Browse Numerical methods

106 questions at your level

DifficultyAllLevel 2Level 3Level 4Level 5

Locating roots by change of sign

39 questions

LessonNot started
The diagram shows part of the graph of y=f(x)y = \text{f}(x), where f(x)\text{f}(x) is a cubic polynomial in xx.123xyExplain why one of the roots of the equation f(x)=0\text{f}(x) = 0 cannot be found by the sign change method.

(2 marks)
●●●●●Level 22 marksStart
Show that the equation x3−2x−5=0x^3 - 2x - 5 = 0 has a root between x=2x = 2 and x=3x = 3.
●●●●●Level 22 marksStart
Show that x3−7x+2=0x^3 - 7x + 2 = 0 has a root between x=0x = 0 and x=1x = 1.
●●●●●Level 22 marksStart
Show that the equation x3+2x−7=0x^{3} + 2x - 7 = 0 has a root between x=1x = 1 and x=2x = 2. (2 marks)
●●●●●Level 22 marksStart

Iterative methods (x = g(x))

33 questions

LessonNot started
(a) Show that x3−7x+2=0x^3 - 7x + 2 = 0 can be written as x=x3+27x = \dfrac{x^3 + 2}{7}.
(b) Using xn+1=xn3+27x_{n+1} = \dfrac{x_n^3 + 2}{7} with x0=0x_0 = 0, find the root near 00 to 3 decimal places.
●●●●●Level 25 marksStart
The iteration xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5} is used with x0=2x_0 = 2 to find a root of x3−2x−5=0x^3 - 2x - 5 = 0. Find x1x_1, x2x_2 and x3x_3, giving each to 4 decimal places.
●●●●●Level 23 marksStart
Show that the equation x2−5x+3=0x^{2} - 5x + 3 = 0 can be rearranged into the form

x=x2+35x = \dfrac{x^{2} + 3}{5} (2 marks)
●●●●●Level 22 marksStart
The iteration xn+1=5xn−2x_{n+1} = \sqrt{5x_{n} - 2} is used with x0=2x_{0} = 2.

Find x1x_{1}, x2x_{2} and x3x_{3}, giving each to 4 decimal places. (3 marks)
●●●●●Level 23 marksStart

The Newton–Raphson method

29 questions

LessonNot started
f(x)=x3−2x−5f(x) = x^3 - 2x - 5. Using the Newton–Raphson method with x0=2x_0 = 2, find x1x_1.
●●●●●Level 22 marksStart
The positive root of x2−7=0x^2 - 7 = 0 is 7\sqrt7. Using Newton–Raphson with x0=3x_0 = 3, find x1x_1 and x2x_2 to 4 decimal places.
●●●●●Level 23 marksStart

The trapezium rule

29 questions

LessonNot started
Use the trapezium rule with 4 strips to estimate ∫04x2+1 dx\displaystyle\int_0^4 \sqrt{x^2 + 1}\,dx, giving your answer to 3 decimal places.
●●●●●Level 23 marksStart
Use the trapezium rule with 5 ordinates to estimate ∫132x dx\displaystyle\int_1^3 2^x\,dx, giving your answer to 3 decimal places.
●●●●●Level 23 marksStart
The table below shows corresponding values of xx and yy for y=5+x4y = \sqrt{5 + x^4}

The values of yy are given to 3 decimal places as appropriate.

x00.40.81.21.6y2.2362.2422.3262.6603.399\begin{array}{c|ccccc} x & 0 & 0.4 & 0.8 & 1.2 & 1.6 \\ \hline y & 2.236 & 2.242 & 2.326 & 2.660 & 3.399 \end{array}
(a) Use the trapezium rule with all the values of yy in the table to find an estimate for

∫01.65+x4 dx\int_0^{1.6} \sqrt{5 + x^4}\,dx

giving your answer to 3 significant figures.

(3 marks)
(b) Use your answer to part (a) to find an estimate for

∫01.6(1+45+x4)dx\int_0^{1.6} \left(1 + 4\sqrt{5 + x^4}\right)dx

(2 marks)
●●●●●Level 25 marksStart