[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-270197-105":3,"detail-sidebar-cat-0-en-105":81,"doc-detail-270197-en":131},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":74,"head_meta":76,"extra_data":78,"updated_unix":80},105,"en","sample-questions-suggested-solutions","Sample Questions - Suggested Solutions","","Practice and worked solutions for four mathematics problems covering proof, algebraic equation solving, and number theory divisibility. The materials include an induction proof for the sum of cubes, a solution approach using substitution to determine roots of a radical equation, and a proof by cases showing that the product of integers not divisible by 3 remains not divisible by 3. The final problem derives geometric angles and the area of a shaded triangle using the law of cosines, law of sines, and trigonometric identities.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/sample-questions-suggested-solutions/270197/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/sample-questions-suggested-solutions/270197.png","ImageObject",300,407,{"name":42,"@type":43},"Asher","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-21","2026-09-15",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",15,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"How is the identity for 13 + 23 + ... + n3 proved?","Question",{"text":63,"@type":64},"It is proved by induction: verify the base case n=1, assume the statement true for n=k, then show it holds for k+1 by algebraic simplification after replacing terms using the induction hypothesis.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"What method is used to solve the equation involving 3 − (3 + √3) and radicals?",{"text":68,"@type":64},"The solution introduces a substitution y = √3 and rewrites the expression in terms of that substitution, then factors the resulting equation to obtain values for the unknown.",{"name":70,"@type":61,"acceptedAnswer":71},"How is the shaded triangle area (LMN) found from the given lengths and angle-bisector condition?",{"text":72,"@type":64},"The solution determines key angles using the law of cosines and sines, computes intermediate segment lengths with trigonometry (including angle-bisector relationships and inverse trig values), and finally applies the triangle area formula using two sides and the included angle.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},270197,1789455097,{"code":4,"msg":82,"data":83},"success",[84,88,92,95,100,105,110,115,120,123,127],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":93,"slug":94},70,"exam",{"id":96,"doc_module":4,"doc_module_name":25,"category_name":97,"show_sort_weight":98,"slug":99},5,"Comic",60,"comic",{"id":101,"doc_module":4,"doc_module_name":25,"category_name":102,"show_sort_weight":103,"slug":104},6,"Technology",50,"technology",{"id":106,"doc_module":4,"doc_module_name":25,"category_name":107,"show_sort_weight":108,"slug":109},7,"Healthcare",40,"healthcare",{"id":111,"doc_module":4,"doc_module_name":25,"category_name":112,"show_sort_weight":113,"slug":114},8,"Research & Report",30,"research-report",{"id":116,"doc_module":4,"doc_module_name":25,"category_name":117,"show_sort_weight":118,"slug":119},9,"Religion & Spirituality",20,"religion-spirituality",{"id":118,"doc_module":4,"doc_module_name":25,"category_name":121,"show_sort_weight":118,"slug":122},"World Cup","world-cup",{"id":124,"doc_module":4,"doc_module_name":25,"category_name":125,"show_sort_weight":124,"slug":126},10,"Lifestyle","lifestyle",{"id":128,"doc_module":4,"doc_module_name":25,"category_name":129,"show_sort_weight":96,"slug":130},19,"General","general",{"code":4,"msg":82,"data":132},{"doc_id":79,"user_id":133,"nickname":42,"user_avatar":134,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":101,"language":140,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":12,"update_tm":80,"read_time":55},687197207639,"https://ap-avatar.wpscdn.com/davatar_a8503ba1806abce46bf441b54a3ca4cd","Sample Questions  \n1. Prove that: 13 + 23 + 33 + ... + n3 = ¼n2(n + 1)2  \n2. Find the possible solutions ofthe given equation  \n􀝔 3 − (3 + √3)􀝔 = 0  \n3. Let m and n be 2 integers each one not divisible by 3. Use proof by cases to show that mn is not divisible by 3.  \n4. Given AB = 16 cm, BC = 10 cm & AC = 14 cm. RA = RC, AB ⊥ PC, AQ is angle bisector of ∠BAC calculate the area of the shaded triangle LMN. (Take tan cos⁻ ¹(5/√28) = ⅕√3); sin [90-cos⁻¹ ((7√129)/86)]= 7√129)/86)  \nSuggested solutions  \n1. Step1 Show it is true for n=1  \n13 = ¼ × 12 × 22 is True Step 2 Assume it is true for n=k  \n13 + 23 + 33 + ... + k3 = ¼k2(k + 1)2 is True (An assumption!) Step3 Now, prove it is true for \"k+1\"  \n13 + 23 + 33 + ... + (k + 1)3 = ¼(k + 1)2(k + 2)2 ?  \nWe know that 13 + 23 + 33 + ... + k3 = ¼k2(k + 1)2 (the assumption above), so we can do a replacement for all but the last term:  \n¼k2(k + 1)2 + (k + 1)3 = ¼(k + 1)2(k + 2)2 Multiply all terms by 4:  \nk2(k + 1)2 + 4(k + 1)3 = (k + 1)2(k + 2)2  \nAll terms have a common factor (k + 1)2, so it can be cancelled:  \nk2 + 4(k + 1) = (k + 2)2 And simplify:  \nk2 + 4k + 4 = k2 + 4k + 4 They are the same! So it is true.  \nSo: 13 + 23 + 33 + ... + (k + 1)3 = ¼(k + 1)2(k + 2)2 is True  \n2. 􀝔 3 − (3 + √3)􀝔 = 0 Let y = √3, 􀝕 2 = 3  \n􀝔 3 − (􀝕2 + y )􀝔 + 􀝕 2 = 0  \n􀝔 3 − (􀝕2􀝔 + yx )􀝔 + 􀝕 2 = 0  \n􀝔 3 − 􀝕 2􀝔 + yx + 􀝕2 =0 􀝔 (􀝔2 − 􀝕2) − 􀝕(􀝔 − 􀝕) = 0  \n􀝔 (􀝔 − 􀝕)(􀝔 + 􀝕) − 􀝕(􀝔 − 􀝕) = 0 (􀝔 − 􀝕)(􀝔(􀝔 + 􀝕) − 􀝕) = 0  \n(􀝔 − 􀝕)(􀝔2 + 􀝔􀝕 − 􀝕) = 0  \n(􀝔 − √3)(􀝔2 + 􀝔√3  − √3) =0 􀝔 − √3 = 0 , 􀝔 = √3  \n(􀝔2 + 􀝔√3 − √3) = 0  \n􀝑􀝏􀝅􀝊􀝃  \n􀝔 =  \n−􀜾 ± √􀜾2 − 4􀜽􀜿  \n2􀜽  \n􀝔 =  \n−√3 ± √(√3)2 − 4(−√3)  \n2  \n−√3 ± √(√3)2 − 4(−√3)  \n􀝔 = − ( 2 )  \n3. There are 3 cases; consider each case in turn.  \nIf m = 3a + 1 and n = 3b + 1, then mn = (3a + 1)(3b + 1) = 9ab + 3a  \n+ 3b + 1, which is not divisible by 3.  \nIf m = 3a + 1 and n = 3b + 2, then mn = (3a + 1)(3b + 2) = 9ab + 6a  \n+ 3b + 2, which is not divisible by 3.  \nIf m = 3a + 2 and n = 3b + 2, then (3a + 2)(3b + 2) = 9ab + 6a + 6b + 4, which is not divisible by 3.  \n4. From ΔABC  \nAC² = AB² + BC² -2×AB×BC×cos (∠ABC) {Law of Cosines}⇒ 14² = 16² + 10² -2×16×10 cos (∠ABC)  \n⇒ 196 = 256 + 100-320 cos (∠ABC)  \n⇒ 320 cos (∠ABC) = 160  \n⇒ cos (∠ABC) = ½  \n⇒ ∠ABC = 60°  \nFrom ΔBPC  \nsin (∠ABC) = PC/BC = PC/10  \n⇒ sin 60 = PC/10  \n⇒ ½√3 = PC/10  \n⇒ PC = 5√3 cm  \ncos (∠ABC) = PB/BC  \n⇒ cos 60 = PB/10  \n⇒ ½ = PB/10  \n⇒ PB = 5 cm  \nFrom ΔAPC  \nAP = AB-PB  \n⇒ AP = 16-5  \n⇒ AP = 11 cm  \nsin (∠PAC) = = PC/AC  \n⇒ sin (∠PAC) = 5√3/11  \ncos (∠BAC) = =AP/AC,⇒ cos (∠BAC) = 11/14  \nLet ∠QAC = ∠QAP = x {AQ is angle bisector of ∠PAC} cos (∠BAC) = cos 2x  \n⇒ cos (∠BAC) = 2 cos² x – 1 (double angle identity for cosine)⇒ 2 cos² x-1 = 11/14  \n⇒ cos² x = 25/28  \n⇒ cos x = ± 5/√28  \n0 \u003C x \u003C 180 {x is an angle inside the triangle}⇒ cos x = 5/√28  \n⇒ ∠QAP = cos⁻¹ (5/√28)  \nFrom ΔARB  \nBR² = AB² + AR² -2×AB×AR×cos (∠BAR)⇒ BR² = 16² + 7² -2×16×7×11/14  \n= 256 + 49-176 = 129  \n⇒ BR = √129 cm  \nAR² = AB² + BR² -2×AB×BR×cos (∠ABR)  \n⇒ 7² = 16² + 129-2×16×√129×cos (∠ABR)  \n⇒ 49 = 256 + 129-32√129 cos (∠ABR)  \n⇒ cos (∠ABR) = (7√129)/86 ⇒ ∠ABR = cos⁻¹ ((7√129)/86)  \nFrom ΔAPN  \ntan ∠QAC = tan cos⁻ ¹ (5/√28)  \n⇒ PN/PA = tan cos⁻ ¹ (5/√28)  \n⇒ PN/11 = tan cos⁻ ¹ (5/√28)= ⅕√3  \n⇒ PN = ⅕×11√3 cm ∠ANP = 90-∠QAP ⇒ ∠ANP = 90-cos⁻ ¹ (5/√28)  \n⇒ ∠LNM = 90-cos⁻ ¹ (5/√28)⇒ sin ∠LNM = sin [90-cos⁻ ¹ (5/√28)]= cos cos⁻ ¹ (5/√28)  \n⇒ sin ∠LNM = 5/√28  \nFrom ΔPBM  \ntan ∠PMB = PM/PB  \n⇒ PM/5 = tan (cos⁻¹ ((7√129)/86))  \n= 5√3/21  \n⇒ PM = 25√3/21 cm ∠PMB = 90-∠ABR  \n⇒ ∠PMB = 90-cos⁻ ¹ ((7√129)/86)  \n⇒ ∠LMN = 90-cos⁻ ¹ ((7√129)/86)⇒ sin ∠LMN = sin [90-cos⁻ ¹ ((7√129)/86)]= cos [cos⁻ ¹ ((7√129)/86)]  \n⇒ sin ∠LMN = 7√129)/86  \nMN = PN-PM  \n⇒ MN = ⅕×11√3-25√3/21 ⇒ MN = 106√3/105 cm  \nFrom ΔLMN  \n∠MLN = 180- (∠LMN + ∠LNM)  \n⇒ ∠MLN = 180- [90-cos⁻ ¹ ((7√129)/86) + 90-cos⁻ ¹ (5/√28)]⇒ ∠MLN = cos⁻ ¹ ((7√129)/86) + cos⁻ ¹ (5/√28)  \n⇒ sin ∠MLN = sin [cos⁻¹ ((7√129)/86) + cos⁻ ¹ (5/√28)]  \n= sin [cos⁻¹ ((7√129)/86)] cos cos⁻ ¹ (5/√28) (note: cos cos⁻¹(x)= ","cbCaidxNyV0OX7Ye","https://ap.wps.com/l/cbCaidxNyV0OX7Ye","pdf",372493,"English","# Sample Questions\n## Suggested solutions","[{\"question\":\"How is the identity for 13 + 23 + ... + n3 proved?\",\"answer\":\"It is proved by induction: verify the base case n=1, assume the statement true for n=k, then show it holds for k+1 by algebraic simplification after replacing terms using the induction hypothesis.\"},{\"question\":\"What method is used to solve the equation involving 3 − (3 + √3) and radicals?\",\"answer\":\"The solution introduces a substitution y = √3 and rewrites the expression in terms of that substitution, then factors the resulting equation to obtain values for the unknown.\"},{\"question\":\"How is the shaded triangle area (LMN) found from the given lengths and angle-bisector condition?\",\"answer\":\"The solution determines key angles using the law of cosines and sines, computes intermediate segment lengths with trigonometry (including angle-bisector relationships and inverse trig values), and finally applies the triangle area formula using two sides and the included angle.\"}]","Sample Questions - Suggested Solutions | PDF"]