Further Maths Further calculus Practice Questions

Free Further Maths Further calculus practice questions with full step-by-step worked solutions. Covers improper-integrals, power-rule, exponential-integration, inverse-tangent-integral. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Evaluate ∫3∞1x2 dx\int_{3}^{\infty}\frac{1}{x^{2}}\,dx.
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Worked solution

  1. Write down the integral to be evaluated

    I=∫3∞1x2 dxI=\int_{3}^{\infty}\frac{1}{x^{2}}\,dx

    Identify the integrand and the limits before choosing a method.

  2. State why the integral is improper

    the interval is unbounded above\text{the interval is unbounded above}

    This integral is improper because the upper limit of integration is infinite.

  3. State the exact value

    I=13I=\frac{1}{3}

    This is the exact answer required.

Answer
13\frac{1}{3}
Question 2
2 markseasy
Which of the following is the mean value of f(x)=3xf(x)=3x over the interval [0,4]\left[0,4\right]?
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Worked solution

  1. Quote the formula for the mean value

    fˉ=1b−a∫abf(x) dx=14∫043x dx\bar{f}=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx=\frac{1}{4}\int_{0}^{4}3x\,dx

    The width of [0,4]\left[0,4\right] is 44, so that is what the integral is divided by.

  2. Integrate the polynomial term by term

    ∫(3x) dx=3x22+c\int\left(3x\right)\,dx=\frac{3x^{2}}{2}+c

    Each power of xx is integrated separately with the power rule.

  3. Note that the integrand is continuous on the interval

    3x is a polynomial3x\ \text{is a polynomial}

    A polynomial is defined everywhere, so this integral is a proper one.

  4. State the exact value

    fˉ=6\bar{f}=6

    This is the exact answer required.

Answer
66
Question 3
4 marksintermediate
Which of the following is the correct partial fraction decomposition of 2x+1(x−1)(x+2)\frac{2x+1}{\left(x-1\right)\left(x+2\right)}?
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Worked solution

  1. Write down the fraction to be decomposed

    2x+1(x−1)(x+2)=2x+1(x−1)(x+2)\frac{2x+1}{\left(x-1\right)\left(x+2\right)}=\frac{2x+1}{\left(x-1\right)\left(x+2\right)}

    The denominator is already a product of irreducible factors.

  2. Factorise the denominator

    x2+x−2=(x−1)(x+2)x^{2}+x-2=\left(x-1\right)\left(x+2\right)

    The denominator must be a product of irreducible factors before it can be split up.

  3. Write the integrand in partial fractions

    2x+1(x−1)(x+2)≡A(x−1)+B(x+2)\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\equiv \frac{A}{\left(x-1\right)}+\frac{B}{\left(x+2\right)}

    Each factor contributes one term for every power up to its multiplicity; a quadratic factor carries a linear numerator.

  4. Multiply through by the denominator

    2x+1≡A(x+2)+B(x−1)2x+1\equiv A\left(x+2\right)+B\left(x-1\right)

    Clearing the fractions leaves an identity that holds for every xx.

  5. Substitute x=1x=1 to find AA

    3=A×(3)⇒A=13=A\times\left(3\right)\quad\Rightarrow\quad A=1

    Choosing the root of a factor kills every other term of the identity at once.

  6. Substitute x=−2x=-2 to find BB

    −3=B×(−3)⇒B=1-3=B\times\left(-3\right)\quad\Rightarrow\quad B=1

    Choosing the root of a factor kills every other term of the identity at once.

  7. Select the correct decomposition

    2x+1(x−1)(x+2)≡1(x−1)+1(x+2)\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\equiv \frac{1}{\left(x-1\right)}+\frac{1}{\left(x+2\right)}

    This is the only option that is identically equal to the original fraction.

Answer
1(x−1)+1(x+2)\frac{1}{\left(x-1\right)}+\frac{1}{\left(x+2\right)}
Question 4
6 markshard
Which of the following is the correct partial fraction decomposition of 1(x+1)2(x+2)\frac{1}{\left(x+1\right)^{2}\left(x+2\right)}?
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Worked solution

  1. Write down the fraction to be decomposed

    1(x+1)2(x+2)=1(x+1)2(x+2)\frac{1}{\left(x+1\right)^{2}\left(x+2\right)}=\frac{1}{\left(x+1\right)^{2}\left(x+2\right)}

    The denominator is already a product of irreducible factors.

  2. Factorise the denominator

    x3+4x2+5x+2=(x+1)2(x+2)x^{3}+4x^{2}+5x+2=\left(x+1\right)^{2}\left(x+2\right)

    The denominator must be a product of irreducible factors before it can be split up.

  3. Write the integrand in partial fractions

    1(x+1)2(x+2)≡A(x+1)+B(x+1)2+C(x+2)\frac{1}{\left(x+1\right)^{2}\left(x+2\right)}\equiv \frac{A}{\left(x+1\right)}+\frac{B}{\left(x+1\right)^{2}}+\frac{C}{\left(x+2\right)}

    Each factor contributes one term for every power up to its multiplicity; a quadratic factor carries a linear numerator.

  4. Multiply through by the denominator

    1≡A(x+1)(x+2)+B(x+2)+C(x+1)21\equiv A\left(x+1\right)\left(x+2\right)+B\left(x+2\right)+C\left(x+1\right)^{2}

    Clearing the fractions leaves an identity that holds for every xx.

  5. Substitute x=−1x=-1 to find BB

    1=B×(1)⇒B=11=B\times\left(1\right)\quad\Rightarrow\quad B=1

    Choosing the root of a factor kills every other term of the identity at once.

  6. Substitute x=−2x=-2 to find CC

    1=C×(1)⇒C=11=C\times\left(1\right)\quad\Rightarrow\quad C=1

    Choosing the root of a factor kills every other term of the identity at once.

  7. Compare coefficients to find the remaining constants

    A=−1A=-1

    Equating the coefficients of each power of xx on both sides of the identity gives enough equations for the constants left over.

  8. Check the decomposition by recombining the fractions

    −1(x+1)+1(x+1)2+1(x+2)≡1(x+1)2(x+2)-\frac{1}{\left(x+1\right)}+\frac{1}{\left(x+1\right)^{2}}+\frac{1}{\left(x+2\right)}\equiv\frac{1}{\left(x+1\right)^{2}\left(x+2\right)}

    Putting the partial fractions back over a common denominator must return the original fraction.

  9. Test the identity at x=2x=2

    LHS=136,RHS=136\text{LHS}=\frac{1}{36},\quad\text{RHS}=\frac{1}{36}

    Both sides of an identity must agree at every value of xx, so a single test value is a fast check on the constants.

  10. Recall the linearity of integration

    ∫(λf+μg) dx=λ∫f dx+μ∫g dx\int\left(\lambda f+\mu g\right)\,dx=\lambda\int f\,dx+\mu\int g\,dx

    Constants come outside and a sum integrates term by term.

  11. Select the correct decomposition

    1(x+1)2(x+2)≡−1(x+1)+1(x+1)2+1(x+2)\frac{1}{\left(x+1\right)^{2}\left(x+2\right)}\equiv -\frac{1}{\left(x+1\right)}+\frac{1}{\left(x+1\right)^{2}}+\frac{1}{\left(x+2\right)}

    This is the only option that is identically equal to the original fraction.

Answer
−1(x+1)+1(x+1)2+1(x+2)-\frac{1}{\left(x+1\right)}+\frac{1}{\left(x+1\right)^{2}}+\frac{1}{\left(x+2\right)}
Question 5
9 markschallenging
Which of the following is the correct partial fraction decomposition of 5x+3(x+1)(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}?
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Worked solution

  1. Write down the fraction to be decomposed

    5x+3(x+1)(x2+9)=5x+3(x+1)(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}=\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}

    The denominator is already a product of irreducible factors.

  2. Factorise the denominator

    x3+x2+9x+9=(x+1)(x2+9)x^{3}+x^{2}+9x+9=\left(x+1\right)\left(x^{2}+9\right)

    The denominator must be a product of irreducible factors before it can be split up.

  3. Write the integrand in partial fractions

    5x+3(x+1)(x2+9)≡A(x+1)+Bx+C(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}\equiv \frac{A}{\left(x+1\right)}+\frac{Bx+C}{\left(x^{2}+9\right)}

    Each factor contributes one term for every power up to its multiplicity; a quadratic factor carries a linear numerator.

  4. Multiply through by the denominator

    5x+3≡A(x2+9)+(Bx+C)(x+1)5x+3\equiv A\left(x^{2}+9\right)+\left(Bx+C\right)\left(x+1\right)

    Clearing the fractions leaves an identity that holds for every xx.

  5. Substitute x=−1x=-1 to find AA

    −2=A×(10)⇒A=−15-2=A\times\left(10\right)\quad\Rightarrow\quad A=-\frac{1}{5}

    Choosing the root of a factor kills every other term of the identity at once.

  6. Compare coefficients to find the remaining constants

    B=15,C=245B=\frac{1}{5},\quad C=\frac{24}{5}

    Equating the coefficients of each power of xx on both sides of the identity gives enough equations for the constants left over.

  7. Check the decomposition by recombining the fractions

    −15(x+1)+x+245(x2+9)≡5x+3(x+1)(x2+9)-\frac{1}{5\left(x+1\right)}+\frac{x+24}{5\left(x^{2}+9\right)}\equiv\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}

    Putting the partial fractions back over a common denominator must return the original fraction.

  8. Test the identity at x=2x=2

    LHS=13,RHS=13\text{LHS}=\frac{1}{3},\quad\text{RHS}=\frac{1}{3}

    Both sides of an identity must agree at every value of xx, so a single test value is a fast check on the constants.

  9. Recall the pp-test at infinity

    ∫1∞1xp dx converges  ⟺  p>1\int_{1}^{\infty}\frac{1}{x^{p}}\,dx\ \text{converges}\iff p>1

    The tail must decay strictly faster than 1x\frac{1}{x}.

  10. Recall the pp-test at a singularity

    ∫011xp dx converges  ⟺  p<1\int_{0}^{1}\frac{1}{x^{p}}\,dx\ \text{converges}\iff p<1

    Near the singularity the blow-up must be slower than 1x\frac{1}{x}.

  11. Note the limit of a decaying exponential

    lim⁡t→∞tme−kt=0(k>0)\lim_{t\to\infty}t^{m}e^{-kt}=0\quad\left(k>0\right)

    An exponential beats every power, so the product tends to zero.

  12. Note the limit of a negative power

    lim⁡t→∞1tm=0(m>0)\lim_{t\to\infty}\frac{1}{t^{m}}=0\quad\left(m>0\right)

    Any positive power of tt in a denominator drives the term to zero.

  13. Note the limit of the inverse tangent

    lim⁡t→∞arctan⁡t=π2\lim_{t\to\infty}\arctan t=\frac{\pi}{2}

    The graph of arctan⁡\arctan has a horizontal asymptote at π2\frac{\pi}{2}.

  14. Recall the logarithm law used to combine two logarithms

    ln⁡A−ln⁡B=ln⁡(AB)\ln A-\ln B=\ln\left(\frac{A}{B}\right)

    A difference of logarithms is the logarithm of a quotient.

  15. Recall the exact values of the inverse tangent

    arctan⁡1=π4,arctan⁡3=π3\arctan1=\frac{\pi}{4},\quad\arctan\sqrt{3}=\frac{\pi}{3}

    Exact values let the answer be given in terms of π\pi.

  16. Select the correct decomposition

    5x+3(x+1)(x2+9)≡−15(x+1)+x+245(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}\equiv -\frac{1}{5\left(x+1\right)}+\frac{x+24}{5\left(x^{2}+9\right)}

    This is the only option that is identically equal to the original fraction.

Answer
−15(x+1)+x+245(x2+9)-\frac{1}{5\left(x+1\right)}+\frac{x+24}{5\left(x^{2}+9\right)}

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