SECTION - AI. i) Very short answer type questions.

ii) Answer ALL questions.iii) Each question carries TWO marks. 10 ×× 2 = 20

1. What are Porins? What role do they play in diffusion?2. Which element is regarded as the 17th essential element?

Name a disease caused by its deficiency?3. What are pleomorphic bacteria? Give an example.4. Explain the terms phenotype and genotype?5. Distinguish between heterochromatin and euchromatin.

Which of the two is transcriptionally active?6. What is Eco R1? How does it function?7. Give different types of cry genes and pests which are con-

trolled by the proteins encoded by these genes? 8. What is meant by “hidden hunger”?

9. Name a microbe used for statin production. How do statinslower blood cholesterol level?

10. Name two amino acids in which sulphur is present?

SECTION - BII. i) Short answer type questions.

ii) Answer any SIX questions.iii) Each question carries FOUR marks. 6 ×× 4 = 24

11. Write short notes on facilitated diffusion?12. Explain the steps involved in the formation of root nodule.13. Explain the structure of chloroplast. Draw a neat labelled

diagram.14. Write short notes on seed dormancy.15. Explain the structure of TMV.16. Explain co-dominance with example.

17. Draw the schematic representation or diagrammaticpresentation of the lac operon.

18. List out the beneficial aspects of transgenic plants.

SECTION - CIII. i) Long answer type questions.

ii) Answer any TWO questions.iii) Each question carries EIGHT marks. 2 ×× 8 = 16

19. Give an account of Glycolysis. Where does it occur? Whatare the end products? Trace the fate of these products inboth aerobic and anaerobic respiration?

20. Explain briefly the various processes of recombinant DNAtechnology.

21. Describe the Tissue culture technique and what are theadvantages of tissue culture over conventional method ofplant breeding in crop improvement programmes?

email: help@eenadupratibha.net÷ªÙÞœüŒî¦ô¢Ù 18 íƇvñ÷J 2020

šú¤ÛûÂ - A

I.- i) ÍA-ú£yõp ú£÷«-ëůì êŸô¢ï£„ ví£øŒoõª.-ii) ÍEo ví£øŒo-õÚÛª ú£÷«-ëů-û¦õª ô¦óŸªÙè….-iii) ví£A ví£øŒoÚÛª ·ôÙè[ª ÷«ô¢ª\õª.- 10 ×× 2 = 20

1. ð¼J-ûÂõª ÍÙç˶ ÔNªæ¨? Nú£-ô¢é àŸô¢uö˺ î¦æ¨ ð§vêŸ êμõ-í£Ùè….- 2.- Ô ÜEá ÷´õ-Ú¥Eo 17÷ Î÷-øŒuÚÛ ÷´õ-ÚÛÙÞ¥ ÞœªJhÙ-à¦ô¢ª? ë¯E

ö˺í£Ù ÷õx ÚÛL¸Þ î¦uCÅE êμõ-í£Ùè….-3.- ñ-ô¢«í£ ò°uÚ©d-J-óŸª-îªõª ÍÙç˶ ÔNªæ¨? Ñë¯--ô¢é ô¦óŸªÙè….-4.- áìªu-ô¢«í£Ù, ë]”øŒu-ô¢«í£Ù Íû¶ í£ë¯-õìª N÷-JÙ-àŸÙè….-5.- šï°æ¨-ôÁ-vÚ•-÷«-æ¨ûÂ, óŸ´vÚ•-÷«-æ¨-ûÂÚÛª Þœõ òÅ˶ë¯-õìª êμõ-í£Ùè….-

Íìª-ö˶-ÜìÙ Kê¦u ÔC vÚ¨óŸ«-êŸt-ÚÛÙÞ¥ ÑÙåªÙC?6.- Eco R1 ÍÙç˶ ÔNªæ¨? ÍC ÔN-ëÅ]ÙÞ¥ í£E-à¶-ú£ªhÙC?

7.- Cry á-ìªu-÷±õª, <è[õ NNëÅ] ô¢Ú¥-õìª êμLí‡, Ð áìªu-÷±-õêÁEóŸªÙ-vAÙ-àŸ-ñè˶ vð»æ©-ìxìª êμõ-í£Ùè….-

8.- -cÚÛE-í‡Ù-àŸE ÎÚÛLe- (hidden hunger) ÍÙç˶ ÔNªæ¨?9.- þ§dæ¨û ÑêŸpAh ÚÁú£Ù Ñí£-óμ«-TÙචú£«¤Ût-@N ›íô¢ªìª êμõ-í£Ùè….-

ô¢ÚÛhÙ-ö˺E Ú•÷±y þ§–ô³E êŸT_Ù-àŸ-è¯-EÚ¨ Ð þ§dæ¨ìªx ÔN-ëÅ]ÙÞ¥ Ñí£-óμ«-Þœ-í£-è[-ê¦ô³?

10.- ú£õpÄô ÚÛL-Tì ·ôÙè[ª ÍîμªiûÁ Î÷«xõ ›íô¢xìª êμõ-í£Ùè….-

šú¤ÛûÂ - B

II. i) ú£yõp ú£÷«-ëůì êŸô¢ï£„ ví£øŒoõª.-ii) Ôî�μjû¦ Îô¢ª ví£øŒo-õÚÛª ú£÷«-ëů-û¦õª ô¦óŸªÙè….-iii) ví£A ví£øŒoÚÛª û¦õªÞœª ÷«ô¢ª\õª.- 6 ×× 4 = 24

11.- ú£ªõ-òÅ¡-êŸô¢ Nú£-ô¢-éšíj õíÆ£³-æ©ÚÛ ô¦óŸªÙè….-12.- î�¶ô¢ª-ñª-è…-šíõª Ôô¢pè˶ Nëů-ìÙ-ö˺E NNëÅ] ë]øŒ-õìª N÷-JÙ-àŸÙè….- 13.- J-êŸ-¸ô-éª÷± Eô¦t-é°Eo NøŒ-D-ÚÛ-JÙ-àŸÙè….- ë¯E í£åÙ Uú‡ òÅ°Þ¥õª

ÞœªJhÙ-àŸÙè….- 14.- NêŸhì ú£ªð§h-÷ú£– Oªë] õíÆ£³-æ©ÚÛ ô¦óŸªÙè….-

15.- TMV E-ô¦t-é°Eo N÷-JÙ-àŸÙè….- 16.- ú£ï£°-ñ--ô¢_-êŸ-ê¦yEo Ñë¯--ô¢-éêÁ N÷-JÙ-àŸÙè….- 17.- ö°Ú ֚í-ô¦û ì÷´ì í£æ°Eo UóŸªÙè….-18.- áìªu-í£-J-÷-JhêŸ îμ³ÚÛ\õ ÷õx ÚÛL¸Þ Ñí£-óμ«-Þ¥õ í£æ¨d-ÚÛìª ô¦óŸªÙè….-

šú¤ÛûÂ - C

III. i) Dô¢+ ú£÷«-ëůì êŸô¢ï£„ ví£øŒoõª.ii) Ôî�μjû¦ ·ôÙè[ª ví£øŒo-õÚÛª ú£÷«-ëů-û¦õª ô¦óŸªÙè….-iii) ví£A ví£øŒoÚÛª ÓE-NªC ÷«ô¢ª\õª.- 2 ×× 8 = 16

19.- ·Þ—xÚ¥-L-ú‡-úÃìª N÷-JÙ-àŸÙè….- ÍC áJ¸Þ ví£ë¶øŒÙ, ÍÙêŸu ÑêŸp-û¦oõªÔN? Ð ÑêŸp-û¦oõª î¦óŸ³-ú£-êŸ, î¦óŸ³-ô¢-ï‡°êŸ ø‹yú£-vÚ¨-óŸªõ ë¯yô¦Ô ÷«ô¢ªpÚÛª ö˺ì-÷±-ê¦ô³?

20.- í£±ìl-ú£Ù-óμ«-áÚÛ è†ÓûÂÔ þ§Ù¸Ú-AÚÛ Nëů-ìÙ-ö˺E NNëÅ] ví£vÚ¨-óŸª-õìªÚÛªxí£hÙÞ¥ N÷-JÙ-àŸÙè….-

21.- ÚÛé-â°õ ÷ô¢ÌÄìÙ Íû¶ þ§Ù¸Ú-AÚÛ Nâ‘°ìÙ ÞœªJÙ# N÷-JÙ-àŸÙè….- ú£Ùví£-ë¯óŸª í£ë]ÌÄ-Aö˺ îμ³ÚÛ\õ ví£á-ììÙ, ú£þ§u-GÅ-÷”CÌÄ Ú¥ô¢u-vÚÛ-÷«õÚÛÙç˶ ÚÛé-â°õ ÷ô¢ÌÄìÙ ÷õx ÷à¶a ö°òÅ°õª ÔNªæ¨?

SENIOR INTER MODEL PAPER - 2020Time: 3 Hours Botany (English Medium) Max. Marks: 60

BOARD OF INTERMEDIATE EDUCATION

SENIOR INTER MODEL PAPER - 2020Botany (Telugu Medium)

Time: 3 Hours Max. Marks: 60

1.�‘Transpiration is necessary evil’. Explain?A: The loss of water in the form of vapour from

aerial parts of plant body is called transpi-ration.Beneficial aspects:H Creates a transpiration pull for absorp-

tion.H Transport minerals from soil to all parts

of the plant body.H Supplies water for photosynthesis.H Leaf surface is cool.H Forest vegetation gives cool and good

climate.Harmful aspects:H Rate of photosynthesis is decreased.H Leaves and flowers show wilting.Transpiration is necessary evil due to itsboth beneficial and harmful aspects.

2. Explain different types of co-factors?

A: Co-factors:H Biocatalyst present in living organisms are

called enzymes. They may be simple (contain protein part only) or holoenzymes(contain protein part and non protein part).

H Protein part of an enzyme is called apoen-zyme and non-protein part is called co-factor.

H Co-factors are of 3 types.Co-enzymes: Organic compounds whichare loosely attached to the apoenzyme.e.g.: NAD, NADP, TPP, CO.A, PP

Prosthetic groups: Organic compoundswhich are tightly attached to the apoenzyme. e.g.: Peroxidase, catalyse

Metal ions: Metal ions loosely attached toapoenzyme and also called metalloenzymes.

e.g. Cu+2 ... cytochrome oxidise, Fe+2 ...catalyse, Mg+2 ... hexo kinase

3. Write the physiological responses of gib-berellins in plants.

A: Gibberellins: GA1, GA2, GA3H A class of growth promoter which induces

internodal elongation (bolting) beforeflowering. e.g.: Cabbage,Carrot, Beetroot

H Discovered by Yabuta and Sumuki.H Helps in seed germination by synthesis of

an enzyme α - amylase.H Promotes parthenocarpy (development of

fruit without fertilization).e.g.: Banana, oranges, guavaH Promotes male flowers.

e.g.: CucumbersH Increases the length of grapes stalks.

H Promotes early seed production in conifers.

H Delays senescence (ageing and death ofplant parts).

4. Explain the conjugation in bacteria?

A: H Transfer of genetic material from onebacterium to another bacterium throughconjugation tube.

H True sexual reproduction is absent inbacteria. But the essence of reproduc-tion takes place in bacteria.

H It was discovered by Lederberg andTatum (1946) in Escheritia coli .

H In bacteria small circular, double strand-ed DNA rings are present in addition tonucleoid are called plasmids or �‘F’ factor.

H ‘F’ factor containing bacteria are calleddonor cells or F+ cells and �‘F’ factorlacking bacteria are called recipient cellsor F

−cells.

H Donor cell and recipient cells attached bysex pile, hence conjugation tube isformed.

H Plasmid DNA is replicated in donor celland a copy of it moves into recipient cellthrough conjugation tube.

H F−

cells changed into F+ cells afterconjugation and express donor cellcharacters.

4 Marks Questions 2 Marks Questions

1. Where does the photolysis of H2Ooccur? What is its significance?

A: Thylakoid lumen or grana region of thechloroplast. The oxygen (O2) evolved inphotolysis is life gas for all organisms inthe universe.

2. What are virulent phages? Give anexample.

A: The viruses which attack the bacteriumEscheritia coli and shows lytic cycle.They cause immediate lysis of the hostcell.

e.g.: T-even phages.3. What is the genetic nature of wrinkled

phenotype of pea seeds?A: rrYY, rrYy, rryy.4. What is down-stream processing?A: Separation, purification and preservation

of gene cloning products before ready touse and release in the market.

5. Expand GMO. How is it different from ahybrid?

A: GMO stands for genetically modifiedorganisms. The organisms in whichgenes have been altered are calledGMO’s. GM crops are resistant to cold,heat, drought, salt and rich in nutrients.Hybrids are produced through artificialcross pollination.

6. Name two semi-dwarf varieties of Ricedeveloped in India?

A: Jaya and Ratna.

- Dr. Chenna kesavulu

Target-2020Senior Inter

Botany

100100

What is transpiration?

FOR ANSWERS Clik on : https://tinyurl.com/ufnhbc2

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email: help@eenadupratibha.net÷ªÙÞœüŒî¦ô¢Ù 18 íƇvñ÷J 2020

4 ÷«ô¢ª\õ ví£øŒoõª1. -- √

⎯6 ÚÛô¢-é©óŸª ú£ÙÜu ÍE Eô¢«-í‡Ù-àŸÙè….-

þ§ëÅ]ì:- √⎯6 ÍÚÛ-ô¢-é©óŸª ú£ÙÜu Íìª-ÚÛªÙë¯Ùp

√⎯6 = ⎯ , q ≠ 0, (p, q) = 1 öË¶ë¯ p, q õª

q

í£ô¢-ú£pô¢ ví£ëůì ú£ÙÜuõª√

⎯6 q = p

Ï-ô¢ª-î�μj-í£±ö° ÷ô¢_Ù à¶óŸªÞ¥(√

⎯6 q)2 = p2 ⇒ 6q2 = p2 .-.-.-.-.-.-.- (1)

⇒ p2 ìª 6 EøË™)-ù£ÙÞ¥ òÅ°T-ú£ªhÙC.-⇒ p ìª 6 EøË™)-ù£ÙÞ¥ òÅ°T-ú£ªhÙC.-p = 6c Í-ìª-ÚÛªÙë¯Ù sc í£²-ô¦gÙÚÛÙzp2 = (6c)2 ⇒ 6q2 = 36c2 ((1) ìªÙ#zq2 = 6c2

⇒ q2 ìª 6 EøË™)-ù£ÙÞ¥ òÅ°T-ú£ªhÙC.-⇒ q ìª 6 EøË™)-ù£ÙÞ¥ òÅ°T-ú£ªhÙC.-ÍÙç˶ p, q õÚÛª 6 Ñ÷ªtè… Ú¥ô¢-é°ÙÚÛÙ Í÷±-꟪ÙC.-ÏC (p, q) = 1 Í-ì-è¯-EÚ¨ Nô¢ªë]ÌÄÙ.-Ð Nô¢ª-ë]ÌÄêŸ √

⎯6 Í-ÚÛ-ô¢-é©óŸª ú£ÙÜu Íìª-ÚÁ-÷è[Ù

÷õx ÷#aÙC. -Ú¥ñæ¨d √⎯6 ÚÛ-ô¢-é©óŸª ú£ÙÜu.-

2. -- Ôëμjû¦ ëÅ]ì-í£²ô¢g ú£ÙÜu óμ³ÚÛ\ íÆ£ªìÙ 4m ö˶ë¯4m + 1 öË¶ë¯ 4m + 3 ô¢«-í£Ùö˺ ÑÙåªÙ-ë]E Eô¢«-í‡Ù-àŸÙè….-

þ§ëÅ]ì: óŸ´Ú¨xè òÅ°Þœ-ô¢ EóŸª÷ªÙ a = bq + r, 0 ≤ r < b

a Ô-ëμjû¦ ëÅ]ì-í£²ô¢g ú£ÙÜu Íìª-ÚÛªÙë¯Ùb = 4 Í-ô³ê¶ r = 0 öË¶ë¯ 1 öË¶ë¯ 2 öË¶ë¯ 3 Í÷±-꟪ÙC.-

r = 0 Í-ô³ê¶ a = 4q + 0 = 4 q

a3 = (4q)3 = 64 q3

= 4.16 q3 = 4m (m = 16q3)

r = 1 Í-ô³ê¶ a = 4q + 1

a3 = (4q + 1)3

= (4q)3 + 3.-(4q)2.1 + 3.4q.12 + 13

= 64q3 + 48q2 + 12q + 1

= 4(16q3 + 12q2 + 3q) + 1

= 4m + 1 (m = 16q3 + 12q2 + 3q)

r = 2 Í-ô³ê¶ a = 4q + 2

a3 = (4q + 2)3

= (4q)3 + 3.(4q)2.2 + 3.4q.-22 + 23

= 64q3 + 96q2 + 48q + 8

= 4(16q3 + 24q2 + 12q + 2)

= 4m (m = 16q3 + 24q2 + 12q + 2)

r = 3 Í-ô³ê¶ a = 4q + 3

a3 = (4q + 3)3 = 64q3 + 144q2 + 108q + 27

= 4(16q3 + 36q2 + 27q + 6) + 3

(... 27 = 24 + 3)

= 4m + 3 (m = 16 q3 + 36 q2 + 27q + r)

∴ Ô-ëμjû¦ ëÅ]ì í£²ô¢g ú£ÙÜu óμ³ÚÛ\ íÆ£ªìÙ 4m ö˶ë¯4m + 1 öË¶ë¯ 4m + 3 ô¢«-í£Ùö˺ ÑÙåªÙC.-

3. -- x2 + y2 = 34 xy Í-ô³ê¶ 2 log (x + y) = 2log2 + 2log 3 + log x + log y ÍE àŸ«í£Ùè….-

þ§ëÅ]ì:- x2 + y2 = 34 xy

Ï-ô¢ª-î�μj-í£±ö° 2xy ÚÛ-L-í‡ê¶x2+ y2 + 2xy = 34xy + 2xy

(x + y)2 = 36 xy

log (x + y)2 = log 36 xy Ï-ô¢ª-î�μj-í£±ö° ú£Ù÷-ô¢_-÷«-û¦õª ô¦óŸªÞ¥2 log (x + y) = log 36 + log x + log y (log xyz

= log x + logy + logz)

= log 22 × 32 + log x + log y

= log 22 + log 32 + log x + log y

= 2 log 2 + 2 log 3 + log x + log y

(log xm = m log x)

Íë]ìí£± ví£øŒoõª

1. Ô-ëμjû¦ ëÅ]ì-í£²ô¢g ú£ÙÜu n ÚÛª n, n + 1 öË¶ë¯ n +2 õö˺ Ôëμjû¦ ÖÚÛæ¨ ÷«vêŸî¶ª 3êÁ EøË™)-ù£ÙÞ¥òÅ°TÙ-àŸ-ñ-è[ª-꟪Ù-ë]E Eô¢«-í‡Ù-àŸÙè….-

1 12. (3.2)x = (0.032)y = 1000 Í-ô³ê¶ ⎯ − ⎯

x y

N-õª÷ ÚÛìª-ÚÁ\Ùè….-3. a, b õª ví£ëů-û¦Ù-Ú¥õª Íô³ê¶ √⎯

a + √⎯b

ÚÛ-ô¢-é©óŸª ú£ÙÜu ÍE Eô¢«-í‡Ù-àŸÙè….-

ÓúÃ. ô¢î¶ªùÃNù£óŸª Eí£±éªõª

ô¢àŸô³êŸ

î¦ú£h÷ ú£ÙÜuõª

2 ÷«ô¢ª\õ ví£øŒoõª1. -- 12, 15, 18õ ÚÛþ§Þœª, Þœþ§-òÅ°-õìª ví£ëůì Ú¥ô¢-

é°ÙÚÛ í£ë]ÌÄ-Aö˺ ÚÛìª-ÚÁ\Ùè….-þ§ëÅ]ì: 12 =- 2 × 2 × 3 =- 22 ' 3

15 =- 3 × 5 =- 3 × 5

18 =- 2 × 3 × 3 =- 2 × 32

Þœþ§òÅ° =- þ§÷«ìu ví£ëůì Ú¥ô¢-é°Ù-Ú¥õ ÚÛEù£’íÆ£«ê¦õ õñÌÄÙ =- 3

ÚÛþ§Þœª =- ví£ëůì Ú¥ô¢-é°Ù-Ú¥ö˺x ví£A-ë¯E ÞœJù£’íÆ£«ê¦õ õñÌÄÙ =- 22 × 32 × 5

=- 4 × 9 × 5 =- 180

2. 5 + 2√⎯3 ÚÛ-ô¢-é©óŸª ú£ÙÜu ÍE àŸ«í£Ùè….-

þ§ëÅ]ì: 5 + 2√⎯3 Í-ÚÛ-ô¢-é©óŸª ú£ÙÜu Íìª-ÚÛªÙë¯Ù

p5 + 2√

⎯3 = ⎯ , q ≠ 0 (p, q) = 1

q

p 1 p2√

⎯3 = ⎯ − 5 ⇒ √

⎯3 = ⎯ (⎯ − 5)q 2 q

ÏC Nô¢ªë]ÌÄÙ.- Ð Nô¢ª-ë]ÌÄêŸ 5 + 2√⎯3 Í-ÚÛ-ô¢-é©óŸª

ú£ÙÜu Íìª-ÚÁ-÷è[Ù ÷õx ÷#aÙC.-Ú¥ñæ¨d 5 + 2√

⎯3 ÖÚÛ ÚÛô¢-é©óŸª ú£ÙÜu.-

3. -- 3 × 5 × 71 + 71 ÖÚÛ ú£ÙóŸ³-ÚÛh- ú£ÙÜu ÍEàŸ«í£Ùè….-

þ§ëÅ]ì: 3 × 5 × 71 + 71 = 71(3 × 5 + 1)

= 71(15 + 1) = 71 × 16 × 1

šíj ú£ÙÜuÚÛª ·ôÙè[ª ÚÛÙç˶ ÓÚÛª\÷ Ú¥ô¢-é°Ù-Ú¥õªÑÙè[åÙ ÷õx ÍC ú£ÙóŸ³ÚÛh ú£ÙÜu.-

4. 2 log 5 + 5 log 3 − 3 log 2 ìª Ö¸Ú ú£Ù÷-ô¢_-÷«-ìÙÞ¥ ô¦óŸªÙè….-

þ§ëÅ]ì:- 2 log 5 + 5 log 3 − 3 log 2

= log 52 + log 35 − log 23

(m log a = log am)

= log 25 + log 243 − log 8

= log 25 × 243 − log 8

(log x + log y = log xy)

= log 6075 − log 8

6075 x= log ( ⎯⎯⎯) ( log x − log y = log ⎯ )8 y

5. 5x = 3x − 2 ú£-Oª-ÚÛ-ô¢-é°Eo þ§CÅÙ-àŸÙè….-þ§ëÅ]ì: 5x = 3x − 2

Ï-ô¢ª-î�μj-í£±ö° ú£Ù÷-ô¢_-÷«-û¦õª ô¦óŸªÞ¥log 5x = log 3x − 2

x log 5 = (x − 2)log 3

= x log 3 − 2 log 3

x log 3 − x log 5 = 2 log 3

x(log 3 − log 5) = 2 log 3

2 log 3 x = ⎯⎯⎯⎯⎯⎯

log 3 − log 5

6. q Ô-ëμjû¦ ÖÚÛ í£²ô¢g ú£ÙÜu Íô³-ì-í£±pè[ª ví£A ëÅ]ìò˶ú‡-ú£ÙÜu 4q + 1 öË¶ë¯ 4q + 3 ô¢«-í£Ùö˺ÑÙåªÙ-ë]E Eô¢«-í‡Ù-àŸÙè….-

þ§ëÅ]ì: a Ô-ëμjû¦ ëÅ]ì ò˶ú‡ í£²ô¢g-ú£ÙÜu Íìª-ÚÛªÙë¯ÙóŸ´Ú¨xè òÅ°Þœ-ô¢ EóŸª÷ªÙ ví£Ú¥ô¢Ù a = bq + r, 0 ≤ r < b

b = 4 Í-ô³ê¶ r = 0 öË¶ë¯ 1öË¶ë¯ 2 öË¶ë¯ 3 Í÷±-꟪ÙCr = 0 Í-ô³ê¶ a = 4q = 2.2q ú£-J-ú£ÙÜur = 1 Í-ô³ê¶ a = 4q + 1

r = 2 Íô³ê¶ a = 4q + 2

= 2(2q + 1) ú£-J-ú£ÙÜur = 3 Íô³ê¶ a = 4q + 3

a ò˶-ú‡-ú£ÙÜu Ú¥ñæ¨d a ô¢«í£Ù 4q + 1 öË¶ë¯ 4q + 3 Í-÷±-꟪ÙC.-

Íô¢-÷«ô¢ª\ ví£øŒoõª

1. ·ôÙè[ª ú£ÙÜuõ Þœþ§òÅ° 27, ÚÛþ§Þœª 162.-ÍÙë]ªö˺ ÖÚÛ ú£ÙÜu 54 Íô³ê¶ ·ôÙèÁ ú£ÙÜu?

á:- 81 sú£«vêŸÙ:- ÚÛþ§Þœª × Þœ-þ§òÅ° = a × b)

62. -- ⎯⎯ ÓEo ë]ø‹Ù-ø‹õ êŸô¦yêŸ ÍÙêŸ-÷ª-÷±-꟪ÙC?

24 × 5

á:- 3

3. -- 62020 íÆ£-LêŸ ú£ÙÜuö˺ ÖÚÛåx þ§–ìÙö˺ ÑÙè˶ÍÙ·Ú ÔC? á:- 6

4. -- a, b ú£Ù-Üuõ Þœþ§òÅ° 1 Íô³ê¶ a, b õªA) ví£-ëůì ú£ÙÜuõª B) í£-ô¢-ú£pô¢ ví£-ëůì ú£ÙÜuõªC) ÚÛ÷õ ví£ëůì ú£ÙÜuõª D) A, B, C õö˺ Ôëμjû¦ ÖÚÛæ¨ á:- D

5. -- Ú¨ÙC-î¦-æ¨ö˺ ú£·ôjÙC.-A) R Q Z W N

B) Q Z W N R

C) N W Z Q R

D) N W Z R Q á:- C6. Ú¨ÙC-î¦-æ¨E áêŸ-í£-ô¢-àŸÙè….-

5i) ÍÙêŸ÷ªó¶ªu ë]ø‹ÙøŒÙ a) ⎯

3

ii) ÍÙ-êŸÙ-Ú¥E Î÷-ô¢hì ë]ø‹ÙøŒÙ b) π3

iii) ÍÙ-êŸÙ-Ú¥E Î÷-ô¢hìÙ Ú¥E ë]ø‹ÙøŒÙ c) ⎯5

A) i-a, ii-b, iii-c B) i-c, ii-a, iii-b

C) i-b, ii-c, iii-a D) i-c, ii-b, iii-a á:- B

7. -- ÚÛEù£’ ví£ëůì ú£ÙÜu, ÚÛEù£’ ú£ÙóŸ³ÚÛh ú£ÙÜuõÚÛþ§Þœª a, Þœþ§òÅ° b Í-ô³ê¶ ab = ? á:- 8

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è†ÓíÆÃúˆúˆÕÓöËÀ, ì«uCMxòÅ°ô¢êŸ ví£òÅ¡ªêŸy ·ôjö˶y÷ªÙvAêŸy ø‹ÜÚÛª àμÙCìì«uCMxö˺E èËμè…¸ÚçËμèÂvšíÆjæËÀ Ú¥Jè¯ô ڥô•p¸ôù£ûË ÎíÆà ÏÙè…óŸ« LNªçËμèÂsè†ÓíÆÃúˆúˆÕÓöËÀz Ú¨ÙC ð¼ú£ªdöË òÅ¡KhÚ¨ ë]ô¢Ý°ú£ªhõªÚÁô¢ªêÁÙC. l îμ³êŸhÙ Ý°Sõª: 54ð¼ú£ªdõªnÝ°Sõª: Íú‡šúdÙæËÀ û¶áô sšíÆjû¦ûËÂq,Õæ©zn30, WEóŸªô û¶áô sšï°àÂÎôÂ, šíÆjû¦ûËÂq,Õæ©zn24. Íô¢|êŸ: ð¼ú£ªdE Íìªú£JÙ# ú£ÙñÙCÅêŸú£òËμbÚÛªdö˺x ò°u#öËôÂq è…vU, HÐ/ HçËμÚÂ, ÓÙHÔ,úˆÔ/ úˆÓÙÔ ÑBhô¢gêŸ, ÍìªòÅ¡÷Ù.ÓÙí‡ÚÛ NëůìÙ: ÚÛÙí£²uæô ò˶úÓ çËμúÃd súˆHæ©z,ÏÙæô¢«yu, îμªè…ÚÛöËÀ çËμúÃd Îëůô¢ÙÞ¥. ë]ô¢Ý°ú£ªhNëůìÙ: ÎûËÂöËμjûËÂ. #÷Jê¶C: ÷«Ja 15, 2020.

îËμòËÀšújæËÀ: https://dfccil.com/

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îËμòËÀšújæËÀ: http:///spa.ac.in/

email: help@eenadupratibha.net÷ªÙÞœüŒî¦ô¢Ù 18 íƇvñ÷J 2020

4 ÷«ô¢ª\õ ví£øŒoõª1.- ÖÚÛ êŸô¢-Þœ-Aö˺ 25 ÷ªÙC Në¯u-ô¢ª–õª Ñû¦oô¢ª.-

î¦Jö˺ 12 ÷ªÙCÚ¨ Þœé¨êŸÙ ÍÙç˶ Ïù£dÙ.- 8÷ªÙCÚ¨ Þœé¨êŸÙ ÍÙç˶ Ïù£dÙ Ú¥F @÷-ø‹ú£YÙ ÍÙç˶Ïù£dÙ ö˶ë]ª.- ví£A Në¯uJ– Þœé¨êŸÙ öË¶ë¯ @÷-ø‹ú£YÙöË¶ë¯ ·ôÙè…Ù-æ¨F Ïù£d-í£-è…ê¶,i) @-÷-ø‹ú£YÙ ÷«vêŸî¶ª Ïù£d-í£-è˶-î¦ô¢ª ÓÙêŸ-÷ªÙC? ii) Þœ-é¨êŸÙ, @÷-ø‹ú£YÙ ·ôÙè…Ù-æ¨F Ïù£d-í£-è˶-î¦ô¢ª

ÓÙë]ô¢ª?þ§ëÅ]ì:- n(M) = 12

n(M − B) = 8

n(M ∪ B) = 25

n(B − M) = ? n(M ∩ B) = ?

i) n(M ∪ B) = n(M) + n(B − M)

25 = 12 + n(B − M)

∴ n(B − M) = 25 − 12 = 13

ÍÙç˶ @÷-ø‹ú£YÙ ÷«vêŸî¶ª Ïù£d-í£-è˶-î¦ô¢ª 13 ÷ªÙC.-ii) n(M ∪ B) = n(M − B) + n(M ∩ B) +

n(B − M)

25 = 8 + n(M ∩ B) + 13

∴ n(M ∩ B) = 25 − 21 = 4

ÍÙç˶ Þœé¨êŸÙ, @÷-ø‹ú£YÙ ·ôÙè…Ù-æ¨F Ïù£d-í£-è˶-î¦ô¢ªìõª-Þœªô¢ª.-

2. A = {x : x ∈ 25 óμ³ÚÛ\ Ú¥ô¢-é°Ù-Ú¥õª}, B ={x : x Í-û¶C 20 ÚÛÙç˶ êŸÚÛª\-î�μjì ví£ëů-û¦Ù-Ú¥õª} Í-ô³ê¶ i) A ∪ B ii) n(A ∩ B) iii) B − A

iv) n(A − B) õìª ÚÛìª-ÚÁ\Ùè….-þ§ëÅ]ì:- A = {x : x ∈ 25 óμ³ÚÛ\ Ú¥ô¢-é°Ù-Ú¥õª}

∴ A = {1, 5, 25} (25 = 1 × 25 ; 5 × 5 = 25)

B = {x : x Í-û¶C 20 ÚÛÙç˶ êŸÚÛª\-î�μjì ví£ëů-û¦Ù-Ú¥õª}

B = { 2, 3, 5, 7, 11, 13, 17, 19}

i) A ∪ B = {1, 5, 25} ∪ {2, 3, 5, 7, 11, 13,17, 19}

= {1, 2, 3, 5, 7, 11, 13, 17, 19, 25}

ii) A ∩ B = {5} ∴ n(A ∩ B) = 1

iii) B − A = {2, 3, 7, 11, 13, 17, 19}

(Bö˺ ÑÙè… Aö˺ ö˶E ÷´õ-Ú¥õªziv) A − B = {1, 25} (Aö˺ ÑÙè… Bö˺ ö˶E

÷´õ-Ú¥õªz ∴ n(A − B) = 2

3. Ú¨ÙC î�μû #vê¦Eo Ñí£-óμ«-TÙ# n(A ∪ B) + n(A ∩ B) = n(A) + n(B) ÍE àŸ«í£Ùè….-

þ§ëÅ]ì:- A = {1, 2, 3, 4, 5, 6, 7}

n(A) = 7

B = {5, 6, 7, 8, 9, 11}

∴ n(B) = 6

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11}

∴ n(A ∪ B) = 10

A ∩ B = {5, 6, 7}

∴ n(A ∩ B) = 3

∴ LHS =- n(A ∪ B) + n(A ∩ B)

= 10 + 3 = 13

RHS = n(A) + n(B) = 7 + 6 = 13

∴ LHS = RHS

ÍÙç˶ n(A ∪ B) + n(A ∩ B)

= n(A) + n(B) ÍE àŸ«í£-ñ-è…ÙC.-4.- A = {x : x ∈ N, x Í-û¶C 3 Þœªé¨-â°õ ú£NªA}

B = {x : x ∈ N, x Í-û¶C 5 Þœª-é¨-â°õ ú£NªA}

C = {x : x ∈ N, x Í-û¶C 3, 5 õ ÚÛ.-þ§.-Þœª.-óμ³ÚÛ\ Þœªé¨-â°õ ú£NªA}

Í-ô³ê¶ i) A ∩ Bìª ÚÛìª-ÚÁ\Ùè….- ii) A ∩ B = C Í-÷±-꟪Ùë¯?

þ§ëÅ]ì:- A = {3, 6, 9, 12, 15, 18, ......}

B = {5, 10, 15, 20, .......}

3, 5 õ ÚÛ.-þ§.-Þœª.- = 15 ÍÙç˶ C ú£-NªA 15 Þœªé¨-â°õ ú£NªA Í÷±-꟪ÙC.-∴ C = {15, 30, 45,.....}

i) A ∩ B = {15, 30, 45,.....}

ii) A ∩ B = C Í-÷±-꟪ÙC.-

ú£Nª-꟪õª

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í£ëÁ-êŸ-ô¢-ÞœAÞœé̈êŸÙ ›íí£ô n- 1

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í‡. î�¶éªÞÁð§öËÀNù£óŸª Eí£±éªõª

ô¢àŸô³êŸ

M B

8 4 13

μ

A B1234

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567

μ

Í-ô¢-÷«ô¢ª\ ví£øŒoõª

2.- A B Í-ô³ê¶ A ∩ B =

A) A B) B C) φ D) μ3.- Ú¨Ù-C-î¦-æ¨ö˺ ÔC øŒŠìu-ú£-NªA?

A) {x : x ∈ N, x > 50}

B) {x : x ∈ Z, x < −- 2}

C) {x : x ∈ N, x < − 2} D) {0}

4.- A − B = φ Í-ô³ê¶ A) A = φ B) B = φC) A = μ D) Ô-D-Ú¥ë]ª

5. √⎯2 Íû¶C Ú¨ÙC ú£Nª-꟪ö˺x ë¶EÚ¨ àμÙCÙC?

A) N B) Z C) Q D) Q'6. n(A) = 7, n(B) = 2 Íô³ê¶ n (A × B) =

A) 2 B) 7 C) 9 D) 14

7. A = {{1, 2, 3}, {4, 5}, {6, 7, 8}} Í-ô³ê¶ Ú¨ÙC-î¦-æ¨ö˺ ÔC ú£êŸuÙ? A) 1 ∈ A B) {6, 7, 8} ∈ A

C) {1, 2, 3} A D) φ ∈ A

8.- A = φ Í-ô³ê¶ n[P(A)] =

A) 1 B) 0 C) 2 D) 4

9. ú£Nª-꟪-õìª ví£î�¶-øŒ-ší-æ¨dì ø‹ú£Y-î�¶êŸh?A) ô¢-šúqöËÀ B) â°Jb Ú¥Ùåô C) J-àŸô“ D) ì«u-åûÂ

ú£÷«-ëů-û¦õª1--D 2--A 3--C 4--A 5--D 6-D 7--B 8--A 9-B.

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12

A B

34

μ1.- Íô³ê¶ Ú¨ÙC-î¦-æ¨ö˺ÔC ú£êŸuÙ?

A) A ∩ B = μ B) A ∪ B = μ

C) A ∪ B = φ D) A ∩ B = φ

1.- A, B õª ·ôÙè[ª ú£Nª-꟪õª Íô³ê¶ A BE î�μûÂ#vêŸÙ ë¯yô¦ àŸ«í£Ùè….-

2.- 2 óμ³ÚÛ\ Þœªé¨-â°õ ú£NªA, 5 óμ³ÚÛ\ Þœªé¨-â°õú£Nª-꟪õª NóŸ³ÚÛh ú£Nª-꟪õª ÍE ô¢N Íû¦oè[ª.-DÙêÁ Oªô¢ª ÔÚ©-òÅ¡-N-ú£ªh-û¦oô¦? þ¼ë¯--ô¢-éÙÞ¥êμõ-í£Ùè….-

þ§ëÅ]ì:- A = {2, 4, 6, 8, 10, 12, .......}

B = {5, 10, 15,.......}

A ∩ B = ({10, 20,....}

A, B ·ôÙè[ª ú£Nª-꟪ö˺x Ñ÷ªtè… ÷´õ-Ú¥õªÑû¦oô³.- Ú¥ñæ¨d NóŸ³ÚÛh ú£Nª-꟪õª ú£Nª-꟪õªÚ¥÷±.- NóŸ³ÚÛh ú£Nª-꟪õª Ú¥î¦-õÙç˶ A ∩ B =φ Ú¥-î¦L. Ú¥ñæ¨d û¶ìª ô¢NêÁ ÔÚ©òÅ¡NÙàŸìª.-

3.- A = {x : x ∈ N, x < 4} Í-ô³ê¶ A ú£-NªAóμ³ÚÛ\ Ñí£-ú£-Nª-꟪-õFo ô¦óŸªÙè….-

þ§ëÅ]ì:- A = {1, 2, 3}

∴ n(A) = 3

AÚ¨ Þœõ Ñí£-ú£-Nª-꟪õ ú£ÙÜu =- 23 =- 8

ÍN φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},{1, 2, 3}

4. n(A − B) = 6, n(A ∩ B) = 2, n(B − A) = 7 Í-ô³ê¶ n(A ∪ B) = ?.-

þ§ëÅ]ì:- n(A ∪ B) = n(A − B) + n(A ∩ B) +

n(B − A)

= 6 + 2 + 7 = 15

5.- í£-J-NªêŸ, Íí£-J-NªêŸ ú£Nª-꟪-õÚÛª ÖÚÁ\ Ñë¯--ô¢éô¦óŸªÙè….-

þ§ëÅ]ì:- í£J-NªêŸ ú£NªA:- ÷´õ-Ú¥õª í£J-Nª-êŸÙÞ¥ÑÙè˶ ú£NªA

A = {x : x ∈ N, x < 10}

Í-í£-J-NªêŸ ú£NªA:- ÷´õ-Ú¥õª Íí£-J-Nª-êŸÙÞ¥ÑÙè˶ ú£NªAA = {x : x ∈ N, x > 10}

6.- A = {1, 2, 3, 4}, B = {x : x ∈ N & 4 < x ≤ 6} õª NóŸ³ÚÛh ú£Nª-꟪õª ÍE vøŒ÷éË Íû¦oè[ª?Oª ú£÷«-ëů-û¦Eo ú£÷ª-J–Ù-àŸÙè….-

þ§ëÅ]ì:- A = {1, 2, 3, 4}

B = {5, 6}

∴ A ∩ B = φ ÍÙç˶ A, B õö˺ Ñ÷ªtè… ÷´õ-Ú¥õª ö˶÷±A, Bõª NóŸ³ÚÛh ú£Nª-꟪õª∴ vøŒ÷éË ú£÷«-ëů-û¦Eo ú£÷ª-J–-ú£ªh-û¦oìª.-

7. A = {a, e, i, o, u}, B = {a, i, u} A ∪ B = A Í-÷±-꟪Ùë¯?ÓÙ-ë]ªÚÛª?

þ§ëÅ]ì:- A ∪ B = {a, e, i, o, u}

ÓÙ-ë]ª-ÚÛÙç˶ B A Ú¥-ñæ¨d (B A Í-ô³ê¶A ∪ B = A Í-÷±-꟪ÙC.-z

8.- X = (x : x ∈ 60E òÅ°TÙචví£ëůì ú£ÙÜuzìªâ°Gê¦ ô¢«í£Ùö˺ ô¦óŸªÙè….-

þ§ëÅ]ì:- X = {2, 3, 5}

1, 2 ÷«ô¢ª\õ ví£øŒoõª

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A

B μþ§ëÅ]ì:-

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Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?Þœé¨ê¦Eo Ïù£dí£è˶î¦ô¢ª ÓÙêŸ÷ªÙC?

â°Jb Ú¥ÙåôÂ

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îËμòËÀšújæËÀ: https://dfccil.com/

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