is a useful sanity check on the exact answer.
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.
State the exact value of the definite integral
$\int_{0}^{\ln{\left(3\right)}}\sinh{\left(x\right)}\cosh{\left(x\right)}\,\mathrm{d}x=\frac{8}{9}$
This is the exact value of the integral.

Answer

\frac{8}{9}

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