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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/gcse/maths/ocr/grade-7-7-10-iteration/flashcards","label":"Iteration","subjectCode":"gcse_maths__ocr","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"4a224a79-a645-3a68-41f4-50d14aaffe8a","title":"Iteration Flashcards","topicId":"17e0e276-380c-4c8f-86de-602032b75bce","topicSlug":"grade-7-7-10-iteration","cards":[{"id":"93fc4970-bc2a-8592-9c99-a51bbad81d49","cardType":"cloze","points":null,"front":"Iteration means [ ].","back":"Iteration means repeatedly applying the same rule.","mcqOptions":[{"text":"repeatedly applying the same rule","correct":true,"explanation":"Iteration generates successive values by applying the same rule repeatedly, with each result normally becoming the input for the next application."},{"text":"applying several different rules once","correct":false,"explanation":"Using different rules once is not iteration. Iteration means repeatedly applying the same rule, usually feeding each calculated value back into that rule."},{"text":"reversing the same rule repeatedly","correct":false,"explanation":"Iteration does not require a rule to be reversed. It means repeatedly applying the same rule to generate a sequence of successive values."},{"text":"repeatedly choosing a different rule","correct":false,"explanation":"Changing the rule at every step contradicts the definition of iteration. The right description is repeatedly applying the same rule."}]},{"id":"b3e13230-ed77-0fa9-d9cf-45560f2f7a5b","cardType":"basic","points":1,"front":"In \$x_3\$, what does the subscript 3 tell you?","back":"The term with subscript 3.","mcqOptions":[{"text":"The term with subscript 3.","correct":true,"explanation":"A subscript identifies a term’s position or label in a sequence, so \$x_3\$ denotes the term with subscript 3."},{"text":"The term raised to power 3.","correct":false,"explanation":"This confuses a subscript with an exponent. The 3 labels the term with subscript 3; a power would be written \$x^3\$."},{"text":"The term multiplied by 3.","correct":false,"explanation":"A subscript does not indicate multiplication. In \$x_3\$, the 3 identifies the term with subscript 3, rather than meaning \$3x\$."},{"text":"The term divided by 3.","correct":false,"explanation":"A subscript labels a term and does not mean division. The notation \$x_3\$ identifies the term with subscript 3."}]},{"id":"08191ee3-73a0-4ee9-92fd-aaa76337bcfb","cardType":"cloze","points":null,"front":"In \$x_3\$, the 3 is a [ ], not a [ ].","back":"In \$x_3\$, the 3 is a subscript, not a power.","mcqOptions":[{"text":"In \$x_3\$, the 3 is a subscript, not a power.","correct":true,"explanation":"The 3 is written below the variable, so it is a subscript labelling a term. An exponent or power would be written above the variable."},{"text":"In \$x_3\$, the 3 is a power, not a subscript.","correct":false,"explanation":"This reverses the meanings of the two positions. In \$x_3\$, the 3 is a subscript; a power of 3 would be written \$x^3\$."},{"text":"In \$x_3\$, the 3 is a coefficient, not a power.","correct":false,"explanation":"A coefficient is placed before a variable, as in \$3x\$. Here the 3 is below the variable, making it a subscript rather than a power."},{"text":"In \$x_3\$, the 3 is a subscript, not a coefficient.","correct":false,"explanation":"Although 3 is a subscript, the intended contrast is with a power, a common notation error. In \$x_3\$, it is a subscript, not a power."}]},{"id":"a8303c9e-2d79-7ea5-896e-9466cdb1377a","cardType":"cloze","points":null,"front":"A starting value, such as \$x_0\$, is used [ ] the recurrence formula.","back":"A starting value, such as \$x_0\$, is used before the recurrence formula.","mcqOptions":[{"text":"before","correct":true,"explanation":"The starting value must be known before the recurrence formula can first be used, because it provides the input needed to calculate the next term."},{"text":"after","correct":false,"explanation":"Placing the starting value after the recurrence formula reverses the process. The starting value is used before the formula to calculate the first new term."},{"text":"instead of","correct":false,"explanation":"The starting value does not replace the recurrence formula. It is used before the formula as the initial input from which later terms are generated."},{"text":"independently of","correct":false,"explanation":"The starting value and recurrence formula are not independent. The starting value is used before the formula and supplies its first input."}]},{"id":"3df45341-6bf0-3972-c05f-c5690f19dc9a","cardType":"basic","points":1,"front":"When using \$x_{n+1}=f(x_n)\$, which value is substituted to find the next term?","back":"The current term \$x_n\$.","mcqOptions":[{"text":"The current term \$x_n\$.","correct":true,"explanation":"The recurrence formula applies \$f\$ to the current term \$x_n\$, producing the next term \$x_{n+1}\$."},{"text":"The starting term \$x_0\$.","correct":false,"explanation":"The starting term is used only to begin the iteration. At each later step, the current term \$x_n\$ is substituted to obtain \$x_{n+1}\$."},{"text":"The next term \$x_{n+1}\$.","correct":false,"explanation":"Substituting the next term would use the unknown value being calculated. The current term \$x_n\$ is substituted into \$f\$ to find \$x_{n+1}\$."},{"text":"The term number \$n\$.","correct":false,"explanation":"The index \$n\$ labels the term but is not the value substituted here. The recurrence uses the current term \$x_n\$ to calculate the next one."}]},{"id":"94581bfa-35d1-67a0-afe1-465f627a3c9d","cardType":"cloze","points":null,"front":"In \$x_{n+1}=2+\\frac{6}{x_n^2}\$, \$x_n\$ is the [ ] value and \$x_{n+1}\$ is the [ ] value.","back":"In \$x_{n+1}=2+\\frac{6}{x_n^2}\$, \$x_n\$ is the current value and \$x_{n+1}\$ is the next value.","mcqOptions":[{"text":"In \$x_{n+1}=2+\\frac{6}{x_n^2}\$, \$x_n\$ is the current value and \$x_{n+1}\$ is the next value.","correct":true,"explanation":"The formula uses the current value \$x_n\$ as its input and produces the next value \$x_{n+1}\$, advancing the sequence by one term."},{"text":"In \$x_{n+1}=2+\\frac{6}{x_n^2}\$, \$x_n\$ is the previous value and \$x_{n+1}\$ is the current value.","correct":false,"explanation":"This labels the terms from the following step rather than the calculation shown. Here \$x_n\$ is the current value and \$x_{n+1}\$ is the next value."},{"text":"In \$x_{n+1}=2+\\frac{6}{x_n^2}\$, \$x_n\$ is the next value and \$x_{n+1}\$ is the current value.","correct":false,"explanation":"This reverses the order of the sequence. The input \$x_n\$ is the current value, while the output \$x_{n+1}\$ is the next value."},{"text":"In \$x_{n+1}=2+\\frac{6}{x_n^2}\$, \$x_n\$ is the starting value and \$x_{n+1}\$ is the final value.","correct":false,"explanation":"The indices do not necessarily represent the beginning and end of the iteration. They identify consecutive terms: \$x_n\$ is current and \$x_{n+1}\$ is next."}]},{"id":"54b3466a-feae-22c3-9ea3-4080657abb75","cardType":"cloze","points":null,"front":"Starting from \$x_0\$, after one use you get [ ]; after three uses you get [ ].","back":"Starting from \$x_0\$, after one use you get \$x_1\$; after three uses you get \$x_3\$.","mcqOptions":[{"text":"Starting from \$x_0\$, after one use you get \$x_1\$; after three uses you get \$x_3\$.","correct":true,"explanation":"Each use of the recurrence increases the subscript by one. Starting at \$x_0\$, one use reaches \$x_1\$ and three uses reach \$x_3\$."},{"text":"Starting from \$x_0\$, after one use you get \$x_0\$; after three uses you get \$x_2\$.","correct":false,"explanation":"This fails to count the first application as a step. Each use increases the subscript, so one use gives \$x_1\$ and three uses give \$x_3\$."},{"text":"Starting from \$x_0\$, after one use you get \$x_1\$; after three uses you get \$x_4\$.","correct":false,"explanation":"The first result is counted correctly, but the third is one step too far. Three applications from \$x_0\$ produce \$x_1\$, \$x_2\$, then \$x_3\$."},{"text":"Starting from \$x_0\$, after one use you get \$x_2\$; after three uses you get \$x_3\$.","correct":false,"explanation":"This skips \$x_1\$ on the first application. A recurrence advances one subscript at a time, so one use gives \$x_1\$, while three uses give \$x_3\$."}]},{"id":"1e1d4c4b-3b1e-1f00-2ca5-6ecaa252c138","cardType":"basic","points":1,"front":"For \$P_{t+1}=1.10(P_t-12)\$, which value is used to calculate \$P_2\$?","back":"\$P_1\$, not \$P_0\$.","mcqOptions":[{"text":"\$P_1\$, not \$P_0\$.","correct":true,"explanation":"To calculate \$P_2\$, set \$t=1\$, giving \$P_2=1.10(P_1-12)\$. The recurrence therefore uses the immediately preceding value \$P_1\$."},{"text":"\$P_0\$, not \$P_1\$.","correct":false,"explanation":"Using \$P_0\$ would calculate \$P_1\$, not \$P_2\$. To find \$P_2\$, the current value must be \$P_1\$."},{"text":"\$P_2\$, not \$P_1\$.","correct":false,"explanation":"The recurrence cannot use the unknown value \$P_2\$ to calculate itself. It uses the preceding value \$P_1\$ to produce \$P_2\$."},{"text":"\$P_3\$, not \$P_2\$.","correct":false,"explanation":"This uses a later term rather than the preceding one. Calculating \$P_2\$ requires setting \$t=1\$, so the substituted value is \$P_1\$."}]},{"id":"1c6f5a9c-d829-0939-9a10-169e0ab253b9","cardType":"basic","points":1,"front":"Given \$P_0=180\$ and \$P_{t+1}=1.10(P_t-12)\$, find \$P_1\$.","back":"\$184.8\$","mcqOptions":[{"text":"\$184.8\$","correct":true,"explanation":"Substituting \$P_0=180\$ into the rule gives \$P_1=1.10(180-12)=1.10\\times168=184.8\$."},{"text":"\$184.68\$","correct":false,"explanation":"This comes from a small arithmetic error after subtracting 12. The correct calculation is \$1.10(180-12)=1.10\\times168=184.8\$."},{"text":"\$198\$","correct":false,"explanation":"This multiplies 180 by 1.10 without first subtracting 12. The iteration rule requires \$1.10(180-12)\$, which gives \$184.8\$."},{"text":"\$172.8\$","correct":false,"explanation":"This subtracts 12 after applying the multiplier, rather than inside the brackets. The required order is \$1.10(180-12)=184.8\$."}]},{"id":"bec8a194-6826-1680-5f1e-593848a6b2ab","cardType":"basic","points":1,"front":"Given \$P_2=190.08\$ and \$P_{t+1}=1.10(P_t-12)\$, find \$P_3\$.","back":"\$195.888\$","mcqOptions":[{"text":"\$195.888\$","correct":true,"explanation":"Using \$P_2=190.08\$, the rule gives \$P_3=1.10(190.08-12)=1.10\\times178.08=195.888\$."},{"text":"\$207.088\$","correct":false,"explanation":"This adds the multiplier effect to the original value without subtracting 12 first. Applying the brackets correctly gives \$1.10(190.08-12)=195.888\$."},{"text":"\$196.08\$","correct":false,"explanation":"This effectively adds 6% to \$190.08\$, rather than applying the 10% increase to the value after subtracting 12. The correct result is \$195.888\$."},{"text":"\$185.688\$","correct":false,"explanation":"This subtracts 12 from the result after multiplying, reversing the intended order of operations. The rule gives \$1.10(190.08-12)=195.888\$."}]},{"id":"7925168d-23b6-5c85-762b-2da5f60acde1","cardType":"basic","points":1,"front":"With \$x_0=2\$ and \$x_{n+1}=2+\\frac{6}{x_n^2}\$, find \$x_1\$.","back":"\$3.5\$","mcqOptions":[{"text":"\$3.5\$","correct":true,"explanation":"Substitute \$x_0=2\$ into the rule: \$x_1=2+\\frac{6}{2^2}=2+\\frac{6}{4}=3.5\$."},{"text":"\$5\$","correct":false,"explanation":"This treats the denominator as \$x_0\$ instead of \$x_0^2\$, giving \$2+6\\div2=5\$. Squaring 2 first gives the correct value, \$3.5\$."},{"text":"\$2.75\$","correct":false,"explanation":"This halves the fraction incorrectly: \$6\\div2^2\$ is \$6\\div4=1.5\$, not \$0.75\$. Therefore \$x_1=2+1.5=3.5\$."},{"text":"\$4\$","correct":false,"explanation":"This uses \$6\\div2^2=2\$, confusing the numerator and denominator calculation. Since \$6\\div4=1.5\$, the iteration gives \$x_1=3.5\$."}]},{"id":"d0748487-fe11-cba7-0220-c431e3392506","cardType":"basic","points":1,"front":"Given \$x_1=3.5\$ and \$x_{n+1}=2+\\frac{6}{x_n^2}\$, find \$x_2\$ to 3 d.p.","back":"\$2.490\$","mcqOptions":[{"text":"\$2.490\$","correct":true,"explanation":"Substituting \$x_1=3.5\$ gives \$x_2=2+\\frac{6}{3.5^2}=2+\\frac{6}{12.25}=2.489795\\ldots\$, which rounds to \$2.490\$ to 3 d.p."},{"text":"\$2.489\$","correct":false,"explanation":"The unrounded value is \$2.489795\\ldots\$, so the fourth decimal place is 7 and the third decimal place must round up. The correct rounded value is \$2.490\$."},{"text":"\$2.857\$","correct":false,"explanation":"This results from using \$3.5\$ rather than squaring it in the denominator. The rule requires \$2+6\\div(3.5^2)=2.489795\\ldots\$, giving \$2.490\$."},{"text":"\$3.714\$","correct":false,"explanation":"This adds \$6\\div3.5\$ instead of \$6\\div3.5^2\$, omitting the square on the denominator. Squaring first gives the correct result, \$2.490\$."}]},{"id":"fce9ef58-6c80-717c-986e-f5a9bf7c66dc","cardType":"basic","points":1,"front":"Given \$x_2=2.489795918\\ldots\$ and \$x_{n+1}=2+\\frac{6}{x_n^2}\$, find \$x_3\$ to 3 d.p.","back":"\$2.968\$","mcqOptions":[{"text":"\$2.968\$","correct":true,"explanation":"Using \$x_2=2.489795918\\ldots\$, the rule gives \$x_3=2+\\frac{6}{x_2^2}=2.9677\\ldots\$, which rounds to \$2.968\$ to 3 d.p."},{"text":"\$2.968\\ldots\$","correct":false,"explanation":"The value is approximately \$2.9677\\ldots\$, but the question requires 3 decimal places. Rounding to that precision gives \$2.968\$, not an unrounded recurring value."},{"text":"\$2.410\$","correct":false,"explanation":"This uses a rough or inaccurate square for \$x_2\$, making the fraction too small. Calculating \$2+6/(2.489795918\\ldots)^2\$ gives \$2.9677\\ldots\$, hence \$2.968\$."},{"text":"\$3.506\$","correct":false,"explanation":"This effectively uses \$6\\div x_2\$ rather than \$6\\div x_2^2\$, producing too large an added fraction. Squaring the denominator gives the correct rounded value \$2.968\$."}]},{"id":"842845da-662c-fb9b-88f0-f54c2b73fc0e","cardType":"cloze","points":null,"front":"If the terms get closer and closer to one value, the sequence [ ].","back":"If the terms get closer and closer to one value, the sequence converges.","mcqOptions":[{"text":"converges","correct":true,"explanation":"A sequence converges when its terms approach one fixed value more and more closely, even if the limiting value is never reached exactly."},{"text":"diverges","correct":false,"explanation":"Divergence means the terms do not approach one limiting value. The described behaviour is convergence, so the sequence converges."},{"text":"terminates","correct":false,"explanation":"A terminating sequence stops after a finite number of terms, whereas getting closer to one value describes convergence. The correct term is converges."},{"text":"cycles","correct":false,"explanation":"A cycling sequence repeatedly returns to a pattern of values rather than approaching one value. Terms getting closer to one value means the sequence converges."}]},{"id":"83ee645b-7172-46cc-5a97-71b73690f577","cardType":"cloze","points":null,"front":"At a fixed point, applying the iteration rule leaves the value [ ]; the current and next values are [ ].","back":"At a fixed point, applying the iteration rule leaves the value unchanged; the current and next values are equal.","mcqOptions":[{"text":"At a fixed point, applying the iteration rule leaves the value unchanged; the current and next values are equal.","correct":true,"explanation":"At a fixed point, the iteration rule maps the value to itself. Therefore the value remains unchanged, and the current and next terms are equal."},{"text":"At a fixed point, applying the iteration rule leaves the value changed; the current and next values are equal.","correct":false,"explanation":"A fixed point is not changed by applying the rule, although the current and next values are equal. The correct description is unchanged and equal."},{"text":"At a fixed point, applying the iteration rule leaves the value unchanged; the current and next values are different.","correct":false,"explanation":"If the value is unchanged by the rule, the next value must be the same as the current value. Thus the correct pair is unchanged and equal."},{"text":"At a fixed point, applying the iteration rule leaves the value changed; the current and next values are different.","correct":false,"explanation":"Both terms describe movement away from the defining property of a fixed point. A fixed point remains unchanged, so the current and next values are equal."}]},{"id":"911c3df3-c15e-eca0-5821-2cd125f2b079","cardType":"basic","points":1,"front":"With \$x_0=0\$ and \$x_{n+1}=\\frac{2}{x_n^2+4}\$, find \$x_1\$.","back":"\$0.5\$","mcqOptions":[{"text":"\$0.5\$","correct":true,"explanation":"Substituting \$x_0=0\$ gives \$x_1=\\frac{2}{0^2+4}=\\frac{2}{4}=0.5\$."},{"text":"\$2\$","correct":false,"explanation":"This ignores the denominator and uses the numerator alone. The rule requires \$2\\div(0^2+4)=2\\div4=0.5\$."},{"text":"\$0.25\$","correct":false,"explanation":"This squares the numerator or treats the fraction incorrectly. With \$x_0=0\$, the denominator is 4, so \$x_1=2\\div4=0.5\$."},{"text":"\$0\$","correct":false,"explanation":"The zero is squared in the denominator, not used as the whole result. Since the denominator is \$0^2+4=4\$, the next value is \$0.5\$."}]},{"id":"9a9aef1e-1f67-c800-2992-94a6a2818759","cardType":"basic","points":1,"front":"Given \$x_2=0.470588235\\ldots\$ and \$x_{n+1}=\\frac{2}{x_n^2+4}\$, find \$x_3\$ to 3 d.p.","back":"\$0.474\$","mcqOptions":[{"text":"\$0.474\$","correct":true,"explanation":"Substituting \$x_2=0.470588235\\ldots\$ gives \$x_3=\\frac{2}{x_2^2+4}=0.47377\\ldots\$, which rounds to \$0.474\$ to 3 decimal places."},{"text":"\$0.471\$","correct":false,"explanation":"This is close to the previous iterate and comes from treating the sequence as nearly unchanged. The formula gives \$0.47377\\ldots\$, which rounds to \$0.474\$."},{"text":"\$0.476\$","correct":false,"explanation":"This results from an inaccurate squaring or division step. Substituting the given iterate produces \$x_3=0.47377\\ldots\$, so the correct 3 d.p. value is \$0.474\$."},{"text":"\$0.529\$","correct":false,"explanation":"This can result from using \$2/x_n^2+4\$ instead of placing the whole denominator below 2. The required calculation is \$2/(x_n^2+4)=0.47377\\ldots\$, giving \$0.474\$."}]},{"id":"8d1b7249-b013-3209-dd8a-628c49eb81fd","cardType":"basic","points":1,"front":"What substitution links a convergent iteration to its equation?","back":"Set \$x_{n+1}=x_n=a\$.","mcqOptions":[{"text":"Set \$x_{n+1}=x_n=a\$.","correct":true,"explanation":"At convergence, consecutive iterates approach the same limiting value. Calling that value \$a\$ means setting both \$x_{n+1}\$ and \$x_n\$ equal to \$a\$."},{"text":"Set \$x_{n+1}=x_n+1=a\$.","correct":false,"explanation":"Adding 1 incorrectly describes a sequence increasing by a fixed amount, not a convergent sequence. Convergence requires consecutive iterates to become equal: \$x_{n+1}=x_n=a\$."},{"text":"Set \$x_{n+1}=x_n=0\$.","correct":false,"explanation":"Zero is not automatically the limiting value; it must satisfy the iteration equation. The general fixed-point substitution is \$x_{n+1}=x_n=a\$, with \$a\$ then determined algebraically."},{"text":"Set \$x_{n+1}=\\frac{1}{x_n}=a\$.","correct":false,"explanation":"Taking a reciprocal imposes an unrelated condition and does not express convergence. A convergent iteration has equal consecutive values, so the substitution is \$x_{n+1}=x_n=a\$."}]},{"id":"f033ccdb-dc40-7836-14b2-22c5b22dd80e","cardType":"basic","points":2,"front":"From \$a=\\frac{2}{a^2+4}\$, what equation follows after rearranging?","back":"\$a^3+4a-2=0\$","mcqOptions":[{"text":"\$a^3+4a-2=0\$","correct":true,"explanation":"Multiplying by \$a^2+4\$ gives \$a(a^2+4)=2\$. Expanding and moving all terms to one side produces \$a^3+4a-2=0\$."},{"text":"\$a^3+4a+2=0\$","correct":false,"explanation":"The constant term changes sign incorrectly when 2 is moved to the left. Rearranging gives \$a^3+4a-2=0\$, because the original right-hand side is positive 2."},{"text":"\$a^2+4a-2=0\$","correct":false,"explanation":"This misses the factor of \$a\$ multiplying the denominator. Multiplying through gives \$a(a^2+4)=a^3+4a\$, leading to \$a^3+4a-2=0\$."},{"text":"\$a^3+2a-4=0\$","correct":false,"explanation":"The coefficients have been rearranged incorrectly. Expanding \$a(a^2+4)\$ gives \$a^3+4a\$, and subtracting 2 yields \$a^3+4a-2=0\$."}]},{"id":"2a5cf5bb-6814-953e-15d7-51f5313c261f","cardType":"cloze","points":null,"front":"A sign change means one output is [ ] and the other is [ ].","back":"A sign change means one output is negative and the other is positive.","mcqOptions":[{"text":"A sign change means one output is negative and the other is positive.","correct":true,"explanation":"A sign change occurs when two function outputs have opposite signs: one is negative and the other is positive. This supports the existence of a root between them for a continuous function."},{"text":"A sign change means one output is positive and the other is positive.","correct":false,"explanation":"Two positive outputs have the same sign, so no sign change has occurred. A sign change requires one negative output and one positive output."},{"text":"A sign change means one output is negative and the other is negative.","correct":false,"explanation":"Two negative outputs also have the same sign. The required condition is that one output is negative and the other is positive."},{"text":"A sign change means one output is zero and the other is positive.","correct":false,"explanation":"An output of zero means a root has already been found, rather than showing a sign change across an interval. A sign change requires opposite non-zero signs: negative and positive."}]},{"id":"fd5897f9-7f2c-8898-7665-bbfd2732025c","cardType":"basic","points":3,"front":"Show that \$x^3+5x-2=0\$ has a solution between \$0\$ and \$1\$.","back":"

\$f(0)=-2\$
\$f(1)=4\$
Opposite signs and continuous, so a root lies between.

","mcqOptions":[{"text":"

\$f(0)=-2\$
\$f(1)=4\$
Opposite signs and continuous, so a root lies between.

","correct":true,"explanation":"Evaluating gives \$f(0)=-2\$ and \$f(1)=4\$, which have opposite signs. Since the polynomial is continuous, the Intermediate Value Theorem guarantees a root between 0 and 1."},{"text":"

\$f(0)=2\$
\$f(1)=4\$
Same signs and continuous, so a root lies between.

","correct":false,"explanation":"The constant term is \$-2\$, so \$f(0)=-2\$, not 2. The correct endpoint values are \$-2\$ and 4, giving opposite signs and therefore a root between them."},{"text":"

\$f(0)=-2\$
\$f(1)=2\$
Opposite signs and continuous, so a root lies between.

","correct":false,"explanation":"At 1, the value is \$1^3+5(1)-2=4\$, not 2. Although the sign conclusion is right, the endpoint calculation must be \$f(1)=4\$."},{"text":"

\$f(0)=-2\$
\$f(1)=4\$
Opposite signs but not continuous, so no conclusion follows.

","correct":false,"explanation":"A polynomial is continuous everywhere, so the opposite endpoint signs do guarantee a root in the interval. The mistake is wrongly denying continuity."}]},{"id":"277277a2-ad60-b07b-c82d-7d62060bf6a3","cardType":"basic","points":2,"front":"Rearrange \$x^3+5x=2\$ into the form \$x=\\cdots\$.","back":"\$x=\\frac{2}{5}-\\frac{x^3}{5}\$","mcqOptions":[{"text":"\$x=\\frac{2}{5}-\\frac{x^3}{5}\$","correct":true,"explanation":"Subtracting \$x^3\$ gives \$5x=2-x^3\$. Dividing every term by 5 produces \$x=\\frac{2}{5}-\\frac{x^3}{5}\$."},{"text":"\$x=\\frac{2}{5}+\\frac{x^3}{5}\$","correct":false,"explanation":"The cubic term must be subtracted when moved from the left side to the right. Thus \$5x=2-x^3\$, giving \$x=\\frac{2}{5}-\\frac{x^3}{5}\$."},{"text":"\$x=2-5x^3\$","correct":false,"explanation":"This reverses the effects of rearranging and dividing by 5. Dividing \$5x=2-x^3\$ by 5 gives \$x=\\frac{2}{5}-\\frac{x^3}{5}\$."},{"text":"\$x=\\frac{2-x^3}{5x}\$","correct":false,"explanation":"Dividing by \$5x\$ leaves the variable on the denominator and does not isolate \$x\$. Dividing the equation by 5 gives the required expression directly."}]},{"id":"a3bcee8c-b473-1b46-edc4-37b6593c2263","cardType":"basic","points":1,"front":"Turn \$x=\\frac{2}{5}-\\frac{x^3}{5}\$ into an iteration formula.","back":"\$x_{n+1}=\\frac{2}{5}-\\frac{x_n^3}{5}\$","mcqOptions":[{"text":"\$x_{n+1}=\\frac{2}{5}-\\frac{x_n^3}{5}\$","correct":true,"explanation":"An iteration formula replaces the current value \$x\$ on the right-hand side with \$x_n\$, while defining the next value as \$x_{n+1}\$."},{"text":"\$x_{n+1}=\\frac{2}{5}-\\frac{x_{n+1}^3}{5}\$","correct":false,"explanation":"Using \$x_{n+1}\$ on both sides makes the formula implicit rather than an iteration from a known value. The right-hand side must use the previous iterate \$x_n\$."},{"text":"\$x_n=\\frac{2}{5}-\\frac{x_{n+1}^3}{5}\$","correct":false,"explanation":"This reverses the roles of the current and next terms. To calculate a new iterate from a known one, use \$x_{n+1}\$ on the left and \$x_n\$ on the right."},{"text":"\$x_{n+1}=\\frac{2}{5}-\\frac{x_n^2}{5}\$","correct":false,"explanation":"The original rearranged equation contains a cube, so changing it to a square changes the function being iterated. The power must remain \$x_n^3\$."}]},{"id":"a8be9b41-dc81-79ba-c67d-7ed23264b8bf","cardType":"basic","points":2,"front":"Using \$x_{n+1}=\\frac{2}{5}-\\frac{x_n^3}{5}\$ with \$x_0=0\$, what is \$x_2\$?","back":"\$0.3872\$","mcqOptions":[{"text":"\$0.3872\$","correct":true,"explanation":"Starting with \$x_0=0\$, first find \$x_1=0.4\$. Then \$x_2=0.4-\\frac{0.4^3}{5}=0.4-0.0128=0.3872\$."},{"text":"\$0.3968\$","correct":false,"explanation":"This subtracts \$0.0032\$, which comes from mishandling the division by 5 after cubing. Since \$0.4^3/5=0.0128\$, the correct result is \$0.3872\$."},{"text":"\$0.3360\$","correct":false,"explanation":"This is obtained by subtracting \$0.064\$ rather than \$0.0128\$, forgetting to divide the cube by 5. The correct calculation gives \$x_2=0.3872\$."},{"text":"\$0.4000\$","correct":false,"explanation":"This uses the first iterate and stops before applying the formula a second time. Since \$x_1=0.4\$, substituting it gives \$x_2=0.3872\$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Iteration","title":"Iteration Flashcards","topbar":{"backHref":"/gcse/maths/ocr/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"17e0e276-380c-4c8f-86de-602032b75bce","walled":"$L39"}],"$L3a","$L3b"]