{"id":92910,"date":"2025-06-01T11:36:23","date_gmt":"2025-06-01T11:36:23","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=92910"},"modified":"2025-06-01T11:36:23","modified_gmt":"2025-06-01T11:36:23","slug":"convert-fa-b-c-a-barb-barb-c-into-canonical-pr","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/","title":{"rendered":"Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr"},"content":{"rendered":"

Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Product of Sum form.<\/p>\n

[amp_mcq option1=”(A + B + C) (A + B + $\\bar{C}$) (A + $\\bar{B}$ + C)” option2=”(A + B + C) ($\\bar{A}$ + B + $\\bar{C}$) ($\\bar{A}$ + $\\bar{B}$ + $\\bar{C}$)” option3=”(A + $\\bar{B}$ + C) (A + $\\bar{B}$ + $\\bar{C}$) ($\\bar{A}$ + B + $\\bar{C}$)” option4=”(A + B + $\\bar{C}$) ($\\bar{A}$ + $\\bar{B}$ + $\\bar{C}$) (A + $\\bar{B}$ + C)” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CISF-AC-EXE – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0.
\nThe given function is $F(A, B, C) = (A + \\bar{B}) (\\bar{B} + C)$.
\nWe can find the minterms where F is 0 by finding where $(A + \\bar{B})=0$ OR $(\\bar{B} + C)=0$.
\n$(A + \\bar{B}) = 0$ if and only if $A=0$ AND $\\bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$).
\n$(\\bar{B} + C) = 0$ if and only if $\\bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$).
\nSo, F=0 for the union of these minterms: $\\{m_2, m_3\\} \\cup \\{m_2, m_6\\} = \\{m_2, m_3, m_6\\}$.
\nThe canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$.
\nThe maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term.
\n$M_2$ from 010: $(A + \\bar{B} + C)$
\n$M_3$ from 011: $(A + \\bar{B} + \\bar{C})$
\n$M_6$ from 110: $(\\bar{A} + \\bar{B} + C)$
\nThe canonical POS form is $(A + \\bar{B} + C)(A + \\bar{B} + \\bar{C})(\\bar{A} + \\bar{B} + C)$.
\nComparing this derived form to the options, Option C is $(A + \\bar{B} + C) (A + \\bar{B} + \\bar{C}) (\\bar{A} + B + \\bar{C})$.
\nOption C has the first two terms correct (M2 and M3). However, the third term in Option C is $(\\bar{A} + B + \\bar{C})$, which is $M_5$ (from 101). The correct third term should be $(\\bar{A} + \\bar{B} + C)$, which is $M_6$ (from 110).
\nThere appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer.
\n<\/section>\n
\n– Canonical POS is a product of maxterms.
\n– Maxterms correspond to the minterms where the function is 0.
\n– For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.
\n<\/section>\n
\nThe original function simplifies to $\\bar{B} + AC$. Its minterms are $\\sum m(0, 1, 4, 5, 7)$, and maxterms are $\\prod M(2, 3, 6)$. This confirms the derived maxterms.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Product of Sum form. [amp_mcq option1=”(A + B + C) (A + B + $\\bar{C}$) (A + $\\bar{B}$ + C)” option2=”(A + B + C) ($\\bar{A}$ + B + $\\bar{C}$) ($\\bar{A}$ + $\\bar{B}$ + $\\bar{C}$)” option3=”(A + $\\bar{B}$ + C) (A + … <\/p>\n

Detailed SolutionConvert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1089],"tags":[1288,1113],"class_list":["post-92910","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1288","tag-information-and-communication-technology","no-featured-image-padding"],"yoast_head":"\nConvert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr<\/title>\n<meta name=\"description\" content=\"The canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0. The given function is $F(A, B, C) = (A + bar{B}) (bar{B} + C)$. We can find the minterms where F is 0 by finding where $(A + bar{B})=0$ OR $(bar{B} + C)=0$. $(A + bar{B}) = 0$ if and only if $A=0$ AND $bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$). $(bar{B} + C) = 0$ if and only if $bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$). So, F=0 for the union of these minterms: ${m_2, m_3} cup {m_2, m_6} = {m_2, m_3, m_6}$. The canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$. The maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term. $M_2$ from 010: $(A + bar{B} + C)$ $M_3$ from 011: $(A + bar{B} + bar{C})$ $M_6$ from 110: $(bar{A} + bar{B} + C)$ The canonical POS form is $(A + bar{B} + C)(A + bar{B} + bar{C})(bar{A} + bar{B} + C)$. Comparing this derived form to the options, Option C is $(A + bar{B} + C) (A + bar{B} + bar{C}) (bar{A} + B + bar{C})$. Option C has the first two terms correct (M2 and M3). However, the third term in Option C is $(bar{A} + B + bar{C})$, which is $M_5$ (from 101). The correct third term should be $(bar{A} + bar{B} + C)$, which is $M_6$ (from 110). There appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer. - Canonical POS is a product of maxterms. - Maxterms correspond to the minterms where the function is 0. - For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr\" \/>\n<meta property=\"og:description\" content=\"The canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0. The given function is $F(A, B, C) = (A + bar{B}) (bar{B} + C)$. We can find the minterms where F is 0 by finding where $(A + bar{B})=0$ OR $(bar{B} + C)=0$. $(A + bar{B}) = 0$ if and only if $A=0$ AND $bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$). $(bar{B} + C) = 0$ if and only if $bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$). So, F=0 for the union of these minterms: ${m_2, m_3} cup {m_2, m_6} = {m_2, m_3, m_6}$. The canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$. The maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term. $M_2$ from 010: $(A + bar{B} + C)$ $M_3$ from 011: $(A + bar{B} + bar{C})$ $M_6$ from 110: $(bar{A} + bar{B} + C)$ The canonical POS form is $(A + bar{B} + C)(A + bar{B} + bar{C})(bar{A} + bar{B} + C)$. Comparing this derived form to the options, Option C is $(A + bar{B} + C) (A + bar{B} + bar{C}) (bar{A} + B + bar{C})$. Option C has the first two terms correct (M2 and M3). However, the third term in Option C is $(bar{A} + B + bar{C})$, which is $M_5$ (from 101). The correct third term should be $(bar{A} + bar{B} + C)$, which is $M_6$ (from 110). There appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer. - Canonical POS is a product of maxterms. - Maxterms correspond to the minterms where the function is 0. - For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:36:23+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"3 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr","description":"The canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0. The given function is $F(A, B, C) = (A + bar{B}) (bar{B} + C)$. We can find the minterms where F is 0 by finding where $(A + bar{B})=0$ OR $(bar{B} + C)=0$. $(A + bar{B}) = 0$ if and only if $A=0$ AND $bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$). $(bar{B} + C) = 0$ if and only if $bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$). So, F=0 for the union of these minterms: ${m_2, m_3} cup {m_2, m_6} = {m_2, m_3, m_6}$. The canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$. The maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term. $M_2$ from 010: $(A + bar{B} + C)$ $M_3$ from 011: $(A + bar{B} + bar{C})$ $M_6$ from 110: $(bar{A} + bar{B} + C)$ The canonical POS form is $(A + bar{B} + C)(A + bar{B} + bar{C})(bar{A} + bar{B} + C)$. Comparing this derived form to the options, Option C is $(A + bar{B} + C) (A + bar{B} + bar{C}) (bar{A} + B + bar{C})$. Option C has the first two terms correct (M2 and M3). However, the third term in Option C is $(bar{A} + B + bar{C})$, which is $M_5$ (from 101). The correct third term should be $(bar{A} + bar{B} + C)$, which is $M_6$ (from 110). There appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer. - Canonical POS is a product of maxterms. - Maxterms correspond to the minterms where the function is 0. - For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/","og_locale":"en_US","og_type":"article","og_title":"Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr","og_description":"The canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0. The given function is $F(A, B, C) = (A + bar{B}) (bar{B} + C)$. We can find the minterms where F is 0 by finding where $(A + bar{B})=0$ OR $(bar{B} + C)=0$. $(A + bar{B}) = 0$ if and only if $A=0$ AND $bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$). $(bar{B} + C) = 0$ if and only if $bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$). So, F=0 for the union of these minterms: ${m_2, m_3} cup {m_2, m_6} = {m_2, m_3, m_6}$. The canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$. The maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term. $M_2$ from 010: $(A + bar{B} + C)$ $M_3$ from 011: $(A + bar{B} + bar{C})$ $M_6$ from 110: $(bar{A} + bar{B} + C)$ The canonical POS form is $(A + bar{B} + C)(A + bar{B} + bar{C})(bar{A} + bar{B} + C)$. Comparing this derived form to the options, Option C is $(A + bar{B} + C) (A + bar{B} + bar{C}) (bar{A} + B + bar{C})$. Option C has the first two terms correct (M2 and M3). However, the third term in Option C is $(bar{A} + B + bar{C})$, which is $M_5$ (from 101). The correct third term should be $(bar{A} + bar{B} + C)$, which is $M_6$ (from 110). There appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer. - Canonical POS is a product of maxterms. - Maxterms correspond to the minterms where the function is 0. - For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:36:23+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"3 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/","url":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/","name":"Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:36:23+00:00","dateModified":"2025-06-01T11:36:23+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0. The given function is $F(A, B, C) = (A + \\bar{B}) (\\bar{B} + C)$. We can find the minterms where F is 0 by finding where $(A + \\bar{B})=0$ OR $(\\bar{B} + C)=0$. $(A + \\bar{B}) = 0$ if and only if $A=0$ AND $\\bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$). $(\\bar{B} + C) = 0$ if and only if $\\bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$). So, F=0 for the union of these minterms: $\\{m_2, m_3\\} \\cup \\{m_2, m_6\\} = \\{m_2, m_3, m_6\\}$. The canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$. The maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term. $M_2$ from 010: $(A + \\bar{B} + C)$ $M_3$ from 011: $(A + \\bar{B} + \\bar{C})$ $M_6$ from 110: $(\\bar{A} + \\bar{B} + C)$ The canonical POS form is $(A + \\bar{B} + C)(A + \\bar{B} + \\bar{C})(\\bar{A} + \\bar{B} + C)$. Comparing this derived form to the options, Option C is $(A + \\bar{B} + C) (A + \\bar{B} + \\bar{C}) (\\bar{A} + B + \\bar{C})$. Option C has the first two terms correct (M2 and M3). However, the third term in Option C is $(\\bar{A} + B + \\bar{C})$, which is $M_5$ (from 101). The correct third term should be $(\\bar{A} + \\bar{B} + C)$, which is $M_6$ (from 110). There appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer. - Canonical POS is a product of maxterms. - Maxterms correspond to the minterms where the function is 0. - For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/convert-fa-b-c-a-barb-barb-c-into-canonical-pr\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CISF-AC-EXE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cisf-ac-exe\/"},{"@type":"ListItem","position":3,"name":"Convert F(A, B, C) = (A + $\\bar{B}$) ($\\bar{B}$ + C) into canonical Pr"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92910","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=92910"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92910\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=92910"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=92910"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=92910"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}