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Domain: set of allowed input values
Range: set of possible output values

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Domain: set of allowed input values
Range: set of possible output values

","correct":true,"explanation":"The domain contains every permitted input to the function, while the range contains every output that the function can produce from those inputs."},{"text":"

Domain: set of possible output values
Range: set of allowed input values

","correct":false,"explanation":"This reverses the two definitions. The domain is the set of allowed input values, while the range is the set of possible output values."},{"text":"

Domain: set of excluded input values
Range: set of impossible output values

","correct":false,"explanation":"Domain and range describe values that are allowed or possible, not values that are excluded or impossible. They refer to permitted inputs and resulting outputs respectively."},{"text":"

Domain: set of possible input values
Range: set of possible input values

","correct":false,"explanation":"This treats both sets as inputs. The domain contains allowed input values, but the range contains the possible output values produced by the function."}]},{"id":"9f1a11cd-d764-4d94-9a04-b51e25c0bb96","cardType":"cloze","points":null,"front":"To find excluded values for a fraction, set the denominator to $\\text{[...]}$.","back":"To find excluded values for a fraction, set the denominator to $\\boxed{\\neq 0}$.","mcqOptions":[{"text":"$$\\neq 0$","correct":true,"explanation":"A fraction is defined only when its denominator is non-zero. Values making the denominator zero are therefore excluded from the function’s domain."},{"text":"$$=0$","correct":false,"explanation":"Setting the denominator equal to zero identifies the values that must be excluded, but the required condition for an allowed denominator is $\\neq 0$."},{"text":"$$>0$","correct":false,"explanation":"A denominator may be positive or negative; it only cannot equal zero. The correct condition for permitted values is therefore $\\neq 0$."},{"text":"$$\\neq 1$","correct":false,"explanation":"There is no general restriction preventing a denominator from equalling one. Division fails only when the denominator is zero, so the condition is $\\neq 0$."}]},{"id":"3a3a3a99-29e1-4831-91fa-08caad7f1a89","cardType":"basic","points":1,"front":"What mathematical condition must be met for values to be included in the domain of $f(x) = \\sqrt{x}$?","back":"The expression inside the square root must be non-negative ($x \\ge 0$).","mcqOptions":[{"text":"The expression inside the square root must be non-negative ($x \\ge 0$).","correct":true,"explanation":"For a real-valued square root, the expression under the root cannot be negative. Therefore $\\sqrt{x}$ is defined for real $x$ only when $x \\ge 0$."},{"text":"The expression inside the square root must be positive ($x > 0$).","correct":false,"explanation":"This unnecessarily excludes zero. Since $\\sqrt{0}=0$ is defined, the expression may equal zero, so the correct condition is $x \\ge 0$."},{"text":"The expression inside the square root must be non-positive ($x \\le 0$).","correct":false,"explanation":"This reverses the inequality and includes negative values, whose square roots are not real. The expression must instead be non-negative, giving $x \\ge 0$."},{"text":"The expression inside the square root must be negative ($x < 0$).","correct":false,"explanation":"Negative inputs do not have real square roots, so this condition describes values outside the real domain. The required condition is $x \\ge 0$."}]},{"id":"618aa3c3-6270-41d2-ac34-029c17d79f30","cardType":"basic","points":1,"front":"What value must be excluded from the domain of $p(x) = \\frac{6}{x-3}$?","back":"$$x = 3$ (because the denominator cannot be zero).","mcqOptions":[{"text":"$$x = 3$ (because the denominator cannot be zero).","correct":true,"explanation":"Substituting $x=3$ makes the denominator $x-3$ equal to zero. Division by zero is undefined, so this value must be excluded."},{"text":"$$x = -3$ (because the denominator cannot be zero).","correct":false,"explanation":"This sign error comes from solving $x-3=0$ incorrectly. Adding 3 to both sides gives $x=3$, which is the excluded value."},{"text":"$$x = 0$ (because the denominator cannot be zero).","correct":false,"explanation":"Substituting zero gives a denominator of $-3$, so the fraction remains defined. The denominator becomes zero when $x=3$, which must be excluded."},{"text":"$$x = 6$ (because the denominator cannot be zero).","correct":false,"explanation":"The numerator does not determine the excluded input. At $x=6$ the denominator is 3; it becomes zero only at $x=3$."}]},{"id":"7e8ebcc9-ad8a-4891-98c6-b0472ce324a1","cardType":"cloze","points":null,"front":"The graph of an inverse function is a reflection of the original function in the line $\\text{[...]}$.","back":"The graph of an inverse function is a reflection of the original function in the line $\\boxed{y=x}$.","mcqOptions":[{"text":"$$y=x$","correct":true,"explanation":"Finding an inverse swaps every point’s coordinates from $(x,y)$ to $(y,x)$. Geometrically, this is precisely a reflection in the line $y=x$."},{"text":"$$y=-x$","correct":false,"explanation":"Reflection in $y=-x$ does not simply swap the input and output coordinates. An inverse graph is obtained by reflection in $y=x$."},{"text":"$$x=0$","correct":false,"explanation":"Reflection in $x=0$, the $y$-axis, changes the sign of each input rather than swapping inputs and outputs. Inverse graphs reflect in $y=x$."},{"text":"$$y=0$","correct":false,"explanation":"Reflection in $y=0$, the $x$-axis, changes the sign of each output rather than interchanging coordinates. The correct mirror line is $y=x$."}]},{"id":"9458b711-2494-4add-97e7-1f96737c1901","cardType":"basic","points":2,"front":"How do the domain and range of $f$ relate to the domain and range of $f^{-1}$?","back":"

The domain of $f$ becomes the range of $f^{-1}$.
The range of $f$ becomes the domain of $f^{-1}$.

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

The domain of $f$ becomes the range of $f^{-1}$.
The range of $f$ becomes the domain of $f^{-1}$.

","correct":true,"explanation":"An inverse function reverses the original mapping, so the original inputs become inverse outputs and the original outputs become inverse inputs."},{"text":"

The domain of $f$ becomes the domain of $f^{-1}$.
The range of $f$ becomes the range of $f^{-1}$.

","correct":false,"explanation":"This incorrectly leaves the sets unchanged. Because an inverse swaps inputs and outputs, the domain and range of the original function exchange roles."},{"text":"

The domain of $f$ becomes the domain of $f^{-1}$.
The range of $f$ becomes the domain of $f^{-1}$.

","correct":false,"explanation":"This assigns both original sets to the inverse domain. Instead, the range of $f$ becomes the inverse domain, while the domain of $f$ becomes the inverse range."},{"text":"

The domain of $f$ becomes the range of $f^{-1}$.
The range of $f$ becomes the range of $f^{-1}$.

","correct":false,"explanation":"This assigns both original sets to the inverse range. Only the domain of $f$ becomes the inverse range; the range of $f$ becomes the inverse domain."}]},{"id":"4a6a5459-028d-4e49-8eda-24f7c770e2a9","cardType":"basic","points":1,"front":"What is the common mistake regarding the notation $f^{-1}(x)$?","back":"Assuming $f^{-1}(x) = \\frac{1}{f(x)}$; it is the inverse function, not the reciprocal.","mcqOptions":[{"text":"Assuming $f^{-1}(x) = \\frac{1}{f(x)}$; it is the inverse function, not the reciprocal.","correct":true,"explanation":"The superscript $-1$ in function notation means the function that reverses $f$. A reciprocal is written $\\frac{1}{f(x)}$ and is generally different."},{"text":"Assuming $f^{-1}(x) = -f(x)$; it is the inverse function, not the negative.","correct":false,"explanation":"Although the superscript may be mistaken for a negative operation, the classic error is interpreting it as a reciprocal. It denotes the inverse function, not $\\frac{1}{f(x)}$."},{"text":"Assuming $f^{-1}(x) = f(-x)$; it is the inverse function, not the reflection.","correct":false,"explanation":"Replacing $x$ by $-x$ reflects a graph in the $y$-axis and does not find its inverse. The common misconception is treating $f^{-1}(x)$ as $\\frac{1}{f(x)}$."},{"text":"Assuming $f^{-1}(x) = f(x-1)$; it is the inverse function, not the translation.","correct":false,"explanation":"Subtracting one from the input translates the graph and is unrelated to inverse notation. The common mistake is confusing $f^{-1}(x)$ with the reciprocal $\\frac{1}{f(x)}$."}]},{"id":"01030d79-7033-41cc-8597-afba9944f689","cardType":"basic","points":2,"front":"What are the algebraic steps to find an inverse function?","back":"

Set $y = f(x)$
Swap $x$ and $y$
Rearrange to make $y$ the subject
Replace $y$ with $f^{-1}(x)$

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

Set $y = f(x)$
Swap $x$ and $y$
Rearrange to make $y$ the subject
Replace $y$ with $f^{-1}(x)$

","correct":true,"explanation":"An inverse reverses inputs and outputs, so $x$ and $y$ are swapped. Solving for the new output and renaming it gives the inverse function."},{"text":"

Set $y = f(x)$
Swap $x$ and $y$
Rearrange to make $x$ the subject
Replace $x$ with $f^{-1}(x)$

","correct":false,"explanation":"After swapping the variables, the new output is $y$, so $y$ must be isolated. It is then replaced with $f^{-1}(x)$, not $x$."},{"text":"

Set $y = f(x)$
Replace $y$ with $\\frac{1}{y}$
Rearrange to make $y$ the subject
Replace $y$ with $f^{-1}(x)$

","correct":false,"explanation":"Taking a reciprocal does not reverse a function’s input-output mapping. The correct process swaps $x$ and $y$, then rearranges to make $y$ the subject."},{"text":"

Set $y = f(x)$
Swap $x$ and $y$
Rearrange to make $x$ the subject
Replace $y$ with $f^{-1}(x)$

","correct":false,"explanation":"This isolates the wrong variable after the swap. To express the inverse as a function of $x$, rearrange for $y$ and then write $f^{-1}(x)$."}]},{"id":"68e4064a-3458-45d1-bcce-d515d7150d30","cardType":"basic","points":1,"front":"Define a one-to-one function.","back":"A function where each output comes from exactly one input.","mcqOptions":[{"text":"A function where each output comes from exactly one input.","correct":true,"explanation":"This is one-to-one because no two different inputs produce the same output. Consequently, each output in the range can be traced back to exactly one input."},{"text":"A function where each input maps to exactly one output.","correct":false,"explanation":"This states the general definition of a function, since every function assigns one output to each input. One-to-one additionally requires each output to come from exactly one input."},{"text":"A function where every possible output has at least one input.","correct":false,"explanation":"This describes an onto function, where every value in the codomain is reached. A one-to-one function instead requires each output in its range to come from exactly one input."},{"text":"A function where each output can come from several different inputs.","correct":false,"explanation":"This describes a many-to-one function, which is the opposite of one-to-one. In a one-to-one function, different inputs must always produce different outputs."}]},{"id":"22aad180-c338-46bd-920e-485e4b0f27ec","cardType":"cloze","points":null,"front":"A function must be [ ] on its domain for its inverse to be a function.","back":"A function must be one-to-one on its domain for its inverse to be a function.","mcqOptions":[{"text":"one-to-one","correct":true,"explanation":"A one-to-one function gives different outputs for different inputs, so reversing the mapping assigns exactly one output to every input of the inverse."},{"text":"many-to-one","correct":false,"explanation":"A many-to-one function sends different inputs to the same output, so its inverse would send one input to several outputs. The original function must be one-to-one."},{"text":"onto","correct":false,"explanation":"Being onto concerns whether every codomain value is reached, not whether outputs are repeated. The condition needed for the inverse to be a function is one-to-one."},{"text":"continuous","correct":false,"explanation":"Continuity does not prevent two different inputs from having the same output. An inverse is a function only when the original function is one-to-one on its domain."}]},{"id":"64a4abfe-d240-443b-8b63-98128d8b412c","cardType":"basic","points":1,"front":"Why do quadratic functions usually need a restricted domain to have an inverse?","back":"Without restriction, they are many-to-one (different inputs can give the same output).","mcqOptions":[{"text":"Without restriction, they are many-to-one (different inputs can give the same output).","correct":true,"explanation":"A full quadratic graph is symmetric about its turning point, so pairs of different inputs can produce the same output. Restricting the domain removes one branch and makes it one-to-one."},{"text":"Without restriction, they are one-to-many (one input can give several different outputs).","correct":false,"explanation":"A quadratic is still a function, so each input gives only one output; it is not one-to-many. The problem is that different inputs can share an output, making it many-to-one."},{"text":"Without restriction, they are undefined at the turning point (the minimum or maximum has no output).","correct":false,"explanation":"A quadratic is defined at its turning point, which gives its minimum or maximum value. Restriction is needed because symmetry makes the unrestricted quadratic many-to-one."},{"text":"Without restriction, they are always decreasing (larger inputs always give smaller outputs).","correct":false,"explanation":"A quadratic generally decreases on one side of its turning point and increases on the other. This creates repeated outputs, so a restricted domain is needed to make it one-to-one."}]},{"id":"5613e042-9ff3-49e1-841b-2da71f3a02b4","cardType":"cloze","points":null,"front":"To find the inverse of a quadratic function, you should first [ ].","back":"To find the inverse of a quadratic function, you should first complete the square.","mcqOptions":[{"text":"complete the square","correct":true,"explanation":"Completing the square rewrites the quadratic in turning-point form. This makes it straightforward to swap variables, isolate the squared expression and select the correct square-root branch."},{"text":"differentiate the function","correct":false,"explanation":"Differentiation can locate the turning point, but it does not rearrange the function into a form suitable for inversion. You should first complete the square."},{"text":"factorise the expression","correct":false,"explanation":"Factorising is useful for finding roots but usually does not make the inverse easy to isolate. Completing the square exposes the squared term and turning point directly."},{"text":"expand the brackets","correct":false,"explanation":"Expanding produces standard quadratic form, which hides the structure needed when taking square roots. Completing the square is the useful first step for finding the inverse."}]},{"id":"321323c3-d74f-48dc-87f8-82d663546ccb","cardType":"basic","points":1,"front":"If $g(x) = x^2 + 10x + 7$ with domain $x \\ge -5$, what is the range of $g^{-1}(x)$?","back":"The range is $x \\ge -5$ (identical to the domain of the original function).","mcqOptions":[{"text":"The range is $x \\ge -5$ (identical to the domain of the original function).","correct":true,"explanation":"The inverse swaps input and output roles, so its range equals the original function’s domain. Since $g$ is restricted to $x \\ge -5$, so is the range of $g^{-1}$."},{"text":"The range is $x \\ge -18$ (identical to the range of the original function).","correct":false,"explanation":"Completing the square gives $g(x)=(x+5)^2-18$, so $x \\ge -18$ is the domain of the inverse, not its range. Its range is $x \\ge -5$."},{"text":"The range is $x \\le -5$ (the opposite inequality to the original domain).","correct":false,"explanation":"This chooses the branch opposite to the stated restriction. The inverse’s range must reproduce the original domain exactly, so the correct inequality is $x \\ge -5$."},{"text":"The range is $x \\ge 5$ (using the positive value of the turning-point input).","correct":false,"explanation":"This loses the negative sign in the turning-point input, which is $-5$, not $5$. The inverse’s range equals the original domain, giving $x \\ge -5$."}]},{"id":"72448acf-f91e-4452-9e8d-b895b0de133a","cardType":"basic","points":2,"front":"How do you decide between the positive or negative square root branch when finding a quadratic inverse?","back":"Check the restricted domain of the original function.","mcqOptions":[{"text":"Check the restricted domain of the original function.","correct":true,"explanation":"The restricted domain identifies which side of the quadratic’s turning point is retained. That determines whether the inverse uses the positive or negative square-root branch."},{"text":"Always choose the positive square root for the inverse function.","correct":false,"explanation":"The positive branch is not automatic; it is used only when it reproduces values in the original restricted domain. The domain determines which square-root sign is valid."},{"text":"Check the sign of the quadratic coefficient in the original function.","correct":false,"explanation":"The quadratic coefficient determines whether the graph opens upwards or downwards, but not which side is retained. The positive or negative branch is chosen from the restricted domain."},{"text":"Check the sign of the constant term in the original function.","correct":false,"explanation":"The constant term affects the graph’s vertical position but does not identify the permitted inverse branch. The branch must match the original function’s restricted domain."}]},{"id":"5dde4d96-9d5e-4e0f-88cf-10390d93a7f2","cardType":"basic","points":1,"front":"What does the composite function $fg(x)$ represent?","back":"$$f(g(x))$; applying $g$ first, then applying $f$ to the result.","mcqOptions":[{"text":"$$f(g(x))$; applying $g$ first, then applying $f$ to the result.","correct":true,"explanation":"Composition is evaluated from the inside out. The input first enters $g$, and the output produced by $g$ then becomes the input of $f$."},{"text":"$$g(f(x))$; applying $f$ first, then applying $g$ to the result.","correct":false,"explanation":"This reverses the order and represents $gf(x)$ rather than $fg(x)$. For $fg(x)$, apply $g$ first and then apply $f$ to its output."},{"text":"$$f(x)g(x)$; multiplying the outputs of $f$ and $g$ for the same input.","correct":false,"explanation":"This is the product of two function values, not a composite function. In $fg(x)$, the output of $g$ is substituted into $f$, giving $f(g(x))$."},{"text":"$$f(x)+g(x)$; adding the outputs of $f$ and $g$ for the same input.","correct":false,"explanation":"This is the sum of two function values rather than composition. The notation $fg(x)$ here means substitute $g(x)$ into $f$, producing $f(g(x))$."}]},{"id":"d2741fb6-a31b-4ab1-bcdb-208bb32160a3","cardType":"cloze","points":null,"front":"When calculating a composite function value like $fg(2)$, you work from [ ].","back":"When calculating a composite function value like $fg(2)$, you work from right to left.","mcqOptions":[{"text":"right to left","correct":true,"explanation":"In $fg(2)=f(g(2))$, the right-hand function $g$ acts on $2$ first. Its result is then used as the input to the left-hand function $f$."},{"text":"left to right","correct":false,"explanation":"This reverses the order of composition and would apply $f$ before $g$. For $fg(2)=f(g(2))$, evaluate $g(2)$ first, so work from right to left."},{"text":"outside to inside","correct":false,"explanation":"Starting with the outer function is impossible until its input has been found. Composite functions are evaluated inside first, which means working from right to left."},{"text":"the left-hand function first","correct":false,"explanation":"The left-hand function receives the result of the right-hand function, so it cannot be evaluated first. For $fg(2)$, calculate $g(2)$ before applying $f$."}]},{"id":"b0388e7b-8557-4bb0-a5e8-b3720fb66daf","cardType":"basic","points":2,"front":"Given $f(x) = \\frac{3}{x-2}$ and $g(x) = x+5$, find $fg(1)$.","back":"

$g(1) = 6$
$f(6) = \\frac{3}{6-2} = \\mathbf{\\frac{3}{4}}$

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

$g(1) = 6$
$f(6) = \\frac{3}{6-2} = \\mathbf{\\frac{3}{4}}$

","correct":true,"explanation":"Composition applies $g$ first, giving $g(1)=6$. Substituting this output into $f$ gives $f(6)=3/(6-2)=3/4$."},{"text":"

$f(1) = -3$
$g(-3) = -3+5 = \\mathbf{2}$

","correct":false,"explanation":"This reverses the order and calculates $gf(1)$ instead of $fg(1)$. Applying $g$ first gives 6, then $f(6)=3/4$."},{"text":"

$g(1) = 6$
$f(6) = \\frac{3}{6+2} = \\mathbf{\\frac{3}{8}}$

","correct":false,"explanation":"This changes the denominator from $x-2$ to $x+2$. Since $g(1)=6$, the correct substitution is $f(6)=3/(6-2)=3/4$."},{"text":"

$g(1) = 6$
$f(6) = \\frac{3}{6}-2 = \\mathbf{-\\frac{3}{2}}$

","correct":false,"explanation":"This treats the denominator brackets incorrectly, reading $3/(x-2)$ as $3/x-2$. The entire expression $6-2$ is the denominator, giving $3/4$."}]},{"id":"6ce13d6d-9de5-4596-9a07-4d069f2e2c1d","cardType":"basic","points":1,"front":"Is $fg(x)$ always equal to $gf(x)$?","back":"No, the order usually changes the result.","mcqOptions":[{"text":"No, the order usually changes the result.","correct":true,"explanation":"Function composition is not generally commutative: $fg(x)$ means $f(g(x))$, whereas $gf(x)$ means $g(f(x))$. Applying the functions in a different order usually changes the result."},{"text":"Yes, the order never changes the result.","correct":false,"explanation":"Function composition is not generally commutative. Since $fg(x)=f(g(x))$ and $gf(x)=g(f(x))$, changing the order usually changes the result."},{"text":"Yes, the same input guarantees the same result.","correct":false,"explanation":"Using the same starting input does not guarantee the same final output because the intermediate values can differ. The order of composition usually changes the result."},{"text":"Yes, the notation shows the functions are interchangeable.","correct":false,"explanation":"The notation indicates which function is applied first; it does not make the functions interchangeable. In general, $f(g(x))$ and $g(f(x))$ produce different results."}]},{"id":"e3b36dbc-58de-4e88-8855-3836f4f35f82","cardType":"cloze","points":null,"front":"In $fg(8)$, the input 8 goes into the function $\\text{[...]}$ first, then the output goes into $f$.","back":"In $fg(8)$, the input 8 goes into the function $\\boxed{g}$ first, then the output goes into $f$.","mcqOptions":[{"text":"$$g$","correct":true,"explanation":"In $fg(8)=f(g(8))$, the inside function is applied first. Therefore, 8 is first substituted into $g$, and its output is then substituted into $f$."},{"text":"$$f$","correct":false,"explanation":"Reading the function letters from left to right reverses the composition. Since $fg(8)$ means $f(g(8))$, the input must go into $g$ first."},{"text":"$$f^{-1}$","correct":false,"explanation":"An inverse function is not introduced merely by composing two functions. The notation $fg(8)$ means $f(g(8))$, so the input goes into $g$ first."},{"text":"$$g^{-1}$","correct":false,"explanation":"Composition does not require undoing $g$. The notation $fg(8)=f(g(8))$ requires applying $g$ itself first, followed by $f$."}]},{"id":"3c8618fe-7887-434f-95e9-b773825cf869","cardType":"cloze","points":null,"front":"In $fg(8)$, the input 8 goes into the function $g$ first, then the output goes into $\\text{[...]}$.","back":"In $fg(8)$, the input 8 goes into the function $g$ first, then the output goes into $\\boxed{f}$.","mcqOptions":[{"text":"$$f$","correct":true,"explanation":"The notation $fg(8)$ means $f(g(8))$. After 8 is substituted into $g$, the resulting output becomes the input of $f$."},{"text":"$$g$","correct":false,"explanation":"Applying $g$ again would calculate $g(g(8))$, not $fg(8)$. After $g$ acts on 8, its output must be substituted into $f$."},{"text":"$$f^{-1}$","correct":false,"explanation":"An inverse is not implied by function composition. In $fg(8)=f(g(8))$, the output from $g$ goes directly into $f$, not into $f^{-1}$."},{"text":"$$g^{-1}$","correct":false,"explanation":"This would attempt to undo the first function rather than complete the stated composition. After applying $g$, its output must be substituted into $f$."}]},{"id":"59168455-5c4e-4b00-b6fb-ae9535da2e12","cardType":"basic","points":2,"front":"What is the inverse of $f(x) = \\frac{4x-7}{3}$?","back":"$$f^{-1}(x) = \\frac{3x+7}{4}$","mcqOptions":[{"text":"$$f^{-1}(x) = \\frac{3x+7}{4}$","correct":true,"explanation":"Writing $y=(4x-7)/3$ and rearranging gives $3y=4x-7$, then $x=(3y+7)/4$. Swapping variables produces the stated inverse."},{"text":"$$f^{-1}(x) = \\frac{3x-7}{4}$","correct":false,"explanation":"The sign is not changed correctly when rearranging: adding 7 to both sides gives $4x=3y+7$, not $3y-7$. Therefore the inverse is $(3x+7)/4$."},{"text":"$$f^{-1}(x) = \\frac{4x+7}{3}$","correct":false,"explanation":"Changing only the sign of the constant does not reverse the function’s operations. Rearranging $y=(4x-7)/3$ correctly gives the inverse $(3x+7)/4$."},{"text":"$$f^{-1}(x) = \\frac{3x+7}{3}$","correct":false,"explanation":"This divides by the original outer denominator instead of the coefficient of the original input. Rearrangement gives $4x=3y+7$, so the inverse is $(3x+7)/4$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Inverse and Composite Functions","title":"Inverse and Composite Functions Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"248c6c46-b271-4dee-8928-ef8a04d6b980","walled":"$L39"}],"$L3a","$L3b"]