12th Chemistry Unit 5 Coordination Chemistry Book Back Answers & Easy Concept Guide (Part 1)

Welcome to our comprehensive 12th Standard Chemistry study resource for the Tamil Nadu State Board (Samacheer Kalvi) curriculum. In this post, we provide Textbook Exercise Solutions and a beginner-friendly Concept Guide for Unit 5: Coordination Chemistry (Book Back Questions Part 1). Even if you are starting from scratch, these step-by-step bilingual explanations (in English and Tamil) will help you master IUPAC Nomenclature, Molar Conductivity, and Valence Bond (VB) Theory to score full marks in your board exams.

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Part 1: 4 Golden Rules for Beginners (கணக்குகளைப் போடும் முன் தெரிந்துகொள்ள வேண்டிய அடிப்படை விதிகள்)

In Coordination Chemistry, a coordination compound is typically written in the format \(\text{Na}_2[\text{Ni(EDTA)}]\). The part inside the square brackets \([\ ]\) is called the Coordination Sphere (அணைவுக்கோளம்), and the part outside is called the Ionisation Sphere or Counter Ion (அயனியாக்கும் கோளம்).

1. Rule 1 (Cation First, Anion Next): எப்போதும் பெயரிடும்போது நேர்மின் அயனியை (Cation, \(+\)) முதலில் எழுத வேண்டும். எதிர்மின் அயனியை (Anion, \(-\)) இரண்டாவதாக எழுத வேண்டும். (நினைவில் வைக்க: ஒரு சேர்மத்தின் இடது பக்கம் எப்போதும் \(+\), வலது பக்கம் எப்போதும் \(-\) ஆக இருக்கும்).
2. Rule 2 (Inside the Square Brackets \([\ ]\)): சதுர அடைப்புக்குறிக்குள் ஒரு Central Metal Atom (மைய உலோக அணு: e.g., \(\text{Ni}, \text{Ag}, \text{Co}, \text{Pt}, \text{Fe}, \text{Cr}\)) மற்றும் அதனுடன் இணைந்திருக்கும் Ligands (ஈனிகள்: e.g., \(\text{CN}^-, \text{NH}_3, \text{Cl}^-\)) இருக்கும். பெயரிடும்போது எப்போதும் Ligand-க்குத்தான் முதல் மரியாதை (முதலில் எழுத வேண்டும்), உலோகத்தை (Metal) அதன் பிறகுதான் எழுத வேண்டும்.
3. Rule 3 (Ending of the Central Metal Name):
   • சதுர அடைப்புக்குறி \([\ ]\) நேர்மின் சுமை (\(+\)) அல்லது மின்சுமையற்றதாக (\(0\)) இருந்தால், உலோகத்தின் பெயரை அப்படியே எழுதலாம் (e.g., \(\text{Co} \rightarrow \text{cobalt}\), \(\text{Pt} \rightarrow \text{platinum}\)).
   • சதுர அடைப்புக்குறி \([\ ]\) எதிர்மின் சுமை (\(-\)) கொண்டதாக இருந்தால், உலோகத்தின் பெயருடன் "-ate" சேர்த்து முடிக்க வேண்டும் (e.g., \(\text{Ni} \rightarrow \text{nickelate}\), \(\text{Ag} \rightarrow \text{argentate}\), \(\text{Fe} \rightarrow \text{ferrate}\), \(\text{Cr} \rightarrow \text{chromate}\)).
4. Rule 4 (Ligand Names and Charges for Oxidation State Calculation): Refer to the table below to memorise the charges and IUPAC names of common ligands.

Ligand Formula	Type of Ligand	IUPAC Name	Charge
\(\text{NH}_3\)	Neutral Ligand (மின்சுமையற்ற ஈனி)	ammine (note double 'm')	\(0\)
\(\text{CO}\)	Neutral Ligand (மின்சுமையற்ற ஈனி)	carbonyl	\(0\)
\(\text{en}\)	Neutral Bidentate Ligand	ethane-1,2-diamine	\(0\)
\(\text{Cl}^-\)	Anionic Ligand (எதிர்மின் ஈனி)	chlorido	\(-1\)
\(\text{F}^-\)	Anionic Ligand (எதிர்மின் ஈனி)	fluorido	\(-1\)
\(\text{OH}^-\)	Anionic Ligand (எதிர்மின் ஈனி)	hydroxido	\(-1\)
\(\text{CN}^-\)	Anionic Ligand (எதிர்மின் ஈனி)	cyanido	\(-1\)
\(\text{NO}_2^-\)	Ambidentate Anionic Ligand (N-donor)	\(\text{nitrito-}\kappa\text{N}\)	\(-1\)
\(\text{ONO}^-\)	Ambidentate Anionic Ligand (O-donor)	\(\text{nitrito-}\kappa\text{O}\)	\(-1\)
\(\text{SO}_4^{2-}\)	Anionic Counter Ion / Ligand	sulphate / sulphato	\(-2\)
\(\text{EDTA}^{4-}\)	Hexadentate Anionic Ligand	\(2,2',2'',2'''\text{-(ethane-1,2-diyldinitrilo)tetraacetato}\)	\(-4\)

Exam Tip: When multiple different ligands are present inside the coordination sphere, always arrange their names in alphabetical order (ignoring prefixes like di, tri, tetra). For complex ligands like \(\text{en}\) (ethane-1,2-diamine) that already contain a numerical prefix in their name, use bis (for 2), tris (for 3), and tetrakis (for 4).

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Part 2: Book Back Questions & Step-by-Step Solutions

Question 1: Write the IUPAC names for the following complexes.

(i) \(\text{Na}_2[\text{Ni(EDTA)}]\)

Textbook Answer: \(\text{Sodium } 2,2',2'',2'''\text{-(ethane-1,2-diyldinitrilo)tetraacetatonickelate(II)}\)

எளிய விளக்கம் (Step-by-Step Concept Explanation):

• Step 1 (Identify Ions): இடது பக்கம் உள்ள \(\text{Na}^+\) நேர்மின் அயனி (Cation). வலது பக்கம் உள்ள \([\text{Ni(EDTA)}]^{2-}\) எதிர்மின் அயனி (Anionic coordination sphere).
• Step 2 (Name Cation & Ligand): முதலில் நேர்மின் அயனியான Sodium-த்தை எழுதுகிறோம். அடுத்து அடைப்புக்குறிக்குள் உள்ள Ligand-ஆன EDTA-வின் IUPAC பெயரை எழுதுகிறோம்: \(2,2',2'',2'''\text{-(ethane-1,2-diyldinitrilo)tetraacetato}\).
• Step 3 (Name Central Metal): அடைப்புக்குறி எதிர்மின் சுமை (\(-\)) என்பதால், \(\text{Ni}\) (Nickel) என்பது nickelate என மாறும்.
• Step 4 (Oxidation State Calculation):
  • Given: Charge of \(\text{Na} = +1\), Charge of \(\text{EDTA} = -4\), Let oxidation state of \(\text{Ni} = x\).
  • Formula: Sum of oxidation states of all atoms/ions in a neutral compound \(= 0\).
  • Substitution & Calculation: \[2(+1) + x + (-4) = 0\] \[+2 + x - 4 = 0\] \[x - 2 = 0 \implies x = +2\]
  • Final Answer with Roman Numeral: Oxidation state of Nickel is \(\text{(II)}\).

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(ii) \([\text{Ag(CN)}_2]^-\)

Textbook Answer: \(\text{Dicyanidoargentate(I) ion}\)

எளிய விளக்கம் (Step-by-Step Concept Explanation):

• Step 1 (Identify Ion Type): இதில் வெளியே எந்த அயனியும் இல்லை, ஒரே ஒரு அடைப்புக்குறி மட்டும்தான் உள்ளது. மேலே மைனஸ் (\(-\)) குறி உள்ளதால் இது ஒரு எதிர்மின் அணைவு அயனி (Anionic complex ion).
• Step 2 (Name Ligand): முதலில் Ligand-ஐ எழுத வேண்டும். இங்கு இரண்டு \(\text{CN}^-\) ஈனிகள் உள்ளதால் Dicyanido என எழுதுகிறோம்.
• Step 3 (Name Central Metal): அடைப்புக்குறிக்கு மேலே மைனஸ் (\(-\)) உள்ளதால், வெள்ளி (\(\text{Ag}\) - Argentum) என்பது argentate என மாறும். மேலே மின்சுமை கொடுக்கப்பட்டுள்ளதால் கடைசியில் ion என முடிக்க வேண்டும்.
• Step 4 (Oxidation State Calculation):
  • Given: Charge of \(\text{CN}^- = -1\), Overall charge on complex \(= -1\), Let \(\text{Ag} = x\).
  • Substitution & Calculation: \[x + 2(-1) = -1\] \[x - 2 = -1 \implies x = 2 - 1 = +1\]
  • Final Answer with Roman Numeral: Oxidation state of Silver is \(\text{(I)}\).

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(iii) \([\text{Co(en)}_3]_2(\text{SO}_4)_3\)

Textbook Answer: \(\text{Tris(ethane-1,2-diamine)cobalt(III) sulphate}\)

எளிய விளக்கம் (Step-by-Step Concept Explanation):

• Step 1 (Identify Ions): இடது பக்கம் உள்ள \([\text{Co(en)}_3]^{3+}\) நேர்மின் பகுதி (\(+\)), வலது பக்கம் உள்ள \(\text{SO}_4^{2-}\) எதிர்மின் பகுதி (\(-\)).
• Step 2 (Name Ligand): முதலில் அடைப்புக்குறிக்குள் உள்ள Ligand-ஆன \(\text{en}\) (ethane-1,2-diamine)-ஐ எழுத வேண்டும். இது ஒரு Bidentate ligand. எனவே 3 முறை வரும்போது சாதாரண 'tri'-க்கு பதில் Tris எனப் பயன்படுத்த வேண்டும்.
• Step 3 (Name Central Metal & Anion): அடைப்புக்குறி நேர்மின் சுமை (\(+\)) என்பதால் \(\text{Co}\)-வை அப்படியே cobalt என எழுதலாம் (cobaltate என எழுதக்கூடாது). கடைசியில் வெளியே உள்ள sulphate-ஐ எழுதுகிறோம்.
• Step 4 (Oxidation State Calculation):
  • Given: Number of \(\text{Co}\) atoms \(= 2\), Charge of \(\text{en} = 0\), Charge of \(\text{SO}_4^{2-} = -2\).
  • Substitution & Calculation: \[2x + 2(3 \times 0) + 3(-2) = 0\] \[2x - 6 = 0 \implies 2x = 6 \implies x = +3\]
  • Final Answer with Roman Numeral: Oxidation state of Cobalt is \(\text{(III)}\).

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(iv) \([\text{Co(ONO)(NH}_3)_5]^{2+}\)

Textbook Answer: \(\text{Pentaamminenitrito-}\kappa\text{O-cobalt(III) ion}\)

எளிய விளக்கம் (Step-by-Step Concept Explanation):

• Step 1 (Alphabetical Order of Ligands): இங்கு \(\text{NH}_3\) (ammine) மற்றும் \(\text{ONO}^-\) (nitrito-\(\kappa\)O) என இரண்டு ஈனிகள் உள்ளன. இவற்றை ஆங்கில அகரவரிசைப்படி (A first, N next) எழுத வேண்டும்.
• Step 2 (Name Ligands): 5 \(\text{NH}_3\) உள்ளதால் Pentaammine. அடுத்து \(\text{ONO}^-\)-வில் ஆக்ஸிஜன் (\(\text{O}\)) தான் உலோகத்துடன் இணைகிறது (Ambidentate ligand), எனவே \(\text{nitrito-}\kappa\text{O}\) என எழுதுகிறோம்.
• Step 3 (Name Central Metal): மேலே \(2+\) நேர்மின் சுமை உள்ளதால் cobalt என எழுதி, கடைசியில் ion என முடிக்கிறோம்.
• Step 4 (Oxidation State Calculation):
  • Given: Charge of \(\text{ONO}^- = -1\), Charge of \(\text{NH}_3 = 0\), Overall charge \(= +2\).
  • Substitution & Calculation: \[x + (-1) + 5(0) = +2\] \[x - 1 = +2 \implies x = +3\]
  • Final Answer with Roman Numeral: Oxidation state of Cobalt is \(\text{(III)}\).

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(v) \([\text{Pt(NH}_3)_2\text{Cl(NO}_2)]\)

Textbook Answer: \(\text{Diamminechloridonitrito-}\kappa\text{N-platinum(II)}\)

எளிய விளக்கம் (Step-by-Step Concept Explanation):

• Step 1 (Alphabetical Order of 3 Ligands): இங்கு \(\text{NH}_3\) (ammine), \(\text{Cl}^-\) (chlorido), \(\text{NO}_2^-\) (nitrito-\(\kappa\)N) என மூன்று ஈனிகள் உள்ளன. அகரவரிசைப்படி \(\text{A} \rightarrow \text{C} \rightarrow \text{N}\) என்ற வரிசையில் எழுத வேண்டும்.
• Step 2 (Name Ligands & Metal): 2 \(\text{NH}_3\) உள்ளதால் Diammine, ஒரு \(\text{Cl}^-\) உள்ளதால் chlorido, ஒரு \(\text{NO}_2^-\) (நைட்ரஜன் வழியாக இணைவதால்) \(\text{nitrito-}\kappa\text{N}\). இந்தச் சேர்மத்திற்கு மேலே எந்த மின்சுமையும் இல்லை (Neutral complex). எனவே \(\text{Pt}\)-ஐ அப்படியே platinum என எழுதுகிறோம்.
• Step 3 (Oxidation State Calculation):
  • Given: Charge of \(\text{NH}_3 = 0\), Charge of \(\text{Cl}^- = -1\), Charge of \(\text{NO}_2^- = -1\).
  • Substitution & Calculation: \[x + 2(0) + (-1) + (-1) = 0\] \[x - 2 = 0 \implies x = +2\]
  • Final Answer with Roman Numeral: Oxidation state of Platinum is \(\text{(II)}\).

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Question 2: Write the formula for the following coordination compounds.

(முதல் கேள்வியில் பார்முலாவைக் கொடுத்துப் பெயர் கேட்டார்கள், இதில் பெயரைக் கொடுத்து பார்முலா கேட்கிறார்கள். இது இன்னும் எளிமையானது!)

(a) Potassium hexacyanoferrate(II)

• Formula: \(\text{K}_4[\text{Fe(CN)}_6]\)
• எளிய விளக்கம்: முதலில் Potassium (\(\text{K}\)) வெளியே இருக்கும். அடைப்புக்குறிக்குள் மைய உலோகம் ferrate (\(\text{Fe}\)) மற்றும் hexacyano (6 \(\text{CN}^-\) ஈனிகள்) இருக்கும்: \(\text{K}_x[\text{Fe(CN)}_6]\).
• Calculation for \(x\): \(\text{Fe}\)-க்கு \(\text{(II)}\) எனக் கொடுத்துள்ளதால் அதன் மதிப்பு \(+2\), ஒரு \(\text{CN}^-\)-க்கு \(-1\). \[x(+1) + 2 + 6(-1) = 0 \implies x + 2 - 6 = 0 \implies x - 4 = 0 \implies x = 4\] எனவே \(\text{K}_4[\text{Fe(CN)}_6]\) என எழுத வேண்டும்.

(b) Pentacarbonyliron(0)

• Formula: \([\text{Fe(CO)}_5]\)
• எளிய விளக்கம்: Iron என்பது \(\text{Fe}\), Pentacarbonyl என்பது 5 \(\text{CO}\) ஈனிகள். இரண்டின் மின்சுமையும் \(0\) என்பதால் நேரடியாக \([\text{Fe(CO)}_5]\) என எழுதிவிடலாம்.

(c) Pentaamminenitrito-\(\kappa\)N-cobalt(III) ion

• Formula: \([\text{Co(NH}_3)_5(\text{NO}_2)]^{2+}\)
• எளிய விளக்கம்: Cobalt (\(\text{Co}\)), Pentaammine என்றால் \((\text{NH}_3)_5\), \(\text{nitrito-}\kappa\text{N}\) என்றால் \((\text{NO}_2)\).
• Charge Calculation: 'ion' என முடிவதால் மேலே என்ன மின்சுமை வரும் எனக் கண்டுபிடிக்க வேண்டும். \(\text{Co} = +3\), \(\text{NH}_3 = 0\), \(\text{NO}_2^- = -1\). \[\text{Net Charge} = +3 + 5(0) + (-1) = +2\] எனவே அடைப்புக்குறிக்கு மேலே \(2+\) போட வேண்டும்.

(d) Hexaamminecobalt(III) sulphate & Sodium tetrafluoridodihydroxidochromate(III)

• Case 1: Hexaamminecobalt(III) sulphate
  • Formula: \([\text{Co(NH}_3)_6]_2(\text{SO}_4)_3\)
  • எளிய விளக்கம்: \(\text{Co} = +3\), \(\text{NH}_3 = 0\) என்பதால் நேர்மின் அணைவு அயனியின் மின்சுமை \([\text{Co(NH}_3)_6]^{3+}\). சல்பேட்டின் மின்சுமை \(\text{SO}_4^{2-}\). மின்சுமைகளைக் குறுக்குப் பெருக்கல் (Criss-cross) செய்தால் \([\text{Co(NH}_3)_6]_2(\text{SO}_4)_3\) கிடைக்கும்.
• Case 2: Sodium tetrafluoridodihydroxidochromate(III)
  • Formula: \(\text{Na}_3[\text{CrF}_4(\text{OH})_2]\)
  • எளிய விளக்கம்: வெளியே \(\text{Na}\) (\(x\) என்க), உள்ளே \(\text{Cr}\) (\(+3\)), 4 \(\text{F}^-\) (\(-4\)), 2 \(\text{OH}^-\) (\(-2\)). \[x(+1) + 3 + 4(-1) + 2(-1) = 0 \implies x + 3 - 4 - 2 = 0 \implies x - 3 = 0 \implies x = 3\] எனவே \(\text{Na}_3[\text{CrF}_4(\text{OH})_2]\) என வரும்.

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Question 3: Arrange the following in order of increasing molar conductivity:

(i) \(\text{Mg}[\text{Cr(NH}_3)(\text{Cl})_5]\)    (ii) \([\text{Cr(NH}_3)_5\text{Cl}]_3[\text{CoF}_6]_2\)    (iii) \([\text{Cr(NH}_3)_3\text{Cl}_3]\)

Textbook Answer:

Molar conductivity increases as the total number of ions produced in solution increases:

\[[\text{Cr(NH}_3)_3\text{Cl}_3] < \text{Mg}[\text{Cr(NH}_3)(\text{Cl})_5] < [\text{Cr(NH}_3)_5\text{Cl}]_3[\text{CoF}_6]_2\]

எளிய விளக்கம் (Concept Explanation):

Key Concept: ஒரு சேர்மத்தை நீரில் கரைக்கும்போது, சதுர அடைப்புக்குறிக்கு \([\ ]\) வெளியே உள்ள பகுதிகள் மட்டுமே அயனிகளாகப் (Ions) பிரியும். அயனிகளின் எண்ணிக்கை (Number of ions) அதிகமாக இருந்தால், மின்சாரத்தைக் கடத்தும் திறனும் (Molar conductivity) அதிகமாக இருக்கும்.

Complex Compound	Ions Produced in Aqueous Solution	Total Number of Ions	Molar Conductivity
(iii) \([\text{Cr(NH}_3)_3\text{Cl}_3]\)	Does not ionise (Neutral complex)	\(0\) ions	Lowest (மிகக் குறைவு)
(i) \(\text{Mg}[\text{Cr(NH}_3)(\text{Cl})_5]\)	\(1\text{ Mg}^{2+} + 1\ [\text{Cr(NH}_3)(\text{Cl})_5]^{2-}\)	\(2\) ions	Intermediate (நடுத்தரம்)
(ii) \([\text{Cr(NH}_3)_5\text{Cl}]_3[\text{CoF}_6]_2\)	\(3\ [\text{Cr(NH}_3)_5\text{Cl}]^{2+} + 2\ [\text{CoF}_6]^{3-}\)	\(5\) ions	Highest (மிக அதிகம்)

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Question 4: Give an example of coordination compound used in medicine and two examples of biologically important coordination compounds.

Textbook Answer:

• Coordination compound used in medicine:
  • Cis-platin (\(\text{cis-}[\text{Pt(NH}_3)_2\text{Cl}_2]\)) is used as an antitumour drug in cancer treatment.
• Biologically important coordination compounds:
  1. Red Blood Corpuscles (RBC): RBC is composed of the heme group, which is an \(\text{Fe}^{2+}\)-porphyrin complex. It plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.
  2. Chlorophyll: A green pigment present in green plants and algae, is a coordination complex containing \(\text{Mg}^{2+}\) as the central metal ion surrounded by a modified porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants convert \(\text{CO}_2\) and water into carbohydrates and oxygen.

எளிய விளக்கம் (One-Mark Quick Points):

• மருத்துவத்தில் (In Medicine): புற்றுநோய் கட்டிகளைக் (Cancer / Tumour) குணப்படுத்தப் பயன்படும் அணைவுச் சேர்மம் Cis-platin.
• இரத்த சிவப்பணுக்கள் (RBC / Hemoglobin): நம் இரத்தத்தில் ஆக்ஸிஜனை நுரையீரலிலிருந்து மற்ற பாகங்களுக்கு எடுத்துச் செல்லும் Heme குரூப்பில் உள்ள மைய உலோக அயனி \(\text{Fe}^{2+}\) (இரும்பு) ஆகும். இதில் உள்ள வளைய அமைப்பு Porphyrin complex எனப்படும்.
• பச்சையம் (Chlorophyll): தாவரங்கள் ஒளிச்சேர்க்கை (Photosynthesis) மூலம் உணவு தயாரிக்க உதவும் பச்சையத்தில் உள்ள மைய உலோக அயனி \(\text{Mg}^{2+}\) (மெக்னீசியம்) ஆகும். இதில் உள்ள வளைய அமைப்பு Corrin ring எனப்படும்.

---

Question 5: Based on VB theory explain why \([\text{Cr(NH}_3)_6]^{3+}\) is paramagnetic, while \([\text{Ni(CN)}_4]^{2-}\) is diamagnetic.

Memory Trick for 3d Series (Atomic Numbers 21 to 30):
"Sun TV Chief Minister Iconic Z"
\(\text{Sc (21)}, \text{Ti (22)}, \text{V (23)}, \mathbf{\text{Cr (24)}}, \text{Mn (25)}, \text{Fe (26)}, \text{Co (27)}, \mathbf{\text{Ni (28)}}, \text{Cu (29)}, \text{Zn (30)}\).
Paramagnetic: Has unpaired electrons (\(\uparrow\)).  |  Diamagnetic: All electrons are paired (\(\uparrow\downarrow\)).

Part A: Why \([\text{Cr(NH}_3)_6]^{3+}\) is Paramagnetic

1. Central Metal Atom & Electronic Configuration:
   • Central metal atom is Chromium (\(\text{Cr}\), Atomic Number \(Z = 24\)).
   • Outer electronic configuration of \(\text{Cr}\) atom: \(3d^5 4s^1\).
2. Oxidation State & Configuration of Metal Ion:
   • Oxidation state of Chromium in \([\text{Cr(NH}_3)_6]^{3+}\) is \(+3\) (\(\text{Cr}^{3+}\)).
   • Outer electronic configuration of \(\text{Cr}^{3+}\) ion (after losing 3 electrons): \(3d^3 4s^0\).
   • Orbital diagram of \(\text{Cr}^{3+}\) ion: \[\underbrace{\begin{array}{|c|c|c|c|c|} \hline \uparrow & \uparrow & \uparrow & \phantom{\uparrow} & \phantom{\uparrow} \\ \hline \end{array}}_{3d^3} \quad \underbrace{\begin{array}{|c|} \hline \phantom{\uparrow} \\ \hline \end{array}}_{4s^0} \quad \underbrace{\begin{array}{|c|c|c|} \hline \phantom{\uparrow} & \phantom{\uparrow} & \phantom{\uparrow} \\ \hline \end{array}}_{4p^0}\]
3. Hybridisation & Ligand Donation:
   • Since two inner \(3d\) orbitals are already vacant, no pairing of \(3d\) electrons is required.
   • Two vacant \(3d\) orbitals, one \(4s\) orbital, and three \(4p\) orbitals hybridise to form six equivalent \(d^2sp^3\) hybridised orbitals (Inner orbital octahedral complex).
   • Six \(\text{NH}_3\) ligands donate six pairs of electrons into these vacant \(d^2sp^3\) hybrid orbitals: \[\underbrace{\begin{array}{|c|c|c|} \hline \uparrow & \uparrow & \uparrow \\ \hline \end{array}}_{3d} \underbrace{\begin{array}{|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow \\ \hline \end{array} \quad \begin{array}{|c|} \hline \uparrow\downarrow \\ \hline \end{array} \quad \begin{array}{|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \hline \end{array}}_{d^2sp^3 \text{ hybridisation (6 pairs from 6 }\text{NH}_3\text{)}}\]
4. Magnetic Property & Spin-Only Magnetic Moment:
   • Due to the presence of \(3\) unpaired electrons (\(n = 3\)) in the \(3d\) orbitals, \([\text{Cr(NH}_3)_6]^{3+}\) is paramagnetic.
   • Spin-only magnetic moment: \[\mu_s = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} = 3.87\text{ BM}\]

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Part B: Why \([\text{Ni(CN)}_4]^{2-}\) is Diamagnetic

1. Central Metal Atom & Electronic Configuration:
   • Central metal atom is Nickel (\(\text{Ni}\), Atomic Number \(Z = 28\)).
   • Outer electronic configuration of \(\text{Ni}\) atom: \(3d^8 4s^2\).
2. Oxidation State & Configuration of Metal Ion:
   • Oxidation state of Nickel in \([\text{Ni(CN)}_4]^{2-}\): \(x + 4(-1) = -2 \implies x = +2\) (\(\text{Ni}^{2+}\)).
   • Outer electronic configuration of \(\text{Ni}^{2+}\) ion: \(3d^8 4s^0\).
   • Orbital diagram of \(\text{Ni}^{2+}\) ion before ligand approach: \[\underbrace{\begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow & \uparrow \\ \hline \end{array}}_{3d^8} \quad \underbrace{\begin{array}{|c|} \hline \phantom{\uparrow} \\ \hline \end{array}}_{4s^0} \quad \underbrace{\begin{array}{|c|c|c|} \hline \phantom{\uparrow} & \phantom{\uparrow} & \phantom{\uparrow} \\ \hline \end{array}}_{4p^0}\]
3. Effect of Strong Field Ligand & Hybridisation:
   • \(\text{CN}^-\) is a strong field ligand (வலிமையான ஈனி). Therefore, it forces the two unpaired electrons in the \(3d\) orbitals to pair up against Hund's rule.
   • This leaves one inner \(3d\) orbital vacant. One \(3d\), one \(4s\), and two \(4p\) orbitals hybridise to form four \(dsp^2\) hybridised orbitals (Square planar geometry).
   • Four \(\text{CN}^-\) ligands donate four pairs of electrons into these vacant \(dsp^2\) hybrid orbitals: \[\underbrace{\begin{array}{|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \hline \end{array}}_{3d\text{ (paired)}} \underbrace{\begin{array}{|c|} \hline \uparrow\downarrow \\ \hline \end{array} \quad \begin{array}{|c|} \hline \uparrow\downarrow \\ \hline \end{array} \quad \begin{array}{|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow \\ \hline \end{array}}_{dsp^2 \text{ hybridisation (4 pairs from 4 }\text{CN}^-\text{)}} \begin{array}{|c|} \hline \phantom{\uparrow} \\ \hline \end{array}_{4p}\]
4. Magnetic Property & Spin-Only Magnetic Moment:
   • Since all electrons are paired and there are no unpaired electrons (\(n = 0\)), \([\text{Ni(CN)}_4]^{2-}\) is diamagnetic (\(\mu_s = 0\text{ BM}\)).

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Part 3: Quick Revision Quiz Answers (1-Mark Self-Check)

1. What will you name first in a coordination compound: Cation or Anion?
   Answer: Cation first, followed by the Anion.
2. Inside the coordination sphere, will you name the Ligand first or the Metal first?
   Answer: Ligand first (in alphabetical order), followed by the Central Metal Atom.
3. Less number of ions in solution have _____ molar conductivity (High / Low)?
   Answer: Low molar conductivity.
4. Which coordination compound is used as an antitumour drug in cancer treatment?
   Answer: Cis-platin (\(\text{cis-}[\text{Pt(NH}_3)_2\text{Cl}_2]\)).

Ranipet District 10th Maths 2nd Mid Term Exam 2024 Original Question Paper with Solutions

இரண்டாம் இடைப்பருவத் தேர்வு - 2024

பத்தாம் வகுப்பு - கணிதம் (விடைகளுடன்)

மதிப்பெண்கள்: 50 நேரம்: 1.30 மணி

பகுதி - அ (7x1=7)

I. சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

1. A என்ற அணியின் வரிசை 2 × 3, B என்ற அணியின் வரிசை 3 × 4 எனில், AB என்ற அணியின் நிரல்களின் எண்ணிக்கை

• அ) 3
• ஆ) 4
• இ) 2
• ஈ) 5

விடை: ஆ) 4

விளக்கம்: A-யின் வரிசை m x n மற்றும் B-யின் வரிசை n x p எனில், AB-யின் வரிசை m x p ஆகும். இங்கு, A-யின் வரிசை 2x3, B-யின் வரிசை 3x4. எனவே, AB-யின் வரிசை 2x4. நிரல்களின் எண்ணிக்கை 4.

2. \( 2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} \) எனில் X என்ற அணியைக் காண்க.

• அ) \( \begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix} \)
• ஆ) \( \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix} \)
• இ) \( \begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix} \)
• ஈ) \( \begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix} \)

விடை: ஆ) \( \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix} \)

விளக்கம்:
\( 2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5-1 & 7-3 \\ 9-5 & 5-7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} \)
\( X = \frac{1}{2} \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix} \)

3. ஒரு கோபுரத்தின் உயரத்திற்கும், அதன் நிழலின் நீளத்திற்கும் உள்ள விகிதம் \( \sqrt{3}:1 \) எனில், சூரியனைக் காணும் ஏற்றக் கோண அளவானது

• அ) 45°
• ஆ) 30°
• இ) 90°
• ஈ) 60°

விடை: ஈ) 60°

விளக்கம்: \( \tan \theta = \frac{உயரம்}{நிழலின் நீளம்} = \frac{\sqrt{3}}{1} = \sqrt{3} \). எனவே, \( \theta = 60° \).

4. ஒரு கோபுரத்தின் உயரம் 60 மீ ஆகும். சூரியனைக் காணும் ஏற்றக் கோணம் 30° லிருந்து 45° ஆக உயரும் போது கோபுரத்தின் நிழலானது x மீ குறைகிறது எனில் x ன் மதிப்பு

• அ) 41.92 மீ
• ஆ) 43.92 மீ
• இ) 43 மீ
• ஈ) 45.6 மீ

விடை: ஆ) 43.92 மீ

விளக்கம்:
ஆரம்பத்தில், \( \tan 30° = \frac{60}{D} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60}{D} \Rightarrow D = 60\sqrt{3} \) மீ.
பின்னர், \( \tan 45° = \frac{60}{D-x} \Rightarrow 1 = \frac{60}{D-x} \Rightarrow D-x = 60 \) மீ.
\( x = D - 60 = 60\sqrt{3} - 60 = 60(\sqrt{3} - 1) = 60(1.732 - 1) = 60(0.732) = 43.92 \) மீ.

5. ஆரம் 5 செ.மீ மற்றும் சாயுயரம் 13 செ.மீ உடைய நேர்வட்டக் கூம்பின் உயரம்

• அ) 12 செ.மீ
• ஆ) 10 செ.மீ
• இ) 13 செ.மீ
• ஈ) 5 செ.மீ

விடை: அ) 12 செ.மீ

விளக்கம்: \( h = \sqrt{l^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \) செ.மீ.

6. 15 செ.மீ உயரமும் 16 செ.மீ விட்டமும் கொண்ட ஒரு நேர்வட்டக் கூம்பின் வளைபரப்பு

• அ) 60π ச.செ.மீ
• ஆ) 68π ச.செ.மீ
• இ) 120π ச.செ.மீ
• ஈ) 136π ச.செ.மீ

விடை: ஈ) 136π ச.செ.மீ

விளக்கம்: h = 15 செ.மீ, விட்டம் = 16 செ.மீ, எனவே ஆரம் r = 8 செ.மீ.
சாயுயரம் \( l = \sqrt{h^2 + r^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \) செ.மீ.
வளைபரப்பு = \( \pi r l = \pi \times 8 \times 17 = 136\pi \) ச.செ.மீ.

7. ஓர் இருபடிச் சமன்பாட்டின் வரைபடம் ஒரு ______ ஆகும்.

• அ) நேர்க்கோடு
• ஆ) வட்டம்
• இ) பரவளையம்
• ஈ) அதிபரவளையம்

விடை: இ) பரவளையம்

விளக்கம்: இருபடிச் சமன்பாட்டின் (quadratic equation) வரைபடம் எப்போதும் ஒரு பரவளையமாக (parabola) இருக்கும்.

பகுதி - ஆ (5x2=10)

II. எவையேனும் 5 வினாக்களுக்கு விடையளி. (வினா எண் 14 கட்டாய வினா)

8. \( a_{ij} = |i - 2j| \) வை 3 x 3 வரிசையைக் கொண்ட அணி A = \( [a_{ij}] \) யினைக் காண்க.

விடை:

\( a_{11} = |1 - 2(1)| = |-1| = 1 \)
\( a_{12} = |1 - 2(2)| = |-3| = 3 \)
\( a_{13} = |1 - 2(3)| = |-5| = 5 \)

\( a_{21} = |2 - 2(1)| = |0| = 0 \)
\( a_{22} = |2 - 2(2)| = |-2| = 2 \)
\( a_{23} = |2 - 2(3)| = |-4| = 4 \)

\( a_{31} = |3 - 2(1)| = |1| = 1 \)
\( a_{32} = |3 - 2(2)| = |-1| = 1 \)
\( a_{33} = |3 - 2(3)| = |-3| = 3 \)

எனவே, அணி A:

\( A = \begin{pmatrix} 1 & 3 & 5 \\ 0 & 2 & 4 \\ 1 & 1 & 3 \end{pmatrix} \)

9. \( A = \begin{pmatrix} 1 & 9 \\ 3 & 4 \\ 8 & -3 \end{pmatrix}, B = \begin{pmatrix} 5 & 7 \\ 3 & 3 \\ 1 & 0 \end{pmatrix} \) ல் A + B = B + A சரிபார்க்க.

விடை:

LHS = A + B

\( A + B = \begin{pmatrix} 1 & 9 \\ 3 & 4 \\ 8 & -3 \end{pmatrix} + \begin{pmatrix} 5 & 7 \\ 3 & 3 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1+5 & 9+7 \\ 3+3 & 4+3 \\ 8+1 & -3+0 \end{pmatrix} = \begin{pmatrix} 6 & 16 \\ 6 & 7 \\ 9 & -3 \end{pmatrix} \) ---(1)

RHS = B + A

\( B + A = \begin{pmatrix} 5 & 7 \\ 3 & 3 \\ 1 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 9 \\ 3 & 4 \\ 8 & -3 \end{pmatrix} = \begin{pmatrix} 5+1 & 7+9 \\ 3+3 & 3+4 \\ 1+8 & 0-3 \end{pmatrix} = \begin{pmatrix} 6 & 16 \\ 6 & 7 \\ 9 & -3 \end{pmatrix} \) ---(2)

(1) மற்றும் (2) லிருந்து, A + B = B + A என்பது சரிபார்க்கப்பட்டது.

10. ஒரு கோபுரம் தரைக்குச் செங்குத்தாக உள்ளது. கோபுரத்தின் அடிப்பகுதியிலிருந்து தரையில் 48 மீ தொலைவில் உள்ள ஒரு புள்ளியிலிருந்து கோபுர உச்சியின் ஏற்றக்கோணம் 30° எனில், கோபுரத்தின் உயரம் காண்க.

விடை:

கோபுரத்தின் உயரம் = h, தொலைவு = 48 மீ, ஏற்றக்கோணம் θ = 30°.

\( \tan \theta = \frac{உயரம்}{அடிப்பக்கம்} \)

\( \tan 30° = \frac{h}{48} \)

\( \frac{1}{\sqrt{3}} = \frac{h}{48} \)

\( h = \frac{48}{\sqrt{3}} = \frac{48 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{48\sqrt{3}}{3} = 16\sqrt{3} \) மீ.

கோபுரத்தின் உயரம் \( 16\sqrt{3} \) மீ ஆகும்.

11. 13 மீ உயரமுள்ள ஒரு மரத்தின் உச்சியிலிருந்து மற்றொரு மரத்தின் உச்சி மற்றும் அடியின் ஏற்றக்கோணம் மற்றும் இறக்கக்கோணம் முறையே 45° மற்றும் 30° எனில் இரண்டாவது மரத்தின் உயரத்தைக் காண்க. (√3 = 1.732)

விடை:

முதல் மரத்தின் உயரம் (AB) = 13 மீ. இரண்டாவது மரத்தின் உயரம் (CD) = H.

மரங்களுக்கு இடையேயுள்ள தொலைவு = x.

படத்திலிருந்து, இறக்கக்கோணம் 30°:

\( \tan 30° = \frac{13}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{13}{x} \Rightarrow x = 13\sqrt{3} \) மீ.

ஏற்றக்கோணம் 45°:

\( \tan 45° = \frac{H-13}{x} \Rightarrow 1 = \frac{H-13}{x} \Rightarrow x = H - 13 \)

சமன்பாடுகளை ஒப்பிட, \( 13\sqrt{3} = H - 13 \)

\( H = 13 + 13\sqrt{3} = 13(1 + \sqrt{3}) = 13(1 + 1.732) = 13(2.732) = 35.516 \) மீ.

இரண்டாவது மரத்தின் உயரம் 35.516 மீ.

12. 704 ச.செ.மீ மொத்தப் புறப்பரப்பு கொண்ட ஒரு கூம்பின் ஆரம் 7 செ.மீ எனில், அதன் சாயுயரம் காண்க.

விடை:

கூம்பின் மொத்தப் புறப்பரப்பு (TSA) = \( \pi r (l+r) = 704 \)

ஆரம் r = 7 செ.மீ

\( \frac{22}{7} \times 7 \times (l+7) = 704 \)

\( 22 (l+7) = 704 \)

\( l+7 = \frac{704}{22} = 32 \)

\( l = 32 - 7 = 25 \) செ.மீ.

சாயுயரம் 25 செ.மீ ஆகும்.

13. ஒரு கோளத்தின் புறப்பரப்பு 154 ச.மீ எனில், அதன் விட்டம் காண்க.

விடை:

கோளத்தின் புறப்பரப்பு = \( 4\pi r^2 = 154 \)

\( 4 \times \frac{22}{7} \times r^2 = 154 \)

\( r^2 = \frac{154 \times 7}{4 \times 22} = \frac{7 \times 7}{4} \)

\( r = \sqrt{\frac{49}{4}} = \frac{7}{2} = 3.5 \) மீ.

விட்டம் = \( 2r = 2 \times 3.5 = 7 \) மீ.

14. \( A = \begin{pmatrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{pmatrix} \) எனில் \( (A^T)^T = A \) சரிபார்க்க. (கட்டாய வினா)

விடை:

கொடுக்கப்பட்ட அணி: \( A = \begin{pmatrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{pmatrix} \)

A-யின் நிரை-நிரல் மாற்று அணி (Transpose of A):

\( A^T = \begin{pmatrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{pmatrix} \)

\( A^T \)-யின் நிரை-நிரல் மாற்று அணி:

\( (A^T)^T = \begin{pmatrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{pmatrix} \)

எனவே, \( (A^T)^T = A \) என்பது சரிபார்க்கப்பட்டது.

பகுதி - இ (5x5=25)

III. எவையேனும் 5 வினாக்களுக்கு விடையளி. (வினா எண் 21 கட்டாய வினா)

15. பின்வருவனவற்றை எடுத்துக்காட்டுகளுடன் வரையறை.
i) மூலைவிட்ட அணி (3 × 3)
ii) திசையிலி அணி (4 x 4)
iii) அலகு அணி (3 x 3)

விடை:

i) மூலைவிட்ட அணி (Diagonal Matrix): ஒரு சதுர அணியில், முதன்மை மூலைவிட்ட உறுப்புகளைத் தவிர மற்ற அனைத்து உறுப்புகளும் பூச்சியம் எனில், அது மூலைவிட்ட அணி எனப்படும்.
எடுத்துக்காட்டு (3x3): \( \begin{pmatrix} 5 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{pmatrix} \)

ii) திசையிலி அணி (Scalar Matrix): ஒரு மூலைவிட்ட அணியில், முதன்மை மூலைவிட்ட உறுப்புகள் அனைத்தும் சமமாக இருப்பின் அது திசையிலி அணி எனப்படும்.
எடுத்துக்காட்டு (4x4): \( \begin{pmatrix} 7 & 0 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 7 \end{pmatrix} \)

iii) அலகு அணி (Identity/Unit Matrix): ஒரு திசையிலி அணியில், முதன்மை மூலைவிட்ட உறுப்புகள் அனைத்தும் 1 ஆக இருப்பின் அது அலகு அணி எனப்படும்.
எடுத்துக்காட்டு (3x3): \( I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \)

16. \( A = \begin{pmatrix} 7 & 8 & 6 \\ 1 & 3 & 9 \\ -4 & 3 & -1 \end{pmatrix}, B = \begin{pmatrix} 4 & 11 & -3 \\ -1 & 2 & 4 \\ 7 & 5 & 0 \end{pmatrix} \) எனில், 2A + B காண்க.

விடை:

\( 2A = 2 \times \begin{pmatrix} 7 & 8 & 6 \\ 1 & 3 & 9 \\ -4 & 3 & -1 \end{pmatrix} = \begin{pmatrix} 14 & 16 & 12 \\ 2 & 6 & 18 \\ -8 & 6 & -2 \end{pmatrix} \)

\( 2A + B = \begin{pmatrix} 14 & 16 & 12 \\ 2 & 6 & 18 \\ -8 & 6 & -2 \end{pmatrix} + \begin{pmatrix} 4 & 11 & -3 \\ -1 & 2 & 4 \\ 7 & 5 & 0 \end{pmatrix} \)

\( = \begin{pmatrix} 14+4 & 16+11 & 12-3 \\ 2-1 & 6+2 & 18+4 \\ -8+7 & 6+5 & -2+0 \end{pmatrix} = \begin{pmatrix} 18 & 27 & 9 \\ 1 & 8 & 22 \\ -1 & 11 & -2 \end{pmatrix} \)

17. \( A = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix}, B = \begin{pmatrix} 1 & 2 \\ -4 & 2 \end{pmatrix}, C = \begin{pmatrix} -7 & 6 \\ 3 & 2 \end{pmatrix} \) எனில், A(B + C) = AB + AC என்பதைச் சரிபார்க்க.

விடை:

LHS = A(B + C)

\( B+C = \begin{pmatrix} 1-7 & 2+6 \\ -4+3 & 2+2 \end{pmatrix} = \begin{pmatrix} -6 & 8 \\ -1 & 4 \end{pmatrix} \)

\( A(B+C) = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} -6 & 8 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} -6-1 & 8+4 \\ 6-3 & -8+12 \end{pmatrix} = \begin{pmatrix} -7 & 12 \\ 3 & 4 \end{pmatrix} \) ---(1)

RHS = AB + AC

\( AB = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} 1-4 & 2+2 \\ -1-12 & -2+6 \end{pmatrix} = \begin{pmatrix} -3 & 4 \\ -13 & 4 \end{pmatrix} \)

\( AC = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} -7 & 6 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} -7+3 & 6+2 \\ 7+9 & -6+6 \end{pmatrix} = \begin{pmatrix} -4 & 8 \\ 16 & 0 \end{pmatrix} \)

\( AB+AC = \begin{pmatrix} -3-4 & 4+8 \\ -13+16 & 4+0 \end{pmatrix} = \begin{pmatrix} -7 & 12 \\ 3 & 4 \end{pmatrix} \) ---(2)

(1) மற்றும் (2) லிருந்து, A(B + C) = AB + AC என்பது சரிபார்க்கப்பட்டது.

18. இரு கப்பல்கள் கலங்கரைவிளக்கத்தின் இரு பக்கங்களிலும் கடலில் பயணம் செய்கின்றன. இரு கப்பல்களிலிருந்தும் கலங்கரைவிளக்கத்தின் உச்சியின் ஏற்றக்கோணங்கள் முறையே 30° மற்றும் 45° ஆகும். கலங்கரைவிளக்கத்தின் உயரம் 200 மீ எனில், இரு கப்பல்களுக்கு இடையே உள்ள தொலைவைக் காண்க. (√3 = 1.732)

விடை:

கலங்கரை விளக்கத்தின் உயரம் (h) = 200 மீ.

முதல் கப்பலுக்கும் கலங்கரை விளக்கத்திற்கும் இடையே உள்ள தொலைவு = x.

\( \tan 30° = \frac{200}{x} \Rightarrow x = \frac{200}{\tan 30°} = 200\sqrt{3} \) மீ.

இரண்டாவது கப்பலுக்கும் கலங்கரை விளக்கத்திற்கும் இடையே உள்ள தொலைவு = y.

\( \tan 45° = \frac{200}{y} \Rightarrow y = \frac{200}{\tan 45°} = 200 \) மீ.

இரு கப்பல்களுக்கு இடையே உள்ள தொலைவு = x + y = \( 200\sqrt{3} + 200 = 200(\sqrt{3} + 1) \)

\( = 200(1.732 + 1) = 200(2.732) = 546.4 \) மீ.

19. 1.6 மீ உயரமுள்ள சிலை ஒன்று பீடத்தின் மேல் அமைந்துள்ளது. தரையிலுள்ள ஒரு புள்ளியிலிருந்து 60° ஏற்றக் கோணத்தில் சிலையின் உச்சி அமைந்துள்ளது. மேலும் அதே புள்ளியிலிருந்து பீடத்தின் உச்சியானது 40° ஏற்றக் கோணத்தில் உள்ளது எனில், பீடத்தின் உயரத்தைக் காண்க. (tan 40° = 0.8391, √3 = 1.732)

விடை:

பீடத்தின் உயரம் = h மீ என்க.

சிலையின் உயரம் = 1.6 மீ.

தரையிலுள்ள புள்ளிக்கும் பீடத்தின் அடிக்கும் உள்ள தொலைவு = x மீ.

கொடுக்கப்பட்ட விவரங்களின்படி, இரண்டு செங்கோண முக்கோணங்கள் உருவாகின்றன.

பீடத்தின் உச்சிக்கான ஏற்றக்கோணம் 40°:

\( \tan 40° = \frac{எதிர்ப்பக்கம்}{அடுத்துள்ள பக்கம்} = \frac{h}{x} \)

\( x = \frac{h}{\tan 40°} \) --- (1)

சிலையின் உச்சிக்கான ஏற்றக்கோணம் 60°:

\( \tan 60° = \frac{பீடத்தின் உயரம் + சிலையின் உயரம்}{x} = \frac{h + 1.6}{x} \)

\( x = \frac{h + 1.6}{\tan 60°} \) --- (2)

சமன்பாடு (1) மற்றும் (2) லிருந்து,

\( \frac{h}{\tan 40°} = \frac{h + 1.6}{\tan 60°} \)

மதிப்புகளைப் பிரதியிட (\( \tan 40° = 0.8391 \), \( \tan 60° = \sqrt{3} = 1.732 \)),

\( \frac{h}{0.8391} = \frac{h + 1.6}{1.732} \)

குறுக்குப் பெருக்கல் செய்ய,

\( 1.732 \times h = 0.8391 \times (h + 1.6) \)

\( 1.732h = 0.8391h + (0.8391 \times 1.6) \)

\( 1.732h = 0.8391h + 1.34256 \)

\( 1.732h - 0.8391h = 1.34256 \)

\( 0.8929h = 1.34256 \)

\( h = \frac{1.34256}{0.8929} \approx 1.5036 \) மீ

எனவே, பீடத்தின் உயரம் தோராயமாக 1.50 மீ ஆகும்.

20. ஓர் உருளையின் ஆரம் மற்றும் உயரங்களின் விகிதம் 5:7 ஆகும். அதன் வளைபரப்பு 5500 ச.செ.மீ எனில் உருளையின் ஆரம் மற்றும் உயரம் காண்க.

விடை:

ஆரம் (r) = 5x, உயரம் (h) = 7x.

உருளையின் வளைபரப்பு (CSA) = \( 2\pi rh = 5500 \)

\( 2 \times \frac{22}{7} \times (5x) \times (7x) = 5500 \)

\( 2 \times 22 \times 5x^2 = 5500 \)

\( 220x^2 = 5500 \)

\( x^2 = \frac{5500}{220} = 25 \Rightarrow x=5 \)

ஆரம் r = 5x = 5(5) = 25 செ.மீ.

உயரம் h = 7x = 7(5) = 35 செ.மீ.

21. (கட்டாய வினா)

அ) 45 செ.மீ உயரமுள்ள ஓர் இடைக்கண்டத்தின் இருபுற ஆரங்கள் முறையே 28 செ.மீ மற்றும் 7 செ.மீ எனில் இடைக்கண்டத்தின் கனஅளவு காண்க.

(அல்லது)

ஆ) பிதாகரஸ் தேற்றம் எழுதி நிறுவுக.

விடை:

அ) இடைக்கண்டத்தின் கனஅளவு:

h = 45 செ.மீ, R = 28 செ.மீ, r = 7 செ.மீ.

கனஅளவு \( V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) \)

\( V = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7) \)

\( V = \frac{22}{7} \times 15 \times (784 + 49 + 196) \)

\( V = \frac{22}{7} \times 15 \times 1029 \)

\( V = 22 \times 15 \times 147 = 48510 \) க.செ.மீ.

ஆ) பிதாகரஸ் தேற்றம்:

தேற்றம்: ஒரு செங்கோண முக்கோணத்தில் கர்ணத்தின் வர்க்கம் மற்ற இரு பக்கங்களின் வர்க்கங்களின் கூடுதலுக்குச் சமம்.

நிரூபணம்:
ΔABC -யில் ∠B = 90°. BD ⊥ AC வரைக.
1. ΔADB மற்றும் ΔABC -யில், ∠A பொதுவானது, ∠ADB = ∠ABC = 90°. எனவே, ΔADB ~ ΔABC (AA விதி).
அதனால், \( \frac{AD}{AB} = \frac{AB}{AC} \Rightarrow AB^2 = AD \cdot AC \) ---(1)
2. ΔBDC மற்றும் ΔABC -யில், ∠C பொதுவானது, ∠BDC = ∠ABC = 90°. எனவே, ΔBDC ~ ΔABC (AA விதி).
அதனால், \( \frac{CD}{BC} = \frac{BC}{AC} \Rightarrow BC^2 = CD \cdot AC \) ---(2)
3. (1) மற்றும் (2) ஐக் கூட்ட,
\( AB^2 + BC^2 = AD \cdot AC + CD \cdot AC = AC(AD+CD) \)
படத்திலிருந்து AD + CD = AC.
\( AB^2 + BC^2 = AC(AC) = AC^2 \).
தேற்றம் நிரூபிக்கப்பட்டது.

பகுதி - ஈ (1x8=8)

IV. விடையளி

22.

அ) \( x^2 - 9x + 20 = 0 \) வரைபடம் வரைந்து தீர்வுகளின் தன்மையைக் கூறுக.

(அல்லது)

ஆ) 4 செ.மீ ஆரமுள்ள வட்டம் வரைந்து அதன் மையத்திலிருந்து 11 செ.மீ தொலைவிலுள்ள ஒரு புள்ளியைக் குறித்து, அப்புள்ளியிலிருந்து வட்டத்திற்கு இரண்டு தொடுகோடுகள் வரைக.

விடை:

அ) வரைபடம்: \( y = x^2 - 9x + 20 \)

1. அட்டவணை தயாரித்தல்:

x	-1	0	2	4	4.5	5	6
y	30	20	6	0	-0.25	0	2

2. வரைபடம் வரைதல்:
மேற்கண்ட புள்ளிகளை வரைபடத்தாளில் குறித்து, அவற்றை ஒரு மென்மையான வளைகோட்டால் இணைத்தால் ஒரு பரவளையம் கிடைக்கும்.

3. தீர்வு:
பரவளையம் x-அச்சை x=4 மற்றும் x=5 ஆகிய இரு வெவ்வேறு புள்ளிகளில் வெட்டுகிறது.

4. தீர்வுகளின் தன்மை:
மூலங்கள் மெய்யானவை மற்றும் சமமற்றவை (Real and Unequal roots).

ஆ) தொடுகோடுகள் வரைதல்:

வரைமுறை:

1. O-வை மையமாகக் கொண்டு 4 செ.மீ ஆரமுள்ள வட்டம் வரைக.
2. மையம் O-விலிருந்து 11 செ.மீ தொலைவில் P என்ற புள்ளியைக் குறிக்க (OP = 11 செ.மீ).
3. OP என்ற கோட்டுத்துண்டிற்கு மையக்குத்துக்கோடு வரைக. அது OP-ஐ M-ல் சந்திக்கட்டும்.
4. M-ஐ மையமாகவும், MO-வை ஆரமாகவும் கொண்டு மற்றொரு வட்டம் வரைக.
5. இந்த புதிய வட்டம், பழைய வட்டத்தை A மற்றும் B ஆகிய இரு புள்ளிகளில் வெட்டும்.
6. PA மற்றும் PB-ஐ இணைக்க.
7. PA மற்றும் PB என்பவையே தேவையான தொடுகோடுகள் ஆகும். (அளந்தால், \( PA = PB \approx 10.25 \) செ.மீ இருக்கும்).

10th Maths - 2nd Mid Term Exam 2024 - Original Question Paper | Ariyalur District

Ariyalur District Second Mid Term Exam 2024
Std: 10 Mathematics - Solutions

PART - A

I. Choose the best answer (7 x 1 = 7)

1. If number of columns and rows are not equal in a matrix then it is said to be a

Answer: b) rectangular matrix
Explanation: A matrix is called a rectangular matrix if the number of rows is not equal to the number of columns (m ≠ n).

2. If A is a 2x3 matrix and B is a 3x4 matrix how many columns does AB have.

Answer: b) 4
Explanation: When multiplying a matrix of order $m \times n$ by a matrix of order $n \times p$, the resulting matrix has the order $m \times p$. Here, A is $2 \times 3$ and B is $3 \times 4$. The resulting matrix AB will have the order $2 \times 4$. Therefore, it has 4 columns.

3. A tangent is perpendicular to the radius at the

Answer: b) point of contact
Explanation: A fundamental theorem in circle geometry states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

4. In figure if PR and QR are tangents to the circle at P and Q and O is the centre of the circle, then $\angle POQ$ is

Note: The question is incomplete as the angle at R ($\angle PRQ$) is not given. Assuming a standard problem structure, we find the relationship. The sum of the angle at the center ($\angle POQ$) and the angle between the tangents ($\angle PRQ$) is 180°. Let's assume $\angle PRQ = 70^{\circ}$ to match one of the options.

Answer: c) $110^{\circ}$
Explanation: In the quadrilateral OPRQ, $\angle OPR = 90^{\circ}$ and $\angle OQR = 90^{\circ}$ (radius is perpendicular to tangent). The sum of angles in a quadrilateral is $360^{\circ}$. $$ \angle POQ + \angle OPR + \angle PRQ + \angle OQR = 360^{\circ} $$ $$ \angle POQ + 90^{\circ} + \angle PRQ + 90^{\circ} = 360^{\circ} $$ $$ \angle POQ + \angle PRQ = 180^{\circ} $$ If we assume $\angle PRQ = 70^{\circ}$, then $\angle POQ = 180^{\circ} - 70^{\circ} = 110^{\circ}$.

5. A tower is 60m high. Its shadow reduces by x meters when the angle of elevation of the sun increases from 30° to 45° then x is equal to

Answer: b) 43.92m
Explanation: Let the height of the tower AB = 60 m. Let the initial and final positions of the shadow be C and D. When the angle is $30^{\circ}$, $\tan 30^{\circ} = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60}{BC} \Rightarrow BC = 60\sqrt{3}$ m. When the angle is $45^{\circ}$, $\tan 45^{\circ} = \frac{AB}{BD} \Rightarrow 1 = \frac{60}{BD} \Rightarrow BD = 60$ m. The reduction in shadow length, $x = BC - BD = 60\sqrt{3} - 60 = 60(\sqrt{3} - 1)$. $x = 60(1.732 - 1) = 60(0.732) = 43.92$ m.

6. If the radius of the base of a cone is tripled and the height is doubled then the volume is

Answer: b) made 18 times
Explanation: Original volume $V_1 = \frac{1}{3}\pi r^2 h$. New radius $r' = 3r$, new height $h' = 2h$. New volume $V_2 = \frac{1}{3}\pi (r')^2 h' = \frac{1}{3}\pi (3r)^2 (2h) = \frac{1}{3}\pi (9r^2)(2h) = 18 \left(\frac{1}{3}\pi r^2 h\right) = 18V_1$.

7. The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is

Answer: d) 3 : 1 : 2
Explanation: Let the radius be $r$. Same diameter means same radius. Same height $h$. For a sphere, height = diameter, so $h = 2r$. $V_{cylinder} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3$. $V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3$. $V_{sphere} = \frac{4}{3}\pi r^3$. Ratio: $V_{cylinder} : V_{cone} : V_{sphere} = 2\pi r^3 : \frac{2}{3}\pi r^3 : \frac{4}{3}\pi r^3$. Multiplying by $\frac{3}{2\pi r^3}$, we get the ratio: $3 : 1 : 2$.

PART - B

II. Answer any five questions only [Q. No. 14 is compulsory] (5 x 2 = 10)

8. If $A = \begin{bmatrix} 0 & 4 & 9 \\ 8 & 3 & 7 \end{bmatrix}, B = \begin{bmatrix} 7 & 3 & 8 \\ 1 & 4 & 9 \end{bmatrix}$ find the value of B - 5A.

Solution: First, calculate 5A: $$ 5A = 5 \begin{bmatrix} 0 & 4 & 9 \\ 8 & 3 & 7 \end{bmatrix} = \begin{bmatrix} 5(0) & 5(4) & 5(9) \\ 5(8) & 5(3) & 5(7) \end{bmatrix} = \begin{bmatrix} 0 & 20 & 45 \\ 40 & 15 & 35 \end{bmatrix} $$ Now, calculate B - 5A: $$ B - 5A = \begin{bmatrix} 7 & 3 & 8 \\ 1 & 4 & 9 \end{bmatrix} - \begin{bmatrix} 0 & 20 & 45 \\ 40 & 15 & 35 \end{bmatrix} $$ $$ = \begin{bmatrix} 7-0 & 3-20 & 8-45 \\ 1-40 & 4-15 & 9-35 \end{bmatrix} = \begin{bmatrix} 7 & -17 & -37 \\ -39 & -11 & -26 \end{bmatrix} $$

$B - 5A = \begin{bmatrix} 7 & -17 & -37 \\ -39 & -11 & -26 \end{bmatrix}$

9. If $A = \begin{bmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{bmatrix}$ then verify $(A^T)^T = A$.

Solution: Given matrix A: $$ A = \begin{bmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{bmatrix} $$ First, find the transpose of A, $A^T$: $$ A^T = \begin{bmatrix} 5 & -\sqrt{17} & 8 \\ 2 & 0.7 & 3 \\ 2 & \frac{5}{2} & 1 \end{bmatrix} $$ Now, find the transpose of $A^T$, which is $(A^T)^T$: $$ (A^T)^T = \begin{bmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{bmatrix} $$ Comparing this with the original matrix A, we see that $(A^T)^T = A$.

Hence, verified.

10. Find the length of the tangent drawn from a point whose distance from the centre of a circle is 5cm and radius of the circle is 3 cm.

Solution: Let O be the center of the circle, P be the external point, and T be the point of contact. Given: Distance from the center, OP = 5 cm. Radius of the circle, OT = 3 cm. The tangent is perpendicular to the radius at the point of contact, so $\triangle OTP$ is a right-angled triangle with the right angle at T. By Pythagoras theorem: $$ OP^2 = OT^2 + PT^2 $$ $$ 5^2 = 3^2 + PT^2 $$ $$ 25 = 9 + PT^2 $$ $$ PT^2 = 25 - 9 = 16 $$ $$ PT = \sqrt{16} = 4 $$

The length of the tangent is 4 cm.

11. A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

The unit in the question paper is 'cm', which is likely a typo for a tower. The solution uses 'm'.

Solution: Let the height of the tower be 'h' meters. Let the distance from the foot of the tower be 'd' = 48 m. The angle of elevation, $\theta = 30^{\circ}$. In the right-angled triangle formed: $$ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{d} $$ $$ \tan 30^{\circ} = \frac{h}{48} $$ $$ \frac{1}{\sqrt{3}} = \frac{h}{48} $$ $$ h = \frac{48}{\sqrt{3}} = \frac{48 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{48\sqrt{3}}{3} = 16\sqrt{3} $$

The height of the tower is $16\sqrt{3}$ m.

12. If the base area of a hemispherical solid is 1386 sq. meters then find its total surface area?

Solution: The base of a hemispherical solid is a circle. Given: Base area = $\pi r^2 = 1386$ sq. m. The Total Surface Area (TSA) of a hemisphere is the sum of its curved surface area ($2\pi r^2$) and its base area ($\pi r^2$). $$ \text{TSA of hemisphere} = 2\pi r^2 + \pi r^2 = 3\pi r^2 $$ We are given $\pi r^2 = 1386$. $$ \text{TSA} = 3 \times (\pi r^2) = 3 \times 1386 $$ $$ \text{TSA} = 4158 $$

The total surface area of the hemisphere is 4158 sq. meters.

13. The volume of a solid right circular cone is 11088 cm³. If its height is 14cm then find the radius of the cone.

The numbers in this question might be unusual. A common textbook problem has a height of 24 cm for this volume. We will solve with the given value.

Solution: Given: Volume of the cone, V = 11088 cm³. Height, h = 14 cm. The formula for the volume of a cone is $V = \frac{1}{3}\pi r^2 h$. $$ 11088 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 14 $$ $$ 11088 = \frac{1}{3} \times 22 \times r^2 \times 2 $$ $$ 11088 = \frac{44}{3} r^2 $$ $$ r^2 = \frac{11088 \times 3}{44} $$ Dividing 11088 by 44: $11088 \div 44 = 252$. $$ r^2 = 252 \times 3 = 756 $$ $$ r = \sqrt{756} = \sqrt{36 \times 21} = 6\sqrt{21} $$

The radius of the cone is $6\sqrt{21}$ cm.

14. (Compulsory) Find the angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of a tower of height 10√3 m.

Solution: Let the height of the tower, h = $10\sqrt{3}$ m. Distance from the foot of the tower, d = 30 m. Let the angle of elevation be $\theta$. $$ \tan \theta = \frac{\text{Height of tower}}{\text{Distance from foot}} = \frac{10\sqrt{3}}{30} $$ $$ \tan \theta = \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{1}{\sqrt{3}} $$ Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, the angle of elevation is $30^{\circ}$.

The angle of elevation is $30^{\circ}$.

PART - C

III. Answer any five questions [Q. No. 21 is compulsory] (5 x 5 = 25)

15. Find X and Y if $X + Y = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}$.

Solution: Let the given equations be: $$ X + Y = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix} \quad \cdots(1) $$ $$ X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} \quad \cdots(2) $$ Adding (1) and (2): $$ (X+Y) + (X-Y) = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} $$ $$ 2X = \begin{bmatrix} 7+3 & 0+0 \\ 3+0 & 5+4 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 3 & 9 \end{bmatrix} $$ $$ X = \frac{1}{2} \begin{bmatrix} 10 & 0 \\ 3 & 9 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ \frac{3}{2} & \frac{9}{2} \end{bmatrix} $$ Subtracting (2) from (1): $$ (X+Y) - (X-Y) = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} $$ $$ 2Y = \begin{bmatrix} 7-3 & 0-0 \\ 3-0 & 5-4 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix} $$ $$ Y = \frac{1}{2} \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix} $$

$X = \begin{bmatrix} 5 & 0 \\ \frac{3}{2} & \frac{9}{2} \end{bmatrix}$, $Y = \begin{bmatrix} 2 & 0 \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix}$

16. Given that $A=\begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix}$, $B=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{bmatrix}$, $C=\begin{bmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{bmatrix}$ verify that $A(B+C) = AB+AC$.

Solution: L.H.S = A(B+C) First, find B+C: $$ B+C = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1+1 & -1+3 & 2+2 \\ 3-4 & 5+1 & 2+3 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{bmatrix} $$ Now, find A(B+C): $$ A(B+C) = \begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{bmatrix} $$ $$ = \begin{bmatrix} (1)(2)+(3)(-1) & (1)(2)+(3)(6) & (1)(4)+(3)(5) \\ (5)(2)+(-1)(-1) & (5)(2)+(-1)(6) & (5)(4)+(-1)(5) \end{bmatrix} $$ $$ = \begin{bmatrix} 2-3 & 2+18 & 4+15 \\ 10+1 & 10-6 & 20-5 \end{bmatrix} = \begin{bmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{bmatrix} $$ R.H.S = AB + AC First, find AB: $$ AB = \begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{bmatrix} = \begin{bmatrix} 1+9 & -1+15 & 2+6 \\ 5-3 & -5-5 & 10-2 \end{bmatrix} = \begin{bmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{bmatrix} $$ Next, find AC: $$ AC = \begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1-12 & 3+3 & 2+9 \\ 5+4 & 15-1 & 10-3 \end{bmatrix} = \begin{bmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{bmatrix} $$ Now, find AB + AC: $$ AB+AC = \begin{bmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{bmatrix} + \begin{bmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{bmatrix} = \begin{bmatrix} 10-11 & 14+6 & 8+11 \\ 2+9 & -10+14 & 8+7 \end{bmatrix} = \begin{bmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{bmatrix} $$ Since L.H.S = R.H.S, the property is verified.

Hence, A(B+C) = AB + AC is verified.

17. Show that in a triangle, the medians are concurrent.

Proof using Coordinate Geometry: Let the vertices of a triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. Let D, E, and F be the midpoints of the sides BC, CA, and AB respectively. The coordinates of the midpoints are: $$ D = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right) $$ $$ E = \left(\frac{x_3+x_1}{2}, \frac{y_3+y_1}{2}\right) $$ $$ F = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) $$ The medians are the line segments AD, BE, and CF. The point that divides each median in the ratio 2:1 is called the centroid (G). Let's find the point G that divides the median AD in the ratio 2:1 using the section formula: $$ G = \left[\frac{1 \cdot x_1 + 2 \left(\frac{x_2+x_3}{2}\right)}{1+2}, \frac{1 \cdot y_1 + 2 \left(\frac{y_2+y_3}{2}\right)}{1+2}\right] $$ $$ G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$ Similarly, for median BE: $$ G = \left[\frac{1 \cdot x_2 + 2 \left(\frac{x_3+x_1}{2}\right)}{1+2}, \frac{1 \cdot y_2 + 2 \left(\frac{y_3+y_1}{2}\right)}{1+2}\right] = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$ And for median CF: $$ G = \left[\frac{1 \cdot x_3 + 2 \left(\frac{x_1+x_2}{2}\right)}{1+2}, \frac{1 \cdot y_3 + 2 \left(\frac{y_1+y_2}{2}\right)}{1+2}\right] = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$ Since all three medians pass through the same point G, they are concurrent.

Thus, the medians of a triangle are concurrent.

18. Two ships are sailing in the sea on either side of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships are 30° and 45° respectively. If the lighthouse is 200m high. Find the distance between the two ships. (√3 = 1.732).

Solution: Let AB be the lighthouse of height h = 200 m. Let the two ships be at positions C and D on either side of the lighthouse. The angles of elevation are $\angle ACB = 30^{\circ}$ and $\angle ADB = 45^{\circ}$. In the right-angled triangle $\triangle ABC$: $$ \tan 30^{\circ} = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{200}{BC} $$ $$ BC = 200\sqrt{3} \text{ m} $$ In the right-angled triangle $\triangle ABD$: $$ \tan 45^{\circ} = \frac{AB}{BD} \Rightarrow 1 = \frac{200}{BD} $$ $$ BD = 200 \text{ m} $$ The distance between the two ships is CD = BC + BD. $$ CD = 200\sqrt{3} + 200 = 200(\sqrt{3} + 1) $$ Using $\sqrt{3} = 1.732$: $$ CD = 200(1.732 + 1) = 200(2.732) = 546.4 $$

The distance between the two ships is 546.4 meters.

19. From a window (h meters high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $\theta_1$ and $\theta_2$ respectively show that the height of the opposite house is $h\left(1+\frac{\cot \theta_2}{\cot \theta_1}\right)$.

There appears to be a typo in the question paper's formula. The correct formula is $h\left(1+\frac{\cot \theta_2}{\cot \theta_1}\right)$ or $h(1+\frac{\tan\theta_1}{\tan\theta_2})$. We will prove the correct formula as derived from the problem description.

Proof: Let W be the window at height $h$ from the ground A, so $AW=h$. Let PQ be the opposite house, with P being the top and Q being the foot. Let the horizontal distance between the houses be $x$. Draw a horizontal line WR from W to the house PQ. From the diagram, $WR = AQ = x$ and $RQ = AW = h$. The angle of elevation to P from W is $\angle PWR = \theta_1$. The angle of depression to Q from W is $\angle RWQ = \theta_2$. In right-angled triangle $\triangle WRQ$: $$ \tan \theta_2 = \frac{RQ}{WR} = \frac{h}{x} \Rightarrow x = \frac{h}{\tan \theta_2} = h \cot \theta_2 \quad \cdots(1)$$ In right-angled triangle $\triangle WRP$: $$ \tan \theta_1 = \frac{PR}{WR} = \frac{PR}{x} \Rightarrow PR = x \tan \theta_1 \quad \cdots(2)$$ Substitute the value of $x$ from (1) into (2): $$ PR = (h \cot \theta_2) \tan \theta_1 = h \frac{\cot \theta_2}{\cot \theta_1} $$ The total height of the opposite house is $H = PQ = PR + RQ$. $$ H = h \frac{\cot \theta_2}{\cot \theta_1} + h $$ $$ H = h\left(1 + \frac{\cot \theta_2}{\cot \theta_1}\right) $$

Hence, the height of the opposite house is $h\left(1 + \frac{\cot \theta_2}{\cot \theta_1}\right)$. (The formula given in the question was slightly different, this is the correct derivation).

20. The internal and external diameter of a hollow hemispherical shell are 6cm and 10cm respectively. If it is melted and recast into solid cylinder of diameter 14cm, then find the height of the cylinder.

Solution: For the hollow hemispherical shell: External diameter = 10 cm $\Rightarrow$ External radius, $R = 5$ cm. Internal diameter = 6 cm $\Rightarrow$ Internal radius, $r = 3$ cm. Volume of the hollow hemisphere = $\frac{2}{3}\pi(R^3 - r^3)$ $$ V_{hemi} = \frac{2}{3}\pi(5^3 - 3^3) = \frac{2}{3}\pi(125 - 27) = \frac{2}{3}\pi(98) \text{ cm}^3 $$ For the solid cylinder: Diameter = 14 cm $\Rightarrow$ Radius, $r_{cyl} = 7$ cm. Let the height be $h$. Volume of the cylinder = $\pi r_{cyl}^2 h = \pi (7)^2 h = 49\pi h \text{ cm}^3$. Since the shell is melted and recast into the cylinder, their volumes are equal. $$ V_{cyl} = V_{hemi} $$ $$ 49\pi h = \frac{2}{3}\pi(98) $$ $$ 49 h = \frac{2 \times 98}{3} $$ $$ h = \frac{2 \times 98}{3 \times 49} = \frac{2 \times 2}{3} = \frac{4}{3} $$

The height of the cylinder is $\frac{4}{3}$ cm or approximately 1.33 cm.

21. (Compulsory) If $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{bmatrix}$ show that $(AB)^T = B^T A^T$.

Solution: L.H.S = $(AB)^T$ First, calculate AB: $$ AB = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{bmatrix} $$ $$ = \begin{bmatrix} (1)(2)+(2)(-1)+(1)(0) & (1)(-1)+(2)(4)+(1)(2) \\ (2)(2)+(-1)(-1)+(1)(0) & (2)(-1)+(-1)(4)+(1)(2) \end{bmatrix} $$ $$ = \begin{bmatrix} 2-2+0 & -1+8+2 \\ 4+1+0 & -2-4+2 \end{bmatrix} = \begin{bmatrix} 0 & 9 \\ 5 & -4 \end{bmatrix} $$ Now, find the transpose $(AB)^T$: $$ (AB)^T = \begin{bmatrix} 0 & 5 \\ 9 & -4 \end{bmatrix} $$ R.H.S = $B^T A^T$ First, find the transposes $B^T$ and $A^T$: $$ B^T = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{bmatrix} $$ Now, calculate $B^T A^T$: $$ B^T A^T = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{bmatrix} $$ $$ = \begin{bmatrix} (2)(1)+(-1)(2)+(0)(1) & (2)(2)+(-1)(-1)+(0)(1) \\ (-1)(1)+(4)(2)+(2)(1) & (-1)(2)+(4)(-1)+(2)(1) \end{bmatrix} $$ $$ = \begin{bmatrix} 2-2+0 & 4+1+0 \\ -1+8+2 & -2-4+2 \end{bmatrix} = \begin{bmatrix} 0 & 5 \\ 9 & -4 \end{bmatrix} $$ Since L.H.S = R.H.S, the property is shown.

Hence, $(AB)^T = B^T A^T$ is shown.

PART - D

IV. Answer any one: (1 x 8 = 8)

22. a) Draw the two tangents from a point which is 5cm away from the centre of a circle of diameter 6 cm. Also measure the lengths of the tangents.

Given: Diameter of the circle = 6 cm, so Radius (r) = 3 cm. Distance of the external point from the centre (d) = 5 cm. Steps of Construction:

1. Draw a circle with center O and radius 3 cm.
2. Mark a point P, 5 cm away from the center O (i.e., OP = 5 cm).
3. Draw the perpendicular bisector of the line segment OP. Let it intersect OP at M. M is the midpoint of OP.
4. With M as the center and MO (or MP) as the radius, draw a new circle.
5. This new circle will intersect the original circle at two points, say A and B.
6. Join PA and PB. These are the required tangents to the circle from point P.

Measurement: On measuring the lengths of the tangents with a ruler, we find that PA = 4 cm and PB = 4 cm. Verification by Calculation: In the right-angled triangle $\triangle OAP$ (since radius OA is perpendicular to tangent PA): By Pythagoras theorem, $OP^2 = OA^2 + PA^2$. $$ 5^2 = 3^2 + PA^2 $$ $$ 25 = 9 + PA^2 $$ $$ PA^2 = 25 - 9 = 16 $$ $$ PA = \sqrt{16} = 4 \text{ cm} $$ The calculated length matches the measured length.

The lengths of the tangents are PA = 4 cm and PB = 4 cm.

22. b) Discuss the nature of solutions of the following quadratic equation. $x^2 + x - 12 = 0$.

Solution: The given quadratic equation is $x^2 + x - 12 = 0$. To determine the nature of the roots (solutions), we use the discriminant, $\Delta = b^2 - 4ac$. First, compare the given equation with the standard form $ax^2 + bx + c = 0$: Here, $a = 1$, $b = 1$, and $c = -12$. Now, calculate the discriminant: $$ \Delta = b^2 - 4ac $$ $$ \Delta = (1)^2 - 4(1)(-12) $$ $$ \Delta = 1 + 48 $$ $$ \Delta = 49 $$ Analysis of the Discriminant: The nature of the roots is determined by the value of $\Delta$:

1. Since $\Delta = 49 > 0$, the roots are real and unequal (distinct).
2. Since $\Delta = 49 = 7^2$, which is a perfect square, the roots are also rational.

We can also find the roots to confirm: Using the quadratic formula, $x = \frac{-b \pm \sqrt{\Delta}}{2a}$: $$ x = \frac{-1 \pm \sqrt{49}}{2(1)} = \frac{-1 \pm 7}{2} $$ The two roots are: $$ x_1 = \frac{-1 + 7}{2} = \frac{6}{2} = 3 $$ $$ x_2 = \frac{-1 - 7}{2} = \frac{-8}{2} = -4 $$ The roots are 3 and -4, which are indeed real, distinct, and rational.

The equation has two real, distinct, and rational roots.

10th Maths - Quarterly Exam 2025 - Questions Paper with Answer Key | Kallakurichi District| English Medium

10th Maths Quarterly Exam 2025-26 | Full Question Paper with Solutions

PART - I (14 x 1 = 14)

Choose the best answer:

1. If there are 1024 relations from a set A = {1,2,3,4,5} to a set B. Then the number of elements in B is

• a) 3
• b) 2
• c) 4
• d) 8

Answer: b) 2

Given, \( n(A) = 5 \). Let \( n(B) = m \).
The number of relations from A to B is \( 2^{n(A) \times n(B)} \).
We are given that this is 1024.
So, \( 2^{5 \times m} = 1024 \).
We know that \( 1024 = 2^{10} \).
Therefore, \( 2^{5m} = 2^{10} \).
Equating the powers, we get \( 5m = 10 \), which gives \( m = 2 \).
Thus, the number of elements in B is 2.

2. Let A = {1, 2, 3, 4}, B = {4, 8, 9, 10}. A function f : A→B given by f = {(1, 4), (2, 8), (3, 9), (4, 10)} is a

• a) Many-one function
• b) one-to-one function
• c) Identify function
• d) Into function

Answer: b) one-to-one function

In the given function f, every element in the domain A has a distinct image in the codomain B. (1→4, 2→8, 3→9, 4→10). Since no two elements of A have the same image in B, it is a one-to-one function.

3. Let \( f(x) = \sqrt{1+x^2} \) then

• a) \( f(xy) = f(x).f(y) \)
• b) \( f(xy) \ge f(x).f(y) \)
• c) \( f(xy) \le f(x).f(y) \)
• d) None of these

Answer: c) \( f(xy) \le f(x).f(y) \)

Let's evaluate both sides.
LHS: \( f(xy) = \sqrt{1+(xy)^2} = \sqrt{1+x^2y^2} \).
RHS: \( f(x) \cdot f(y) = \sqrt{1+x^2} \cdot \sqrt{1+y^2} = \sqrt{(1+x^2)(1+y^2)} = \sqrt{1+x^2+y^2+x^2y^2} \).
Since \(x^2 \ge 0\) and \(y^2 \ge 0\), the term \(x^2+y^2\) is always non-negative.
Therefore, \(1+x^2+y^2+x^2y^2 \ge 1+x^2y^2\).
Taking the square root of both sides, we get \( \sqrt{1+x^2+y^2+x^2y^2} \ge \sqrt{1+x^2y^2} \).
This means \( f(x)f(y) \ge f(xy) \), which is equivalent to \( f(xy) \le f(x)f(y) \).

4. If A = \( 2^{65} \) and B = \( 2^{64} + 2^{63} + 2^{62} + \dots + 2^0 \). which of the following is true?

• a) B is \( 2^{64} \) more than A
• b) A and B are equal
• c) B is larger than A by 1
• d) A is larger than B by 1

Answer: d) A is larger than B by 1

B is a geometric series with first term \( a = 2^0 = 1 \), common ratio \( r = 2 \), and number of terms \( n = 65 \).
The sum of a GP is \( S_n = a(r^n - 1)/(r-1) \).
So, \( B = 1(2^{65} - 1)/(2-1) = 2^{65} - 1 \).
Given, \( A = 2^{65} \).
Therefore, \( A = B + 1 \), which means A is larger than B by 1.

5. The value of \( (1^3+2^3+3^3+\dots+15^3) - (1+2+3+\dots+15) \) is

• a) 14400
• b) 14200
• c) 14280
• d) 14520

Answer: c) 14280

We know the formulas: \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \) and \( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \).
Here, \( n = 15 \).
Let \( S = 1+2+\dots+15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120 \).
The sum of cubes is \( S^2 = 120^2 = 14400 \).
The required value is \( S^2 - S = 14400 - 120 = 14280 \).

6. \( \frac{x}{x^2 - 25} - \frac{8}{x^2+6x+5} \) gives

• a) \( \frac{x^2 - 7x + 40}{(x-5)(x+5)} \)
• b) \( \frac{x^2 + 7x + 40}{(x-5)(x+5)(x+1)} \)
• c) \( \frac{x^2 - 7x + 40}{(x^2-25)(x+1)} \)
• d) \( \frac{x^2 + 10}{(x^2-25)(x+1)} \)

Answer: c) \( \frac{x^2 - 7x + 40}{(x^2-25)(x+1)} \)

First, factor the denominators: \( x^2 - 25 = (x-5)(x+5) \) and \( x^2+6x+5 = (x+5)(x+1) \).
The expression becomes: \( \frac{x}{(x-5)(x+5)} - \frac{8}{(x+5)(x+1)} \).
The LCM of the denominators is \( (x-5)(x+5)(x+1) \).
\( = \frac{x(x+1) - 8(x-5)}{(x-5)(x+5)(x+1)} \)
\( = \frac{x^2+x - 8x+40}{(x^2-25)(x+1)} \)
\( = \frac{x^2 - 7x + 40}{(x^2-25)(x+1)} \)

7. The number of excluded values for the expression \( \frac{x^3 - x^2 - 10x + 8}{x^4 + 8x^2 - 9} \)

• a) 1
• b) 2
• c) 3
• d) 4

Answer: b) 2

Excluded values are the roots of the denominator. Set \( x^4 + 8x^2 - 9 = 0 \).
Let \( y = x^2 \). The equation becomes \( y^2 + 8y - 9 = 0 \).
Factoring gives \( (y+9)(y-1) = 0 \).
So, \( y = -9 \) or \( y = 1 \).
Substituting back \( x^2 = y \):
Case 1: \( x^2 = -9 \). This has no real solutions.
Case 2: \( x^2 = 1 \). This gives \( x = 1 \) and \( x = -1 \).
There are 2 real excluded values.

8. If in triangle ABC and EDF, \( \frac{AB}{DE} = \frac{BC}{FD} \), then they will be similar, when

• a) ∠B = ∠E
• b) ∠A = ∠D
• c) ∠B = ∠D
• d) ∠A = ∠F

Answer: c) ∠B = ∠D

For two triangles to be similar by SAS (Side-Angle-Side) similarity criterion, the angle included between the two pairs of proportional sides must be equal.
In ΔABC, the angle between sides AB and BC is ∠B.
In ΔEDF, the angle between sides DE and FD is ∠D.
Therefore, for the triangles to be similar, we must have \( \angle B = \angle D \).

9. In a given figure ST||QR, PS=2cm and SQ=3cm. Then the ratio of the area of ΔPQR to area of ΔPST is

• a) 25:4
• b) 25:7
• c) 25:11
• d) 25:13

Answer: a) 25:4

Since ST || QR, by basic proportionality theorem, ΔPST is similar to ΔPQR (ΔPST ~ ΔPQR).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta PST)} = \left(\frac{PQ}{PS}\right)^2 \)
Given, PS = 2 cm and SQ = 3 cm. So, PQ = PS + SQ = 2 + 3 = 5 cm.
Ratio of areas = \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
The ratio is 25:4.

10. If slope of the line PQ is \( 1/\sqrt{3} \) then slope of the perpendicular bisector of PQ is

• a) \( \sqrt{3} \)
• b) \( -\sqrt{3} \)
• c) \( 1/\sqrt{3} \)
• d) 0

Answer: b) \( -\sqrt{3} \)

The perpendicular bisector is a line that is perpendicular to PQ. If two lines are perpendicular, the product of their slopes is -1.
Let the slope of PQ be \( m_1 = 1/\sqrt{3} \).
Let the slope of the perpendicular bisector be \( m_2 \).
Then, \( m_1 \times m_2 = -1 \).
\( (1/\sqrt{3}) \times m_2 = -1 \).
\( m_2 = -\sqrt{3} \).

11. The area of Quadrilateral formed by points (a, a), (-a, a), (-a, -a), (a, -a) is 64 sq.units. What is the value of a

• a) 4
• b) 6
• c) 8
• d) 12

Answer: a) 4

The given vertices form a square centered at the origin. The length of the side from (a, a) to (-a, a) is \( \sqrt{(a-(-a))^2 + (a-a)^2} = \sqrt{(2a)^2} = 2a \).
Area of a square = side\(^2\).
Area = \( (2a)^2 = 4a^2 \).
Given, Area = 64 sq. units.
\( 4a^2 = 64 \implies a^2 = 16 \implies a = 4 \) (assuming a > 0).

12. (2, 1) is the point of intersection of two lines

• a) x -y-3=0; 3x-y-7=0
• b) x+y=3; 3x+y=7
• c) 3x+y=3; x+y=7
• d) x+3y-3=0; x-y-7=0

Answer: b) x+y=3; 3x+y=7

We substitute the point (2, 1) into the equations of each option.
For option (b):
First line: \( x+y = 2+1 = 3 \). (Satisfied)
Second line: \( 3x+y = 3(2)+1 = 6+1 = 7 \). (Satisfied)
Since (2, 1) satisfies both equations, it is their point of intersection.

13. The value of \( \sin^2\theta + \frac{1}{1 + \tan^2\theta} \) is equal to

• a) tan²θ
• b) 1
• c) cot²θ
• d) 0

Answer: b) 1

We use the trigonometric identity \( 1 + \tan^2\theta = \sec^2\theta \).
The expression becomes \( \sin^2\theta + \frac{1}{\sec^2\theta} \).
Since \( \frac{1}{\sec^2\theta} = \cos^2\theta \), the expression is \( \sin^2\theta + \cos^2\theta \).
Using the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \).

14. If \( x = a\tan\theta \) and \( y = b\sec\theta \) then

• a) \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
• b) \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• c) \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
• d) \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 \)

Answer: a) \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)

From the given equations, we have \( \tan\theta = \frac{x}{a} \) and \( \sec\theta = \frac{y}{b} \).
We use the identity \( \sec^2\theta - \tan^2\theta = 1 \).
Substituting the expressions for \( \sec\theta \) and \( \tan\theta \):
\( \left(\frac{y}{b}\right)^2 - \left(\frac{x}{a}\right)^2 = 1 \)
\( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)

PART - II (10 x 2 = 20)

Answer any 10 questions. Question No.28 is compulsory.

15. Define identity function.

An identity function is a function that maps every element of a set to itself. For a set A, the identity function \( f: A \rightarrow A \) is defined by \( f(x) = x \) for all \( x \in A \).

16. If A = {1, 3, 5} B = {2, 3} then (i) find AxB and BxA.

Given, A = {1, 3, 5} and B = {2, 3}.
(i) A x B (Cartesian Product):
\( A \times B = \{(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)\} \)
(ii) B x A (Cartesian Product):
\( B \times A = \{(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)\} \)

17. Find K if f o f(k) = 5 where f(k) = 2k-1.

Given \( f(k) = 2k-1 \).
\( f \circ f(k) = f(f(k)) = f(2k-1) \)
\( = 2(2k-1) - 1 \)
\( = 4k - 2 - 1 = 4k - 3 \)
We are given \( f \circ f(k) = 5 \).
So, \( 4k - 3 = 5 \)
\( 4k = 8 \)
\( k = 2 \)

18. Find the least positive value of x such that 5x ≡ 4 (mod 6).

The congruence \( 5x \equiv 4 \pmod{6} \) means that \( 5x - 4 \) is a multiple of 6.
So, \( 5x - 4 = 6n \) for some integer \( n \).
We need to find the least positive integer \( x \) that satisfies this.
Let's test positive values for x:
If \( x=1 \), \( 5(1) - 4 = 1 \), which is not a multiple of 6.
If \( x=2 \), \( 5(2) - 4 = 10 - 4 = 6 \), which is a multiple of 6.
Therefore, the least positive value of x is 2.

19. If \( 1+2+3+\dots+n = 666 \) then find n.

The sum of the first n natural numbers is given by the formula \( \frac{n(n+1)}{2} \).
Given, \( \frac{n(n+1)}{2} = 666 \).
\( n(n+1) = 666 \times 2 = 1332 \)
\( n^2 + n - 1332 = 0 \)
We need to find two consecutive integers whose product is 1332. We can factorize 1332: \( 1332 = 2 \times 666 = 2 \times 2 \times 333 = 4 \times 9 \times 37 = 36 \times 37 \).
So, \( n(n+1) = 36 \times 37 \).
Therefore, \( n = 36 \).

20. Simplify: \( \frac{x}{x-y} + \frac{y}{y-x} \)

We can rewrite the second term by factoring out -1 from the denominator:
\( \frac{y}{y-x} = \frac{y}{-(x-y)} = -\frac{y}{x-y} \)
Now the expression becomes:
\( \frac{x}{x-y} - \frac{y}{x-y} \)
Since the denominators are the same, we can combine the numerators:
\( \frac{x-y}{x-y} = 1 \) (provided \( x \ne y \)).

21. Find the sum and and product of the roots of the quadratic equation \( 3y^2-y-4=0 \).

For a quadratic equation \( ax^2+bx+c=0 \),
Sum of roots \( (\alpha + \beta) = -b/a \)
Product of roots \( (\alpha\beta) = c/a \)
In the given equation \( 3y^2-y-4=0 \), we have \( a=3, b=-1, c=-4 \).
Sum of roots = \( -(-1)/3 = 1/3 \).
Product of roots = \( -4/3 \).

22. If the difference between a number and its reciprocal is 24/5, find the number.

Let the number be \( x \). Its reciprocal is \( 1/x \).
Given, \( x - \frac{1}{x} = \frac{24}{5} \).
Multiplying the entire equation by \( 5x \) to clear fractions:
\( 5x^2 - 5 = 24x \)
\( 5x^2 - 24x - 5 = 0 \)
Factoring the quadratic equation:
\( 5x^2 - 25x + x - 5 = 0 \)
\( 5x(x - 5) + 1(x - 5) = 0 \)
\( (5x + 1)(x - 5) = 0 \)
The possible values for the number are \( x = 5 \) or \( x = -1/5 \).

23. In the figure DE||AC and DC||AP, Prove that \( \frac{BE}{EC} = \frac{BC}{CP} \).

Proof:
Step 1: In ΔABC, we are given that DE || AC.
By the Basic Proportionality Theorem (Thales' Theorem), if a line is parallel to one side of a triangle, it divides the other two sides proportionally.
Therefore, \( \frac{BE}{EC} = \frac{BD}{DA} \) --- (1)
Step 2: In ΔABP, we are given that DC || AP.
Again, by the Basic Proportionality Theorem,
\( \frac{BC}{CP} = \frac{BD}{DA} \) --- (2)
Step 3: From equations (1) and (2), we can see that both ratios are equal to \( \frac{BD}{DA} \).
Therefore, we can equate them:
\( \frac{BE}{EC} = \frac{BC}{CP} \)
Hence Proved.

24. Find the value of 'a' for which the given points are collinear (2, 3) (4, a) and (6, -3).

If three points are collinear, the area of the triangle formed by them is 0.
The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Setting the area to 0:
\( \frac{1}{2} |2(a - (-3)) + 4(-3 - 3) + 6(3 - a)| = 0 \)
\( |2(a + 3) + 4(-6) + 6(3 - a)| = 0 \)
\( |2a + 6 - 24 + 18 - 6a| = 0 \)
\( |-4a| = 0 \)
\( 4a = 0 \implies a = 0 \)

25. A cat is located at the point (-6, -4) in xy plane. A bottle of milk is kept at (5, 11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take.

The shortest path between two points is a straight line. We need to find the equation of the line passing through A(-6, -4) and B(5, 11).
First, find the slope (m):
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - (-4)}{5 - (-6)} = \frac{15}{11} \)
Now, use the point-slope form of the equation: \( y - y_1 = m(x - x_1) \)
Using point A(-6, -4):
\( y - (-4) = \frac{15}{11}(x - (-6)) \)
\( y + 4 = \frac{15}{11}(x + 6) \)
\( 11(y + 4) = 15(x + 6) \)
\( 11y + 44 = 15x + 90 \)
\( 15x - 11y + 90 - 44 = 0 \)
\( 15x - 11y + 46 = 0 \)
This is the equation of the path.

26. Find the equation of a straight line which is parallel to the line 3x-7y=12 and passing through the point (6, 4).

The equation of a line parallel to \( ax + by + c = 0 \) is of the form \( ax + by + k = 0 \).
So, the equation of the line parallel to \( 3x - 7y = 12 \) is \( 3x - 7y + k = 0 \).
This line passes through the point (6, 4). We substitute these values to find k:
\( 3(6) - 7(4) + k = 0 \)
\( 18 - 28 + k = 0 \)
\( -10 + k = 0 \implies k = 10 \)
The required equation is \( 3x - 7y + 10 = 0 \).

27. Prove that \( \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sec\theta + \tan\theta \). (Assuming the standard form of this question)

LHS = \( \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} \)
Multiply the numerator and denominator inside the square root by the conjugate of the denominator, which is \( (1 + \sin\theta) \):
LHS = \( \sqrt{\frac{(1 + \sin\theta)(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}} \)
= \( \sqrt{\frac{(1 + \sin\theta)^2}{1 - \sin^2\theta}} \)
Using the identity \( \cos^2\theta = 1 - \sin^2\theta \):
= \( \sqrt{\frac{(1 + \sin\theta)^2}{\cos^2\theta}} \)
= \( \frac{1 + \sin\theta}{\cos\theta} \)
= \( \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \)
= \( \sec\theta + \tan\theta \) = RHS
Hence Proved.

28. (Compulsory) If in an A.P, a = -10, d = 2, how many terms are needed to reach the term 0?

Let the A.P be \( T_1, T_2, T_3, \dots \)
Given the first term \( a = -10 \) and the common difference \( d = 2 \).
We want to find which term, \( T_n \), is equal to 0.
The formula for the n-th term of an A.P is \( T_n = a + (n-1)d \).
Set \( T_n = 0 \):
\( 0 = -10 + (n-1)2 \)
\( 10 = (n-1)2 \)
\( 5 = n-1 \)
\( n = 6 \)
So, the 6th term of the A.P is 0. 6 terms are needed to reach the term 0.

PART - III (10 x 5 = 50)

Answer any 10 questions. Q.No.42 is compulsory.

29. Let A = The set of all natural numbers less than 8, B=The set of all prime numbers less than 8, C=The set of even prime number. Verify that (A∩B) x C = (AxC) ∩ (BxC).

First, let's write the sets in roster form:
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2} (2 is the only even prime number)

LHS: (A∩B) x C
A∩B = {2, 3, 5, 7}
(A∩B) x C = {2, 3, 5, 7} x {2} = {(2, 2), (3, 2), (5, 2), (7, 2)} --- (1)

RHS: (AxC) ∩ (BxC)
A x C = {1, 2, 3, 4, 5, 6, 7} x {2} = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
B x C = {2, 3, 5, 7} x {2} = {(2, 2), (3, 2), (5, 2), (7, 2)}
(AxC) ∩ (BxC) = {(2, 2), (3, 2), (5, 2), (7, 2)} --- (2)

From (1) and (2), we see that LHS = RHS. Hence verified.

30. A function f: [-5, 9] → R is defined as follows:
\( f(x) = \begin{cases} 6x + 1 & -5 \le x < 2 \\ 5x^2 - 1 & 2 \le x < 6 \\ 3x - 4 & 6 \le x \le 9 \end{cases} \).
Find
i) f(-3)+f(2)
ii) f(7)-f(1)
iii) 2f(4)+f(8)
iv) \( \frac{2f(-2)-f(6)}{f(4)+f(-2)} \)

i) f(-3) + f(2)
f(-3) is in the interval \( -5 \le x < 2 \), so we use \( 6x+1 \). f(-3) = 6(-3) + 1 = -17.
f(2) is in the interval \( 2 \le x < 6 \), so we use \( 5x^2-1 \). f(2) = 5(2)² - 1 = 19.
f(-3) + f(2) = -17 + 19 = 2.

ii) f(7) - f(1)
f(7) is in the interval \( 6 \le x \le 9 \), so we use \( 3x-4 \). f(7) = 3(7) - 4 = 17.
f(1) is in the interval \( -5 \le x < 2 \), so we use \( 6x+1 \). f(1) = 6(1) + 1 = 7.
f(7) - f(1) = 17 - 7 = 10.

iii) 2f(4) + f(8)
f(4) is in the interval \( 2 \le x < 6 \), so we use \( 5x^2-1 \). f(4) = 5(4)² - 1 = 79.
f(8) is in the interval \( 6 \le x \le 9 \), so we use \( 3x-4 \). f(8) = 3(8) - 4 = 20.
2f(4) + f(8) = 2(79) + 20 = 158 + 20 = 178.

iv) \( \frac{2f(-2)-f(6)}{f(4)+f(-2)} \)
f(-2) is in \( -5 \le x < 2 \), so f(-2) = 6(-2) + 1 = -11.
f(6) is in \( 6 \le x \le 9 \), so f(6) = 3(6) - 4 = 14.
f(4) = 79 (from part iii).
The expression is \( \frac{2(-11) - 14}{79 + (-11)} = \frac{-22 - 14}{68} = \frac{-36}{68} = -\frac{9}{17} \).

31. If f(x) = 2x+3, g(x) = 1-2x and h(x)=3x. Prove that fo(goh)=(fog)oh.

To prove the associative property of function composition, we will evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately.

LHS = f o (g o h)

First, find g o h(x):
\( g \circ h(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x \)

Now, find f o (g o h)(x):
\( f \circ (g \circ h)(x) = f(g \circ h(x)) = f(1 - 6x) \)
\( = 2(1 - 6x) + 3 \)
\( = 2 - 12x + 3 \)
\( = 5 - 12x \) --- (1)

RHS = (f o g) o h

First, find f o g(x):
\( f \circ g(x) = f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 \)
\( = 2 - 4x + 3 \)
\( = 5 - 4x \)

Now, find (f o g) o h(x):
\( (f \circ g) \circ h(x) = (f \circ g)(h(x)) = (f \circ g)(3x) \)
\( = 5 - 4(3x) \)
\( = 5 - 12x \) --- (2)

From (1) and (2), we see that LHS = RHS. Hence, \( f \circ (g \circ h) = (f \circ g) \circ h \) is proved.

32. Find the HCF of 396, 504, 636.

We will use Euclid's Division Algorithm to find the HCF.

Step 1: Find HCF of 504 and 396.
\( 504 = 396 \times 1 + 108 \)
\( 396 = 108 \times 3 + 72 \)
\( 108 = 72 \times 1 + 36 \)
\( 72 = 36 \times 2 + 0 \)
The remainder is 0. So, the HCF of 504 and 396 is 36.

Step 2: Find HCF of 636 and the result from Step 1 (which is 36).
\( 636 = 36 \times 17 + 24 \)
\( 36 = 24 \times 1 + 12 \)
\( 24 = 12 \times 2 + 0 \)
The remainder is 0. So, the HCF of 636 and 36 is 12.

Therefore, the HCF of 396, 504, and 636 is 12.

33. The sum of first n, 2n and 3n terms of an A.P. are S₁, S₂, and S₃ respectively. Prove that S₃ = 3(S₂ - S₁).

Let the first term of the A.P be 'a' and the common difference be 'd'. The formula for the sum of the first 'k' terms of an A.P is \( S_k = \frac{k}{2}[2a + (k-1)d] \).

Using this formula, we can write the expressions for S₁, S₂, and S₃:
\( S_1 = \frac{n}{2}[2a + (n-1)d] \) --- (1)
\( S_2 = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] \) --- (2)
\( S_3 = \frac{3n}{2}[2a + (3n-1)d] \) --- (3)

Now, let's evaluate the RHS of the equation we need to prove: 3(S₂ - S₁)

First, calculate S₂ - S₁:
\( S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] \)
Take \( \frac{n}{2} \) as a common factor:
\( = \frac{n}{2} \left( 2[2a + (2n-1)d] - [2a + (n-1)d] \right) \)
\( = \frac{n}{2} [ (4a + (4n-2)d) - (2a + (n-1)d) ] \)
\( = \frac{n}{2} [ 4a - 2a + (4n-2)d - (n-1)d ] \)
\( = \frac{n}{2} [ 2a + (4n-2 - n+1)d ] \)
\( = \frac{n}{2} [ 2a + (3n-1)d ] \)

Now, multiply by 3:
\( 3(S_2 - S_1) = 3 \times \frac{n}{2} [ 2a + (3n-1)d ] \)
\( = \frac{3n}{2} [ 2a + (3n-1)d ] \)
This is exactly the expression for S₃ from equation (3).

Thus, LHS (S₃) = RHS (3(S₂ - S₁)). Hence proved.

34. Find the values of m and n if the polynomial is a perfect square: 36x⁴ - 60x³ + 61x² - mx + n.

We use the long division method to find the square root of the polynomial. For it to be a perfect square, the remainder must be zero.

                 6x²  - 5x   + 3
               _________________________
      6x²      | 36x⁴ - 60x³ + 61x² - mx + n
               |-(36x⁴)
               |_________________________
      12x²-5x  |      - 60x³ + 61x²
               |    -(- 60x³ + 25x²)
               |      ___________________
      12x²-10x+3|             36x² - mx + n
                |           -(36x² - 30x + 9)
                |             ________________
                |                    (-m+30)x + (n-9)
                

Explanation of Steps:

1. The square root of \(36x^4\) is \(6x^2\). Write this in the quotient and divisor. Subtract \( (6x^2)^2 = 36x^4 \).
2. Bring down the next two terms: \(-60x^3 + 61x^2\). Double the current quotient to get the new divisor's first part: \(2 \times 6x^2 = 12x^2\).
3. Divide the first term of the new dividend (\(-60x^3\)) by the first term of the new divisor (\(12x^2\)) to get \(-5x\). This is the next term in the quotient and divisor.
4. Multiply \(-5x\) by the new divisor \( (12x^2 - 5x) \) to get \(-60x^3 + 25x^2\). Subtract this.
5. Bring down the last two terms: \(-mx + n\). The new dividend is \(36x^2 - mx + n\). Double the current quotient \((6x^2 - 5x)\) to get \(12x^2 - 10x\).
6. Divide \(36x^2\) by \(12x^2\) to get \(+3\). This is the last term in the quotient and divisor.
7. Multiply \(+3\) by the new divisor \((12x^2 - 10x + 3)\) to get \(36x^2 - 30x + 9\). Subtract this.

The remainder is \( (-m+30)x + (n-9) \).
Since the polynomial is a perfect square, the remainder must be 0.
\( (-m+30)x + (n-9) = 0x + 0 \)

Equating the coefficients of x and the constant terms to zero:
\( -m + 30 = 0 \implies m = 30 \)
\( n - 9 = 0 \implies n = 9 \)

Therefore, the values are m = 30 and n = 9.

35. Solve: pq x² - (p+q)² x + (p+q)² = 0.

This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \), where:
A = pq
B = \( -(p+q)^2 \)
C = \( (p+q)^2 \)

We use the quadratic formula to solve for x: \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)

Substitute the values of A, B, and C:
\( x = \frac{-(-(p+q)^2) \pm \sqrt{(-(p+q)^2)^2 - 4(pq)((p+q)^2)}}{2(pq)} \)
\( x = \frac{(p+q)^2 \pm \sqrt{(p+q)^4 - 4pq(p+q)^2}}{2pq} \)

Factor out \( (p+q)^2 \) from inside the square root:
\( x = \frac{(p+q)^2 \pm \sqrt{(p+q)^2 [(p+q)^2 - 4pq]}}{2pq} \)

Simplify the term inside the brackets: \( (p+q)^2 - 4pq = p^2 + 2pq + q^2 - 4pq = p^2 - 2pq + q^2 = (p-q)^2 \)

\( x = \frac{(p+q)^2 \pm \sqrt{(p+q)^2 (p-q)^2}}{2pq} \)
\( x = \frac{(p+q)^2 \pm (p+q)(p-q)}{2pq} \)

Now we find the two possible solutions for x:
Case 1 (with '+'):
\( x = \frac{(p+q)^2 + (p+q)(p-q)}{2pq} = \frac{(p+q)[(p+q) + (p-q)]}{2pq} \)
\( x = \frac{(p+q)[2p]}{2pq} = \frac{p+q}{q} \)

Case 2 (with '-'):
\( x = \frac{(p+q)^2 - (p+q)(p-q)}{2pq} = \frac{(p+q)[(p+q) - (p-q)]}{2pq} \)
\( x = \frac{(p+q)[p+q-p+q]}{2pq} = \frac{(p+q)[2q]}{2pq} = \frac{p+q}{p} \)

The solutions are \( x = \frac{p+q}{p} \) and \( x = \frac{p+q}{q} \).

36. State and prove Basic proportionality theorem.

Statement (Thales' Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: In ΔABC, a line DE is parallel to BC (DE || BC), intersecting AB at D and AC at E.

To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \)

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

Proof:

In ∆ABC, D is a point on AB and E is a point on AC.

To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)

Construction: Draw a line DE || BC

No.	Statement	Reason
1.	∠ABC = ∠ADE = ∠1	Corresponding angles are equal because DE || BC
2.	∠ACB = ∠AED = ∠2	Corresponding angles are equal because DE || BC
3.	∠DAE = ∠BAC = ∠3	Both triangles have a common angle
	∆ABC ~ ∆ADE	By AAA similarity
	\(\frac{AB}{AD} = \frac{AC}{AE}\)	Corresponding sides are proportional
	\(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\)	Split AB and AC using the points D and E.
	\(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\)	On simplification
	\(\frac{DB}{AD} = \frac{EC}{AE}\)	Cancelling 1 on both sides
	\(\frac{AD}{DB} = \frac{AE}{EC}\)	Taking reciprocals. Hence proved.

37. Find the area of the quadrilateral whose vertices are at (-9, -2), (-8, -4), (2, 2) and (1, -3).

To find the area of the quadrilateral, we use the shoelace formula. First, let's arrange the vertices in counter-clockwise order to ensure a positive area. A rough plot shows the order A(-9, -2), B(-8, -4), D(1, -3), C(2, 2) is counter-clockwise.

Let the vertices be:
\( (x_1, y_1) = (-9, -2) \)
\( (x_2, y_2) = (-8, -4) \)
\( (x_3, y_3) = (1, -3) \)
\( (x_4, y_4) = (2, 2) \)

The formula for the area is: \( \text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \)

Let's calculate the two parts:

Part 1: \( (x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) \)
\( = (-9)(-4) + (-8)(-3) + (1)(2) + (2)(-2) \)
\( = 36 + 24 + 2 - 4 = 58 \)

Part 2: \( (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \)
\( = (-2)(-8) + (-4)(1) + (-3)(2) + (2)(-9) \)
\( = 16 - 4 - 6 - 18 = -12 \)

Now, substitute these values back into the area formula:
\( \text{Area} = \frac{1}{2} |58 - (-12)| \)
\( = \frac{1}{2} |58 + 12| \)
\( = \frac{1}{2} |70| = 35 \)

The area of the quadrilateral is 35 square units.

38. Find the equation of a straight line through the point of intersection of the lines 8x+3y=18, 4x+5y=9 and bisecting the line segment joining the points (5, -4) and (-7, 6).

Step 1: Find the point of intersection of the two lines.

Line 1: \( 8x + 3y = 18 \) --- (1)
Line 2: \( 4x + 5y = 9 \) --- (2)
Multiply equation (2) by 2: \( 8x + 10y = 18 \) --- (3)
Subtract equation (1) from (3):
\( (8x + 10y) - (8x + 3y) = 18 - 18 \)
\( 7y = 0 \implies y = 0 \)
Substitute \(y = 0\) into equation (2):
\( 4x + 5(0) = 9 \implies 4x = 9 \implies x = 9/4 \)
So, the point of intersection is P(9/4, 0).

Step 2: Find the midpoint of the line segment joining (5, -4) and (-7, 6).

Midpoint M = \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
M = \( \left( \frac{5+(-7)}{2}, \frac{-4+6}{2} \right) = \left( \frac{-2}{2}, \frac{2}{2} \right) = (-1, 1) \)

Step 3: Find the equation of the line passing through P(9/4, 0) and M(-1, 1).

Using the two-point form \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \):
Slope (m) = \( \frac{1 - 0}{-1 - 9/4} = \frac{1}{-13/4} = -\frac{4}{13} \)
Using point M(-1, 1) and the slope:
\( y - 1 = -\frac{4}{13}(x - (-1)) \)
\( y - 1 = -\frac{4}{13}(x + 1) \)
\( 13(y - 1) = -4(x + 1) \)
\( 13y - 13 = -4x - 4 \)
\( 4x + 13y - 13 + 4 = 0 \)
\( 4x + 13y - 9 = 0 \)

The required equation of the straight line is 4x + 13y - 9 = 0.

39. A mobile phone is put to use when the battery power is 100%. The percent of battery power 'y' (in decimal) remaining after using the mobile phone for x hours is assumed as y = -0.25x+1.
(i) Find the number of hours elapsed if the battery power is 40%.
(ii) How much time does it take so that the battery has no power?

The given relation is \( y = -0.25x + 1 \), where 'y' is the battery power in decimal and 'x' is the time in hours.

(i) Find x when battery power is 40%.

40% battery power means \( y = 0.40 \).
Substitute this value into the equation:
\( 0.40 = -0.25x + 1 \)
\( 0.25x = 1 - 0.40 \)
\( 0.25x = 0.60 \)
\( x = \frac{0.60}{0.25} = \frac{60}{25} = \frac{12}{5} = 2.4 \)
So, 2.4 hours have elapsed.

(ii) Find x when the battery has no power.

No power means the battery is at 0%, so \( y = 0 \).
Substitute this value into the equation:
\( 0 = -0.25x + 1 \)
\( 0.25x = 1 \)
\( x = \frac{1}{0.25} = 4 \)
It takes 4 hours for the battery to have no power.

40. A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

This is a problem of geometric progression (GP).

The starting salary (first term, a) = ₹60,000.
The annual increase is 5%. This means each year the salary becomes 105% of the previous year's salary.
The common ratio (r) = \( 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \).

"Salary after 5 years" refers to the salary at the beginning of the 6th year. We need to find the 6th term (\( T_6 \)) of the GP.

The formula for the n-th term of a GP is \( T_n = ar^{n-1} \).
For the 6th year (\(n=6\)):
\( T_6 = 60000 \times (1.05)^{6-1} \)
\( T_6 = 60000 \times (1.05)^5 \)

Now, let's calculate \( (1.05)^5 \):
\( (1.05)^2 = 1.1025 \)
\( (1.05)^4 = (1.1025)^2 = 1.21550625 \)
\( (1.05)^5 = 1.21550625 \times 1.05 = 1.2762815625 \)

Salary after 5 years = \( 60000 \times 1.2762815625 \)
= 76576.89375

Rounding his salary after 5 years will be ₹76,577.

41. Prove that sin²A cos²B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1.

We start with the Left Hand Side (LHS) of the equation:
LHS = \( \sin^2A \cos^2B + \cos^2A \sin^2B + \cos^2A \cos^2B + \sin^2A \sin^2B \)

Let's rearrange and group the terms with common factors:

Group 1 (terms with \( \sin^2A \)):
\( (\sin^2A \cos^2B + \sin^2A \sin^2B) \)
Factor out \( \sin^2A \):
\( = \sin^2A (\cos^2B + \sin^2B) \)
Using the identity \( \cos^2\theta + \sin^2\theta = 1 \), this simplifies to:
\( = \sin^2A (1) = \sin^2A \)

Group 2 (terms with \( \cos^2A \)):
\( (\cos^2A \sin^2B + \cos^2A \cos^2B) \)
Factor out \( \cos^2A \):
\( = \cos^2A (\sin^2B + \cos^2B) \)
Using the same identity, this simplifies to:
\( = \cos^2A (1) = \cos^2A \)

Now, add the results of the two groups:
LHS = (Result of Group 1) + (Result of Group 2)
LHS = \( \sin^2A + \cos^2A \)

Using the fundamental Pythagorean identity \( \sin^2A + \cos^2A = 1 \).
Therefore, LHS = 1.

Since LHS = 1 and RHS = 1, the identity is proved.

42. (Compulsory) If \( f(x) = x^2 - 1 \), \( g(x) = x^3 - 1 \) verify that \( f(x) \times g(x) = \text{LCM}(f(x), g(x)) \times \text{GCD}(f(x), g(x)) \).

Step 1: Factorize f(x) and g(x)
\( f(x) = x^2 - 1 = (x-1)(x+1) \)
\( g(x) = x^3 - 1 = (x-1)(x^2+x+1) \)

Step 2: Find GCD (Greatest Common Divisor)
The common factor is \( (x-1) \).
\( \text{GCD}(f(x), g(x)) = (x-1) \)

Step 3: Find LCM (Least Common Multiple)
The LCM is the product of the GCD and the remaining uncommon factors.
\( \text{LCM}(f(x), g(x)) = (x-1)(x+1)(x^2+x+1) \)

Step 4: Verify the property
LHS: \( f(x) \times g(x) = (x^2-1)(x^3-1) = (x-1)(x+1)(x-1)(x^2+x+1) \)

RHS: \( \text{LCM} \times \text{GCD} = [(x-1)(x+1)(x^2+x+1)] \times [(x-1)] \)
\( = (x-1)(x+1)(x-1)(x^2+x+1) \)

Since LHS = RHS, the property is verified.

PART - IV (2 x 8 = 16)

Answer all of the following.

43. a) Construct a triangle similar to a given triangle PQR with its sides equal to 7/3 of the corresponding sides of the triangle PQR (scale factor 7/3 > 1).

Steps of Construction:

Since the scale factor is \(\frac{7}{3}\), which is greater than 1, the new triangle will be larger than the original triangle PQR.

1. Draw a triangle PQR with any suitable measurements.
2. Draw a ray QX making an acute angle with QR, on the side opposite to vertex P.
3. Locate 7 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7\) on the ray QX such that the distances between them are equal (\(QQ_1 = Q_1Q_2 = \dots = Q_6Q_7\)).
4. Join \(Q_3\) (the 3rd point, as 3 is the denominator) to R.
5. Draw a line through \(Q_7\) parallel to \(Q_3R\). This line will intersect the extended line segment QR at a point R'.
6. Draw a line through R' parallel to PR. This line will intersect the extended line segment QP at a point P'.
7. \(\triangle P'QR'\) is the required similar triangle, with each side being \(\frac{7}{3}\) times the corresponding side of \(\triangle PQR\).

43. b) (OR) Draw a triangle ABC of base BC=8cm, ∠A=60° and the bisector of ∠A meets BC at D such that BD=6cm.

Steps of Construction:

44. a) A company initially started with 40 worker to complete the work by 150 days. Later it decided to fastern up the work increasing the number of workers as shown below.

Number of workers (x)	40	50	60	75
Number of days (y)	150	120	100	80

(i) Graph the above data and identify the type of variation.
(ii) From the graph find the number of days required to complete the work if the company decides to opt for 120 worker?
(iii) If the work has to be completed by 200 days how many workers are required?

44. b) (OR) A garment shop announces a flat 50% discount on every purchase of items for their customers. Draw the graph for the relation between the marked price and the discount. Hence find
(i) the marked price when a customer gets a discount of ₹3250 (from graph)
(ii) the discount when the marked price is ₹2500.

Original Question Paper

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