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muRj;Njh;T rpwg;Gf; ifNaL

ntw;wpf;F top



epidtpw; nfhs;s Ntz;ba Kf;fpa Fwpg;Gfs; ,ay; 1: fzq;fSk; rhh;GfSk; ghpkhw;W tpjpfs; : i) 𝐴 ∪ 𝐵 = 𝐵 ∪ 𝐴 ii) 𝐴 ∩ 𝐵 = 𝐵 ∩ 𝐴 Nrh;g;G tpjpfs; : i) 𝐴 ∪ 𝐵 ∪ 𝐶 = (𝐴 ∪ 𝐵) ∪ 𝐶 ii) 𝐴 ∩ 𝐵 ∩ 𝐶 = (𝐴 ∩ 𝐵) ∩ 𝐶 gq;fPl;L tpjpfs;: i) 𝐴 ∪ 𝐵 ∩ 𝐶 = 𝐴 ∪ 𝐵 ∩ (𝐴 ∪ 𝐶) ii) 𝐴 ∩ 𝐵 ∪ 𝐶 = 𝐴 ∩ 𝐵 ∪ (𝐴 ∩ 𝐶)

fz tpj;jpahrj;jpw;fhd b khh;fdpd; tpjpfs;: i) 𝐴\ 𝐵 ∪ 𝐶 = 𝐴\𝐵 ∩ 𝐴\𝐶 ii) 𝐴\(𝐵 ∩ 𝐶) = (𝐴\𝐵) ∪ (𝐴\𝐶) fz epug;gpf;fhd b khh;fdpd; tpjpfs;: ′

= 𝐴′ ∩ 𝐵′ ii) 𝐴 ∩ 𝐵



= 𝐴′ ∪ 𝐵′

UD Y

i) 𝐴 ∪ 𝐵

fzq;fspd; Nrh;g;gpd; Mjp vz;izf; fz;lwpAk; #j;jpuq;fs;: i) 𝑛 𝐴 ∪ 𝐵 = 𝑛 𝐴 + 𝑛 𝐵 − 𝑛(𝐴 ∩ 𝐵) ii) 𝑛 𝐴 ∪ 𝐵 ∪ 𝐶 = 𝑛 𝐴 + 𝑛 𝐵 + 𝑛 𝐶 − 𝑛 𝐴 ∩ 𝐵 − 𝑛 𝐵 ∩ 𝐶 − 𝑛 𝐴 ∩ 𝐶 + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)

,ay; 2: nka;naz;fspd; njhlh;thpirfSk; njhlh;fSk;

Kjy; 𝑛 ,ay; vz;fspd; th;f;fq;fspd; $Ljy; 𝑛 𝑛+1 (2𝑛+1) = 12 + 22 + 32 + … + 𝑛 2 = 6 vi) Kjy; 𝑛 ,ay; vz;fspd; fdq;fspd; $Ljy; 𝑛(𝑛 + 1) 2 = 13 + 23 + 33 + ⋯ + 𝑛 3 = 2 vii) Kjy; 𝑛 xw;iwg;gil ,ay; vz;fspd; $Ljy; = 1 + 3 + 5 + 7 + … + (2𝑛 − 1) cWg;Gfs; tiu = 𝑛2 viii) Kjy; 𝑛 xw;iwg;gil ,ay; vz;fspd; $Ljy; (filrp cWg;G 𝑙 nfhLf;fg;gl;lhy;) v)

ST

i) xU $l;Lj; njhlh;thpirapy; 𝑛tJ cWg;G fhZk; tha;ghL 𝑡𝑛 = 𝑎 + (𝑛 − 1)𝑑 ii) xU $l;Lj; njhlh; thpirapy; 𝑛 cWg;Gfs; tiu $Ljy; fhZk; tha;g;ghL 𝑛 𝑛 𝑆𝑛 = 2 2𝑎 + 𝑛 − 1 𝑑 𝑜𝑟 2 [𝑎 + 1] iii) xU ngUf;Fj; njhlh; thpirapy; 𝑛 cWg;Gfs; tiu $Ljy; fhZk; tha;g;ghL 𝑎 [𝑟 𝑛 − 1] 𝑎 [1−𝑟 𝑛 ] 𝑆𝑛 = 𝑟−1 = 1−𝑟 (if 𝑟 ≠ 1), 𝑆𝑛 = 𝑛𝑎 ( if 𝑟 = 1) iv) Kjy; 𝑛 ,ay; vz;fspd; $Ljy; 𝑛(𝑛 + 1) = 1 + 2+ 3 + …+ 𝑛 = 2

= 1 + 3 + 5 + 7 + ……….. + 𝑙 =

𝑙+1 2 2

OZ Y

,ay; 3: ,aw;fzpjk; 2

i) 𝑝 𝑥 = 𝑎𝑥 + 𝑏𝑥 + 𝑐 vd;w gy;YWg;Gf; −𝑏 Nfhitapy; G+r;rpaq;fspd; $Ljy; = 𝑎 , G+r;rpaq;fspd; 𝑐

DO

ngUf;fw;gyd; = 𝑎 2 ii) 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 vd;w ,Ugbr; rkd;ghl;bd; %yq;fspd; jd;ikf;fhl;b ∆= 𝑏2 − 4𝑎𝑐 iii) 𝑏2 − 4𝑎𝑐 = 0 vdpy;> %yq;fs; nka;naz;fs;> NkYk; rkkhdit. iv) 𝑏2 − 4𝑎𝑐 > 0 vdpy;> %yq;fs; nka;naz;fs;> Mdhy; rkky;y. v) 𝑏2 − 4𝑎𝑐 < 0 vdpy;> %yq;fs; fw;gid vz;fs;.

𝜶 kw;Wk; 𝜷 Mfpatw;iw nfhz;l rpy KbTfs; gpd;tUkhW jug;gl;Ls;sd. (i) 𝛼 − 𝛽 = (𝛼 + 𝛽)2 − 4𝛼𝛽 (ii) 𝛼 2 + 𝛽 2 = [(𝛼 + 𝛽)2 − 2𝛼𝛽 ] (iii) 𝛼 2 − 𝛽 2 = 𝛼 + 𝛽 𝛼 − 𝛽 = (𝛼 + 𝛽)[ (𝛼 + 𝛽)2 − 4𝛼𝛽 ] only if 𝛼 ≥ 𝛽 3 3 (iv) 𝛼 + 𝛽 = (𝛼 + 𝛽)3 − 3𝛼𝛽(𝛼 + 𝛽) (v) 𝛼 3 − 𝛽 3 = (𝛼 − 𝛽)3 + 3𝛼𝛽(𝛼 − 𝛽) (vi) 𝛼 4 + 𝛽 4 = 𝛼 2 + 𝛽 2 2 − 2𝛼 2 𝛽 2 = [(𝛼 + 𝛽)2 − 2𝛼𝛽 ]2 − 2(𝛼𝛽)2 4 4 (vii) 𝛼 − 𝛽 = 𝛼 + 𝛽 𝛼 − 𝛽 (𝛼 2 + 𝛽 2 )

,ay; 4: mzpfs;

i) xU mzpapy; 𝑚 epiufSk; 𝑛 epuy;fSk; cs;sd. vdpy;> mjd; thpir 𝑚 × 𝑛 MFk;. ii) ,uz;L mzpfspd; thpirfs; rkkhf ,Ug;gpd; me;j mzpfisf; $l;lNth my;yJ fopf;fNth ,aYk;. iii) mzp A-apd; thpir 𝑚 × 𝑛 kw;Wk; B-d; thpir 𝑛 × 𝑝 vdpy;> ngUf;fw;gyd; mzp AB d; thpir 𝑚 × 𝑝

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iv) nghJthf mzpfspd; ngUf;fy; ghpkhw;Wg; gz;G cilajy;y. mjhtJ AB ≠ BA v) mzpfspd; ngUf;fy; Nrh;g;Gg; gz;G cilaJ. mjhtJ (AB)C = A(BC) vi) (𝐴𝑇 )𝑇 = 𝐴, (𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵𝑇 kw;Wk; (𝐴𝐵)𝑇 = 𝐵𝑇 𝐴𝑇

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muRj;Njh;T rpwg;Gf; ifNaL

ntw;wpf;F top



,ay; 5: Maj;njhiy tbtpay;

𝑥 1 +𝑥 2 +𝑥 3

𝐺=

3

iii) gphpT #j;jpuk; (cl;Gwkhf)=

,

𝑦1 +𝑦2 +𝑦3 3

𝑙𝑥 2 +𝑚 𝑥 1 𝑙+𝑚

,

𝑙𝑦 2 +𝑚𝑦 1 𝑙+𝑚

iv) (𝑥1 , 𝑦1 ), 𝑥2 , 𝑦2 , (𝑥3 , 𝑦3 ) Mfpa Gs;spfis cr;rpfshf nfhz;l Kf;Nfhzj;jpd; gug;gsT 1 = 2 {(x1 y2 + x2 y3 + x3 y1) – (x2 y1 +x3 y2 + x1 y3)} r.m v) (𝑥1 , 𝑦1 ), 𝑥2 , 𝑦2 , (𝑥3 , 𝑦3 ) kw;Wk; (𝑥4 , 𝑦4 ) Mfpa Gs;spfis cr;rpfshf nfhz;l ehw;fuj;jpd; gug;gsT =

1 2

{(x1 y2 + x2 y3 + x3y4 + x4 y1)

vi) 𝑥 -mr;rpd; rkd;ghL 𝑦 = 0 vii) 𝑦 -mr;rpd; rkd;ghL 𝑥 = 0 viii) 𝑥 -mr;rpw;F ,izahd Nfhl;bd; rkd;ghL 𝑦 = 𝑘

2

1

2

1

xvi) 𝑥 - ntl;Lj;Jz;L 𝑎 kw;Wk; 𝑦 -ntl;Lj; Jz;L 𝑏 𝑥 𝑦 nfhz;l Nfhl;bd; rkd;ghL + = 1 𝑎 𝑏 xvii) fpilepiyf; Nfhl;bd; rha;T G+r;rpakhFk;. Neh;f;Fj;Jf;Nfhl;bd; rha;T tiuaWf;f ,ayhjJ. xviii) ,U NfhLfspd; rha;Tfs; rkk; vdpy; mf;NfhLfs; xd;Wf;nfhd;W ,izahFk;. xix) Neh;f;Fj;jw;w ,U Neh;f;NfhLfspd; rha;Tfspd; ngUf;fw;gyd; −1 (𝑚1 × 𝑚2 = −1) vdpy;> mf;NfhLfs; xd;Wf;nfhd;W nrq;Fj;jhFk;

ST

– (x2 y1 +x3 y2 +x4y3 + x1 y4)} r.m

ix) 𝑦 - mr;rpw;F ,izahd Nfhl;bd; rkd;ghL 𝑥 = 𝑘 x) 𝑎 𝑥 + 𝑏𝑦 + 𝑐 = 0 vd;w Nfhl;bw;F ,izahd Nfhl;bd; rkd;ghL 𝑎𝑥 + 𝑏𝑦 + 𝑘 = 0 xi) 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 vd;w Nfhl;bw;F nrq;Fj;jhd Nfhl;bd; rkd;ghL 𝑏𝑥 − 𝑎𝑦 + 𝑘 = 0 xii) Mjpg;Gs;sp topr; nry;Yk; Nfhl;bd; rkd;ghL 𝑦 = 𝑚𝑥 xiii) rha;T 𝑚 kw;Wk; 𝑦- ntl;Lj; Jz;L 𝑐 nfhz;l Nfhl;bd; rkd;ghL 𝑦 = 𝑚𝑥 + 𝑐 xiv) rha;T 𝑚 nfhz;L 𝑥1 , 𝑦1 topahfr; nry;Yk; Nfhl;bd; rkd;ghL 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) xv) (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ) Mfpa Gs;spfs; topahfr; nry;Yk; 𝑦 −𝑦 𝑥−𝑥 Nfhl;bd; rkd;ghL 𝑦 −𝑦1 = 𝑥 −𝑥1

UD Y

i) (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ) Mfpa Gs;spfis ,izf;Fk; Neh;f;Nfhl;Lj; Jz;bd; eLg;Gs;sp 𝑥 +𝑥 𝑦 +𝑦 = 12 2 , 12 2 ii) 𝐴(𝑥1 , 𝑦1 ), 𝐵(𝑥2 , 𝑦2 ), kw;Wk; C(𝑥3 , 𝑦3 ). Kf;Nfhzk; 𝐴𝐵𝐶d; eLf;Nfhl;L ikak;

,ay; 7: Kf;Nfhztpay; 2

2

4) 𝑐𝑜𝑠 𝜃 = 1 − 𝑠𝑖𝑛2 𝜃 5) cosθ = 1 − 𝑠𝑖𝑛2 𝜃

6) 𝑠𝑒𝑐 2 𝜃 − 𝑡𝑎𝑛2 𝜃 = 1 7) 𝑠𝑒𝑐 2 𝜃 = 1 + 𝑡𝑎𝑛2 𝜃 8) 𝑠𝑒𝑐𝜃 = 1 + 𝑡𝑎𝑛2 𝜃 9) 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃 − 1 10) tanθ = sec 2 θ − 1

OZ Y

1) 𝑠𝑖𝑛 𝜃 + 𝑐𝑜𝑠 𝜃 = 1 2) 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠 2 𝜃 3) 𝑠𝑖𝑛𝜃 = 1 − 𝑐𝑜𝑠 2 𝜃

2

11) 𝑐𝑜𝑠𝑒𝑐 2 𝜃 = 1 + 𝑐𝑜𝑡 2 𝜃 12)𝑐𝑜𝑠𝑒𝑐𝜃 = 1 + 𝑐𝑜𝑡 2 𝜃 13) 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 𝑐𝑜𝑡 2 𝜃 = 1 14) 𝑐𝑜𝑡 2 𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 1 15) cotθ = 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 1

,ay; 8: mstpay; r.m

DO

i) xU cUisapd; tisgug;G 𝐶𝑆𝐴 = 2𝜋𝑟ℎ r.m ii) xU cUisapd; nkhj;jg; gug;gsT 𝑇𝑆𝐴 = 2𝜋𝑟(ℎ + 𝑟) iii) xU cUisapd; fd msT 𝑉 = 𝜋𝑟 2 ℎ f.m iv) xU $k;gpd; rhAauk; 𝑙 = 𝑟 2 + ℎ2 v) xU $k;gpd; cauk; ℎ = 𝑙 2 − 𝑟 2 vi) xU $k;gpd; Muk; 𝑟 = 𝑙 2 − ℎ2 vii) xU $k;gpd; tisgug;G 𝐶𝑆𝐴 = 𝜋𝑟𝑙 r.m viii) xU $k;gpd; nkhj;jg; gug;gsT 𝑇𝑆𝐴 = 𝜋𝑟(𝑙 + 𝑟) r.m 1 ix) xU $k;gpd; fd msT 𝑉 = 3 𝜋𝑟 2 ℎ f.m

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muRj;Njh;T rpwg;Gf; ifNaL 1

x) ,ilf;fz;lj;jpd; fd msT 𝑉 = 3 𝜋ℎ(𝑅2 + 𝑟 2 + 𝑅𝑟)

ntw;wpf;F top



f.m

UD Y

xi) xU Nfhsj;jpd; tisgug;G 𝐶𝑆𝐴 = 4𝜋𝑟 2 r.m 4 xii) xU Nfhsj;jpd; fd msT 𝑉 = 3 𝜋𝑟 3 f.m

xiii) xU miuf;Nfhsj;jpd; tisgug;G 𝐶𝑆𝐴 = 2𝜋𝑟 2 r.m xiv) xU miuf;Nfhsj;jpd; nkhj;jg;gug;gsT 𝑇𝑆𝐴 = 3𝜋𝑟 2 r.m 4 xv) xU miuf;Nfhsj;jpd; fd msT 𝑉 = 3 𝜋𝑟 3 f.m

OZ Y

xix) xU cs;sPlw;w Nfhsj;jpd; fd msT 4 𝑉 = 3 𝜋(𝑅3 − 𝑟 3 ) f.m

ST

xvi) xU cs;sPlw;w cUisapd; tisgug;G 𝐶𝑆𝐴 = 2𝜋ℎ(𝑅 + 𝑟) r.m xvii) xU cs;sPlw;w cUisapd; nkhj;jg;gug;gsT 𝑇𝑆𝐴 = 2𝜋 (𝑅 + 𝑟) (𝑅 − 𝑟 + ℎ) r.m xviii) xU cs;sPlw;w cUisapd; fd msT 𝑉 = 𝜋ℎ (𝑅 + 𝑟) (𝑅 − 𝑟) f.m

,ay; 11: Gs;spapay;

i) tPr;R = L – S 𝐿−𝑆 ii) tPr;Rf;nfO = 𝐿+𝑆 iii)

iv) njhFf;fg;gl;l tptuq;fspd; jpl;ltpyf;fk; 1.

njhFf;fg;glhj tptuq;fspd; jpl;ltpyf;fk;

1. 𝜎 =

𝑛



∑ 𝑑2

∑𝑥 2 𝑛

(𝑑 = 𝑥 − 𝑥 )

𝑛

DO

2. 𝜎 =

∑𝑥2

3. 𝜎 = 4. 𝜎 =

∑ 𝑑2 𝑛

∑ 𝑑2 𝑛

3.





∑𝑑 2 𝑛

∑𝑑 2 𝑛

(𝑑 = 𝑥 − 𝐴)

×𝑐

(𝑑 =

𝑥−𝐴 𝑐

)

𝜎=

𝜎=

∑ 𝑓𝑑 2 ∑ 𝑓𝑑 2 ∑𝑓

∑ 𝑓𝑑 2

(𝑑 = 𝑥 − 𝑥 ) 2. 𝜎 =

∑𝑓



∑ 𝑓𝑑 2 ∑𝑓

×𝑐

(𝑑 =

∑𝑓

𝑥−𝐴 𝑐



∑ 𝑓𝑑 2 ∑𝑓

(𝑑 = 𝑥 − 𝐴)

)

v) nfhLf;fg;gl;l Gs;sp tptuj;jpy; cs;s xt;nthU vz;ZlDk; (kjpg;G)VNjDk; xU Fwpg;gpl;l vz;izf; $l;bdhNyh my;yJ fopj;jhNyh fpilf;Fk; Gjpa tptuj;jpd;jpl;ltpyf;fk; khwhJ. vi) nfhLf;fg;gl;l tptuj;jpYs;s xt;nthU vz;izAk; (kjpg;G) xU khwpyp k My; ngUf;f my;yJ tFf;f fpilf;Fk; Gjpa kjpg;Gfspd; jpl;l tpyf;fkhdJ> gioa jpl;ltpyf;fj;ij khwpyp k My; ngUf;f my;yJ tFf;f fpilf;Fk; vz;zhf ,Uf;Fk;. vii) Kjy; n ,ay; vz;fspd; jpl;ltpyf;fk; 𝜎 = viii) khWghl;Lf; nfO C. V =

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𝜎 𝑥

𝑛 2 −1 12

× 100

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Doozy Study_SSLC MATHS FORMULAS.pdf

i) xU $l;Lj; njhlh ;thpirapy; ntJ cWg;G fhZk; ... ii) xU $l;Lj; njhlh ; thpirapy; n cWg;Gfs; tiu .... 2θ − tan. 2θ = 1. 7) sec. 2θ = 1 + tan. 2θ. 8) secθ = 1 + tan. 2θ. 9) tan.

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DOOZY STUDY NMMS Question and Answer A (15).pdf ...
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N26349A GCSE Maths Paper 3 Higher Tier.indd - GCSE Maths Tutor
Triangle ABC is similar to triangle DEF. Angle BAC = angle EDF. In triangle ABC, AB = 81 cm, BC = 70 cm, AC = 18 cm. In triangle DEF, DE = 63 cm. (a) Calculate the length of DF. ..................... cm. (2). (b) Calculate the size of angle BAC. Give

N26350A GCSE Maths Paper 3 Higher Terminal ... - GCSE Maths Tutor
Formulae: Higher Tier. You must not write on this formulae page. Anything you write on this formulae page will gain NO credit. Volume of a prism = area of cross section × length. Volume of sphere πr3. Volume of cone πr2h. Surface area of sphere =

preambleof_TGTBIOSCIENCE-MATHS-SOCIAL-mainsresult.pdf ...
liquidar una hipoteca, sufragar la carrera universitaria de sus hijos o el deseo de. mantener cierto estilo de vida, hay quienes terminan olvidando si el camino que han. emprendido les ayuda o no a alcanzar sus sueños, en cuyo caso su entusiasmo ini

Maths- Average.pdf
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maths normal.pdf
stepsize, Lagrange's interpolation formula, Truncation error, Numerical differentiation, Numerical. integration, Newton-Cotes quadrature formula, Gauss's ...

maths kannada.pdf
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maths normal.pdf
Surface Area and Volume : Cube, Cuboids, Cone, Cylinder and Sphere, Conversion of solid from one. shape to another, frustum of a Cone. Trigonometry : Angles and their measurements, Trigonometric ratios of acute angles, Angles and lengths of. arc, tri

maths em.pdf
4) If R= represents the. identity function , find the values of a, b, c and d. 5) Let A= B=N and f:A B , f(x)=x2 then. find the range of f . Identify the type of function .

Maths-Distance.pdf
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Maths Formulas.pdf
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