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Mathematics Formula Handbook by Competishun

Published by Willington Island, 2022-10-08 09:13:52

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MATHEMATICS FORMULA BOOKLET w.jeebooks INDEX S.No. Topic Page No. 1. Straight Line 2–3 2. Circle 4 3. Parabola 5 4. Ellips 5 –6 5. Hyperbola 6–7 6. Limit of Function 8–9 7. Method of Differentiation 9 – 11 8. Application of Derivatves 11 – 13 9. Indefinite Intedration 14 – 17 10. Definite Integration 17 – 18 11. Fundamental of Mathematics 19 – 21 12. Quadratic Equation 22 – 24 13. Sequence & Series 24 – 26 14. Binomial Theorem 26 – 27 15. Permutation & Combinnation 28 – 29 16. Probability 29 – 30 17. Complex Number 31 – 32 18. Vectors 32 – 35 19. Dimension 35 – 40 20. Solution of Triangle 41 – 44 21. Inverse Trigonometric Functions 44 – 46 22. Statistics 47 – 49 23. Mathematical Reasoning 49 – 50 24. Sets and Relation 50 – 51 WWW.JEEBOOKS.IN Page # 1

MATHEMATICS FORMULA BOOKLET w.jeebooks STRAIGHT LINE 1. Distance Formula: d  (x1 – x2 )2  (y1 – y2 )2 . 2. Section Formula : x= mx2 x1 ;y= my2 y1 . mn mn 3. Centroid, Incentre & Excentre: Centroid G  x1  x2  x3 , y1  y2  y3  , 3 3  Incentre I  ax1 bx2  cx3 , ay1 by 2 cy3   abc abc    ax1  bx 2  cx3 , ay1  by2 cy3   abc abc  Excentre I1 4. Area of a Triangle: 1 x1 y1  ABC = 2 2 2 x3 y3 1 5. Slope Formula: yy Line Joining two points (x1 y1) & (x2 y2), m = x1  x2 6. Condition of collinearity of three points: 1 y1 x2 2 =0 x3 y3 1 7. Angle between two straight lines : tan  = m1  m2 . Page # 2  m1m2 WWW.JEEBOOKS.IN

8. ax + by + c = 0 and ax + by + c = 0 two lines ab c 1. parallel if =  . a b c w.jeebooks 2. Distance between two parallel lines = c1c 2 . a 2 b2 3 Perpendicular : If aa + bb = 0. 9. A point and line: 1. Distance between point and line = a x1  b y1  c . a2  b2 2. Reflection of a point about a line: x x1  y y1   2 ax 1  by 1  c a b a2 b 2 3. Foot of the perpendicular from a point on the line is xx1  yy1   ax1by1c a b a2 b2 10. Bisectors of the angles between two lines: ax  by  c = ± ax  by  c a2  b2 a2 b2 11. Condition of Concurrency : a1 b1 c1 of three straight lines aix+ biy + ci = 0, i = 1,2,3 is a 2 b2 c2 = 0. a 3 b3 c3 12. A Pair of straight lines through origin: ax² + 2hxy + by² = 0 If  is the acute angle between the pair of straight lines, then tan  2 h2 ab = ab . WWW.JEEBOOKS.IN Page # 3

CIRCLE 1. Intercepts made by Circle x2 + y2 + 2gx + 2fy + c = 0 on the Axes: (a) 2 g2 c on x -axis (b) 2 f 2c on y - aixs 2. Parametric Equations of a Circle: x = h + r cos  ; y = k + r sin  3. Tangent : (a) Slope form : y = mx ± a 1 m2 (b) Point form : xx + yy = a2 or T = o 11 (c) Parametric form : x cos  + y sin  = a. 4. Pair of Tangents from a Point: SS = T². 1 w.jeebooks 5. Length of a Tangent : Length of tangent is S1 6. Director Circle: x2 + y2 = 2a2 for x2 + y2 = a2 7. Chord of Contact: T = 0 2 LR 1. Length of chord of contact = R2  L2 2. Area of the triangle formed by the pair of the tangents & its chord of R L3 contact = R2  L2 3. Tangent of the angle between the pair of tangents from (x1, y1) =  2RL  L2  R2 4. Equation of the circle circumscribing the triangle PT T is : 12 (x  x1) (x + g) + (y  y1) (y + f) = 0. 8. Condition of orthogonality of Two Circles: 2 g g + 2 f f = c + c . 12 12 1 2 9. Radical Axis : S1  S2 = 0 i.e. 2 (g1  g2) x + 2 (f1  f2) y + (c1  c2) = 0. 10. Family of Circles: S1 + K S2 = 0, S + KL = 0. WWW.JEEBOOKS.IN Page # 4

PARABOLA 1. Equation of standard parabola : y2 = 4ax, Vertex is (0, 0), focus is (a, 0), Directrix is x + a = 0 and Axis is y = 0. Length of the latus rectum = 4a, ends of the latus rectum are L(a, 2a) & L’ (a,  2a). w.jeebooks 2. Parametric Representation: x = at² & y = 2at 3. Tangents to the Parabola y² = 4ax: a 1. Slope form y = mx + (m  0) 2. Parametric form ty = x + at2 m 3. Point form T = 0 4. Normals to the parabola y² = 4ax : y  y1 =  y1 (x  x1) at (x1, y1) ; y = mx  2am  am3 at (am2 2am) ; 2a y + tx = 2at + at3 at (at2, 2at). ELLIPSE 1. Standard Equation : x2 y2 = 1, where a > b & b² = a² (1  e²).  a2 b2 Eccentricity: e = 1 b2 , (0 < e < 1), Directrices : x =± a. a2 e Focii : S  (± a e, 0). Length of, major axes = 2a and minor axes = 2b Vertices : A ( a, 0) & A  (a, 0) .  Latus Rectum : = 2b2  2a 1 e2 a 2. Auxiliary Circle : x² + y² = a² 3. Parametric Representation : x = a cos  & y = b sin  4. Position of a Point w.r.t. an Ellipse: The point P(x1, y1) lies outside, inside or on the ellipse according as; x12  y12 1 > < or = 0. a2 b2 WWW.JEEBOOKS.IN Page # 5

5. The line y = mx + c meets the ellipse x2  y2 = 1 in two points real, a2 b2 w.jeebooks coincident or imaginary according as c² is < = or > a²m² + b². 6. Tangents: Slope form: y = mx ± a2m2  b2 , Point form : xx1  yy1 1, a2 b2 Parametric form: xcos  ysin  1 ab 7. Normals:  a2x b2y  = a²  b², ax. sec   by. cosec  = (a²  b²), y = mx  a2 b2 m x1 y1 . a2 b2m2 8. Director Circle: x² + y² = a² + b² HYPERBOLA 1. Standard Equation: x2  y2  1 , where b2 = a2 (e2  1). Standard equation of the hyperbola is a2 b2 a Focii : S  (± ae, 0) Directrices : x = ± e Vertices : A (± a, 0) Latus Rectum (  ) : = 2b2 = 2a (e2  1). a 2. Conjugate Hyperbola : x2  y2 1 & x2  y2 1 are conjugate hyperbolas of each. a2 b2 a2 b2 3. Auxiliary Circle : x2 + y2 = a2. 4. Parametric Representation : x = a sec & y = b tan  WWW.JEEBOOKS.IN Page # 6

5. A Point 'P' w.r.t. A Hyperbola : S1  x12  y12 1 >, = or < 0 according as the point (x1, y1) lies inside, on a2 b2 w.jeebooks or outside the curve. 6. Tangents : (i) Slope Form : y = m x  a2m2 b2 (ii) Point Form : at the point (x1, y1) is xx1  yy1  1. a2 b2 (iii) Parametric Form : xsec  ytan 1. ab 7. Normals : a2x b2y (a) at the point P (x1, y1) is  = a2 + b2 = a2 e2. x1 y1 (b) at the point P (a sec , b tan ) is ax  by = a2 + b2 = a2 e2. sec tan (c) Equation of normals in terms of its slope 'm' are y = mx   a2  b2 m . a2  b2m2 8. Asymptotes : x  y 0 and x  y 0 . ab ab Pair of asymptotes : x 2  y2 0 . a 2 b2 9. Rectangular Or Equilateral Hyperbola : xy = c2, eccentricity is 2 .  Vertices : (± c, ±c) ; Focii :  2c, 2c . Directrices : x + y =  2c Latus Rectum (l ) :  = 2 2 c = T.A. = C.A. Parametric equation x = ct, y = c/t, t  R – {0} Equation of the tangent at P (x1, y1) is xy = 2 & at P (t) is x  t + t y = 2 c. x1 y1 Equation of the normal at P (t) is x t3  y t = c (t4  1). Chord with a given middle point as (h, k) is kx + hy = 2hk. WWW.JEEBOOKS.IN Page # 7

LIMIT OF FUNCTION 1. Limit of a function f(x) is said to exist as x  a when, w.jeebooks Limit f (a  h) = Limit f (a + h) = some finite value M. h0 h0 (Left hand limit) (Right hand limit) 2. Indeterminant Forms: 0 ,  , 0 , º, 0º,and 1. 0 3. Standard Limits: Limit sinx Limit tanx Limit tan1x x0 x = x0 x = x0 x = Limit sin1x = Limit ex 1 = Limit n(1 x) x x x =1 x0 x0 x0 Limit (1 + x)1/x = Limit 1 1 x = e, Limit ax 1 = log a, a > 0,   x e x0 x  x0 x Limit xn  an = nan – 1. x a x a 4. Limits Using Expansion (i) ax 1 x lna  x2ln2 a  x3ln3 a  .........a  0 1! 2! 3! (ii) ex  1 x x2  x3  ......  1! 2! 3! (iii) ln (1+x) = x  x2  x3  x4  .........for  1  x  1 234 (iv) sin x x  x3  x5  x7  ..... 3! 5! 7! WWW.JEEBOOKS.IN Page # 8

(v) cosx 1 x2  x4  x6  ..... 2! 4! 6! w.jeebooks(vi) tan x = x  x3  2x5  ...... 3 15 (vii) for |x| < 1, n  R (1 + x)n n(n  1) n(n  1)(n  2) x3 + ............ = 1 + nx + x2 + 1. 2 1. 2 . 3 5. Limits of form 1, 00, 0 Also for (1) type of problems we can use following rules. lim (1 + x)1/x = e, lim [f(x)]g(x) , x0 xa lim [f ( x)1] g(x) where f(x)  1 ; g(x) as x  a = exa 6. Sandwich Theorem or Squeeze Play Theorem: If f(x)  g(x)  h(x)  x & Limit f(x) =  = Limit h(x) then Limit g(x) =  . xa xa x a METHOD OF DIFFERENTIATION 1. Differentiation of some elementary functions d 2. d (ax) = ax n a 1. (xn) = nxn – 1 dx dx 4. d (logax) = 1 d1 dx x n a 3. dx (n |x|) = x d d 5. (sin x) = cos x 6. (cos x) = – sin x dx d dx 7. (sec x) = sec x tan x dx 8. d (cosec x) = – cosec x cot x d dx 9. (tan x) = sec2 x dx 10. d (cot x) = – cosec2 x dx WWW.JEEBOOKS.IN Page # 9

2. dd d 2. (k f(x)) = k f(x) 1. (f ± g) = f(x) ± g(x) dx dx dx w.jeebooksd 3. (f(x) . g(x)) = f(x) g(x) + g(x) f(x) dx d  f(x)  = g(x) f(x)  f(x) g(x) d 4. dx g(x) g2 (x) 5. (f(g(x))) = f(g(x)) g(x) dx Derivative Of Inverse Trigonometric Functions. d sin–1 x 1 dcos–1 x 1 =, = – , for – 1 < x < 1. dx 1 x2 dx 1 x2 d tan–1 x 1 dcot –1 x 1 dx = 1 x2 , dx = – 1 x2 (x  R) d sec –1 x 1 dcos ec –1x =, dx | x | x2  1 dx 1 = – , for x  (– , – 1)  (1, ) | x | x2 1 3. Differentiation using substitution Following substitutions are normally used to simplify these expression.  (i) x2  a2 by substituting x = a tan , where – 2 <  2  (ii) a2  x2 by substituting x = a sin , where – 2   2 (iii) x2  a2 by substituting x = a sec , where [0, ],   2 xa (iv) a  x by substituting x = a cos , where (0, ]. WWW.JEEBOOKS.IN Page # 10

4. Parametric Differentiation If y = f() & x = g() where is a parameter, then dy  dy/d . dx dx/d w.jeebooks 5. Derivative of one function with respect to another dy dy/dx f'(x) Let y = f(x); z = g(x) then  . dz dz/dx g'(x) f(x) g(x) h(x) .6 If F(x) = l(x) m(x) n(x) , where f, g, h, l, m, n, u, v, w are differentiable u(x) v(x) w(x) f'(x) g' ( x) h'(x) f(x) g(x) h(x) + m(x) n(x) + l'(x) m'(x) n'(x) functions of x then F (x) = l(x) v(x) w(x) u(x) v(x) w(x) u(x) f(x) g(x) h(x) l(x) m(x) n(x) u'(x) v'(x) w'(x) APPLICATION OF DERIVATIVES 1. Equation of tangent and normal Tangent at (x , y ) is given by (y – y) = f(x1) (x – x) ; when, f(x1) is real. 11 1 1 And normal at (x1 , y1) is 1 real. (y – y1) = – f(x1) (x – x1), when f(x1) is nonzero 2. Tangent from an external point Given a point P(a, b) which does not lie on the curve y = f(x), then the equation of possible tangents to the curve y = f(x), passing through (a, b) can be found by solving for the point of contact Q. f(h)  b f(h) = h  a WWW.JEEBOOKS.IN Page # 11

w.jeebooks f(h)  b And equation of tangent is y – b = h  a (x – a) 3. Length of tangent, normal, subtangent, subnormal (i) PT = | k | 1 1 = Length of Tangent m2 p(h,k) (ii) PN = | k | 1 m2 = Length of Normal k NM T (iii) TM = = Length of subtangent m (iv) MN = |km| = Length of subnormal. 4. Angle between the curves Angle between two intersecting curves is defined as the acute angle between their tangents (or normals) at the point of intersection of two curves (as shown in figure). tan  = m1  m2 1 m1m2 5. Shortest distance between two curves Shortest distance between two non-intersecting differentiable curves is always along their common normal. (W herever defined) 6. Rolle’s Theorem : If a function f defined on [a, b] is (i) continuous on [a, b] (ii) derivable on (a, b) and (iii) f(a) = f(b), then there exists at least one real number c between a and b (a < c < b) such that f(c) = 0 WWW.JEEBOOKS.IN Page # 12

7. Lagrange’s Mean Value Theorem (LMVT) : If a function f defined on [a, b] is (i) continuous on [a, b] and (ii) derivable on (a, b) then there exists at least one real numbers between a and b (a < c < b) such w.jeebooks f(b)  f(a) that b  a = f(c) 8. Useful Formulae of Mensuration to Remember : 1. Volume of a cuboid = bh. 2. Surface area of cuboid = 2(b + bh + h). 3. Volume of cube = a3 4. Surface area of cube = 6a2 1 5. Volume of a cone = 3 r2 h. 6. Curved surface area of cone = r ( = slant height) 7. Curved surface area of a cylinder = 2rh. 8. Total surface area of a cylinder = 2rh + 2r2. 4 9. Volume of a sphere = 3 r3. 10. Surface area of a sphere = 4r2. 1 11. Area of a circular sector = r2 , when  is in radians. 2 12. Volume of a prism = (area of the base) × (height). 13. Lateral surface area of a prism = (perimeter of the base) × (height). 14. Total surface area of a prism = (lateral surface area) + 2 (area of the base) (Note that lateral surfaces of a prism are all rectangle). 1 15. Volume of a pyramid = 3 (area of the base) × (height). 1 16. Curved surface area of a pyramid = 2 (perimeter of the base) × (slant height). (Note that slant surfaces of a pyramid are triangles). WWW.JEEBOOKS.IN Page # 13

INDEFINITE INTEGRATION 1. If f & g are functions of x such that g(x) = f(x) then, w.jeebooksd f(x) dx = g(x) + c  {g(x)+c} = f(x), where c is called the constant of dx integration. 2. Standard Formula:  ax  b n1   a n 1 + c, n  1 (i) (ax + b)n dx =  dx 1 (ii) = n (ax + b) + c ax  b a 1 (iii) eax+b dx = eax+b + c a  1 apxq (iv) apx+q dx = p n a + c; a > 0 (v) 1 (vi) sin (ax + b) dx =  cos (ax + b) + c (vii) a (viii) (ix) 1 (x) cos (ax + b) dx = sin (ax + b) + c a 1 tan(ax + b) dx = n sec (ax + b) + c a 1 cot(ax + b) dx = n sin(ax + b)+ c a 1 sec² (ax + b) dx = tan(ax + b) + c a  cosec²(ax + b) dx =  1 cot(ax + b)+ c a WWW.JEEBOOKS.IN Page # 14

(xi) secx dx = n (secx + tanx) + c OR n tan    2x  +c 4 w.jeebooks(xii) cosec x dx = n (cosecx  cotx) + c x OR n tan + c OR  n (cosecx + cotx) + c 2 (xiii)  dx x = sin1 + c a2  x2 a  dx 1 x (xiv) a2  x2 = tan1 +c a a  dx 1 x (xv) |x| x2 a2 = sec1 + c aa dx = n x  x2  a2 + c x2 a2   (xvi) d x = n x  x2  a2 + c x2  a2   (xvii) (xviii) d x 1 ax = n +c a2  x2 ax 2a (xix) d x 1 xa = n +c x2  a2 xa 2a (xx) a2  x2 dx = x a2  x2 a2 x + sin1 + c 2 2a (xxi)  x2  a2 dx = x x2  a2 a2  x  x2  a2  2 n  a +c + 2  (xxii) x2  a2 dx = x x2  a2 a2  x  x2  a2  2 n   +c  a  2 WWW.JEEBOOKS.IN Page # 15

3. If we subsitute f(x) = t, then f (x) dx = dt 4. Integration by Part : w.jeebooks g(x) dx –  d f(x) g(x) dx  dx  dx     f(x) g(x) dx = f(x) 5. Integration of type dx dx , ax2  bx  c dx  ax2  bx  c, ax2  bx  c Make the substitution x  b  t 2a 6. Integration of type  px q c dx,  px  q dx, (px  q) ax2  bx  c dx ax2  bx  ax2  bx  c Make the substitution x + b = t , then split the integral as some of two 2a integrals one containing the linear term and the other containing constant term. 7. Integration of trigonometric functions (i) dx dx a  b sin2 x OR  a  bcos2 x dx OR a sin2 x  b sin x cos x  c cos2 x put tan x = t. (ii) dx OR dx  a  b sin x  a  bcos x  dx x OR a  b sin x  c cos x put tan = t 2  a.cos x  b.sin x  c d (iii) .cos x  m.sin x  n dx. Express Nr  A(Dr) + B dx (Dr) + c & proceed. WWW.JEEBOOKS.IN Page # 16

8.  x2  1 dx where K is any constant. x 4  Kx2  1 w.jeebooks Divide Nr & Dr by x² & put x  1 = t. x 9. Integration of type dx dx  (ax  b) px  q OR (ax2  bx  c) px  q ; put px + q = t2. 10. Integration of type  dx 1 (ax  b) px2  qx  r , put ax + b = t ;  dx 1 (ax2  b) px2  q , put x = t DEFINITE INTEGRATION Properties of definite integral bb ba  1. f(x) dx = f(t) dt  2. f(x) dx = – f(x) dx aa ab b cb   3. f(x) dx = f(x) dx + f(x) dx a ac aa a  4. f(x) dx = (f(x)  f(x)) dx = 2 f ( x) dx , f(–x)  f(x) a 0  0   0 , f(–x)  –f(x) bb  5. f(x) dx = f(a  b  x) dx aa WWW.JEEBOOKS.IN Page # 17

aa  6. f(x) dx = f(a  x) dx 00 w.jeebooks2a a a 2 f(x)dx f(x) (f(x)  f(2a  x)) dx    7. dx = = , f(2a – x)  f(x) 00  0  0 , f(2a – x)  –f(x) 8. If f(x) is a periodic function with period T, then nT T anT T    f(x) dx = n f(x) dx, n  z, f(x) dx = n f(x) dx, n  z, a  R 00 a0 nT T anT a    f(x) dx = (n – m) f(x) dx, m, n  z, f(x) dx = f(x) dx, n  z, a  R mT 0 nT 0 bnT a  f(x) dx = f(x) dx, n  z, a, b  R anT a 9. If (x)  f(x)   (x) for a  x  b, b bb   then (x) dx  f(x) dx  (x) dx a aa b 10. If m  f(x)  M for a  x  b, then m (b – a)  f(x) dx  M (b – a) a b 11. If f(x)  0 on [a, b] then f(x) dx  0 a h(x) dF(x) Leibnitz Theorem : If F(x) = f(t) dt , then dx = h(x) f(h(x)) – g(x) f(g(x)) g( x ) WWW.JEEBOOKS.IN Page # 18

FUNDAMENTAL OF MATHEMATICS Intervals : w.jeebooksIntervals are basically subsets of R and are commonly used in solving inequalities or in finding domains. If there are two numbers a, b  R such that a < b, we can define four types of intervals as follows : Symbols Used (i) Open interval : (a, b) = {x : a < x < b} i.e. end points are not included. ( ) or ] [ (ii) Closed interval : [a, b] = {x : a  x  b} i.e. end points are also in c lu d e d . [] This is possible only when both a and b are finite. (iii) Open-closed interval : (a, b] = {x : a < x  b} ( ] or ] ] (iv) Closed - open interval : [a, b) = x : a  x < b} [ ) or [ [ The infinite intervals are defined as follows : (i) (a, ) = {x : x > a} (ii) [a, ) = {x : x  a} (, b] = {x : x  b} (iii) (– , b) = {x : x < b} (iv) (v) (– ) = {x : x  R} Properties of Modulus : For any a, b  R |a|  0, |a| = |–a|, |a|  a, |a|  –a, |ab| = |a| |b|, a |a| |a + b|  |a| + |b|, |a – b|  ||a| – |b|| =, b |b| Trigonometric Functions of Sum or Difference of Two Angles: (a) sin (A ± B) = sinA cosB ± cosA sinB  2 sinA cosB = sin(A+B) + sin(AB) and and 2 cosA sinB = sin(A+B)  sin(AB) (b) cos (A ± B) = cosA cosB  sinA sinB  2 cosA cosB = cos(A+B) + cos(AB) and 2sinA sinB = cos(AB)  cos(A+B) (c) sin²A  sin²B = cos²B  cos²A = sin (A+B). sin (A B) (d) cos²A  sin²B = cos²B  sin²A = cos (A+B). cos (A  B) cot A cot B  1 (e) cot (A ± B) = cot B  cot A (f) tan (A + B + C) = tan A  tanB  tanCtan A tanB tan C . 1 tan A tanB  tanB tan C tan C tan A WWW.JEEBOOKS.IN Page # 19

Factorisation of the Sum or Difference of Two Sines or Cosines: (a) sinC + sinD = 2 sin CD CD cos 22 w.jeebooks(b)sinC  sinD = 2 cos CDCD sin 22 (c) cosC + cosD = 2 cos CD cos CD 22 CD CD (d) cosC  cosD =  2 sin 2 sin 2 Multiple and Sub-multiple Angles :  (a) cos 2A = cos²A  sin²A = 2cos²A  1 = 1  2 sin²A; 2 cos² 2  = 1 + cos , 2 sin² = 1  cos . 2 2 tan A 1tan2 A (b) sin 2A = 1 tan2 A , cos 2A = 1tan2 A (c) sin 3A = 3 sinA  4 sin3A where n   (d) cos 3A = 4 cos3A  3 cosA 3 tan A  tan3 A (e) tan 3A = 1 3 tan2A Important Trigonometric Ratios: (a) sin n  = 0 ; cos n  = 1 ; tan n  = 0,  3 1 5 (b) sin 15° or sin 12 = 2 2 = cos 75° or cos 12 ;  3 1 5 cos 15° or cos = = sin 75° or sin ; 12 2 2 12 tan 15° = 3 1 = 2 3 = cot 75° ; tan 75° 3 1 3 1 = = 2 3 = cot 15° 3 1  5 1  5 1 (c) sin or sin 18° = & cos 36° or cos 5 = 4 10 4 WWW.JEEBOOKS.IN Page # 20

Range of Trigonometric Expression: – a2  b2 a sin  + b cos   a2  b2 Sine and Cosine Series :  sin  + sin (+) + sin ( + 2 ) +...... + sin   n 1 w.jeebooks sin n    n  1  2  2  = sin   sin 2  cos  + cos (+) + cos ( + 2 ) +...... + cos   n  1 sin n cos    n 1 2  2 = sin  2 Trigonometric Equations Principal Solutions: Solutions which lie in the interval [0, 2) are called Principal solutions. General Solution : (i) sin  = sin    = n  + (1)n  where     ,  , n  . 2 2  (ii) cos  = cos    = 2 n  ±  where   [0, ], n  . (iii) tan  = tan    = n +  where     ,  , n  .  2 2  (iv) sin²  = sin² , cos²  = cos² , tan²  = tan²    = n  ±  WWW.JEEBOOKS.IN Page # 21

QUADRATIC EQUATIONS 1. Quadratic Equation : a x2 + b x + c = 0, a  0 w.jeebooks b  b2  4ac x = , The expression b2  4 a c  D is called 2a discriminant of quadratic equation. If ,  are the roots, then (a)  +  =  b c (b)  = aa A quadratic equation whose roots are  & , is (x ) (x  ) = 0 i.e. x2  ( +  ) x +  = 0 2. Nature of Roots: Consider the quadratic equation, a x2 + b x + c = 0 having ,  as its roots; D  b2  4 a c D=0 D0 Roots are equal =  =  b/2a Roots are unequal a, b, c  R & D > 0 a, b, c  R & D < 0 Roots are real Roots are imaginary  = p + i q,  = p  i q a, b, c  Q & a, b, c  Q & D is a perfect square D is not a perfect square  Roots are rational  Roots are irrational  i.e.  = p + q ,  = p  q a = 1, b, c   & D is a perfect square  Roots are integral. WWW.JEEBOOKS.IN Page # 22

3. Common Roots: Consider two quadratic equations a1 x2 + b1 x + c1 = 0 & a2 x2 + b2 x + c2 = 0. (i) If two quadratic equations have both roots common, then w.jeebooks a1 = b1 = c1 . a2 b2 c2 (ii) If only one root  is common, then  = c1 a2  c 2 a1 = b1 c 2  b2 c1 a1 b2  a2 b1 c1 a2  c 2 a1 4. Range of Quadratic Expression f (x) = a x2 + b x + c. Range in restricted domain: Given x  [x1, x2] (a) b If   [x1, x2] then, 2a     f(x)  min f (x1) , f (x2 ) , max f (x1) , f (x2 ) b [x , x] (b) If   1 2 then, 2a f(x)    f (x1) , f(x2) ,  D  , max  f (x1) , f (x2 ) ,  D    min  4a   4a         WWW.JEEBOOKS.IN Page # 23

5. Let f (x) = ax² + bx + c, where a > 0 & a, b, c  R. (i) Conditions for both the roots of f (x) = 0 to be greater than a specified number‘x0’ are b²  4ac  0; f (x0) > 0 & ( b/2a) > x0. w.jeebooks (ii) Conditions for both the roots of f (x) = 0 to be smaller than a specified number ‘x0’ are b²  4ac  0; f (x0) > 0 & ( b/2a) < x0. (iii) Conditions for both roots of f (x) = 0 to lie on either side of the number ‘x0’ (in other words the number ‘x0’ lies between the roots of f (x) = 0), is f (x0) < 0. (iv) Conditions that both roots of f (x) = 0 to be confined between the numbers x and x, (x < x) are b²  4ac  0; f (x ) > 0 ; f (x ) > 0 & 1 2 1 2 12 x < ( b/2a) < x. 1 2 (v) Conditions for exactly one root of f (x) = 0 to lie in the interval (x , x ) 12 i.e. x < x < x is f (x ). f (x ) < 0. 1 2 12 SEQUENCE & SERIES An arithmetic progression (A.P.) : a, a + d, a + 2 d,..... a + (n  1) d is an A.P. Let a be the first term and d be the common difference of an A.P., then nth term = t = a + (n – 1) d n The sum of first n terms of A.P. are n n Sn = 2 [2a + (n – 1) d] = 2 [a + ] rth term of an A.P. when sum of first r terms is given is tr = Sr – Sr – 1. Properties of A.P. (i) If a, b, c are in A.P.  2 b = a + c & if a, b, c, d are in A.P.  a + d = b + c. (ii) Three numbers in A.P. can be taken as a  d, a, a + d; four numbers in A.P. can be taken as a  3d, a  d, a + d, a + 3d; five numbers in A.P. are a  2d, a  d, a, a + d, a + 2d & six terms in A.P. are a  5d, a  3d, a  d, a + d, a + 3d, a + 5d etc. (iii) Sum of the terms of an A.P. equidistant from the beginning & end = sum of first & last term. Arithmetic Mean (Mean or Average) (A.M.): If three terms are in A.P. then the middle term is called the A.M. between the other two, so if a, b, c are in A.P., b is A.M. of a & c. WWW.JEEBOOKS.IN Page # 24

If a, b are any two given numbers & a, A1, A2,...., An, b are in A.P. then A1, A2,... An are the ba n A.M.’s between a & b. A1 = a + n  1 , w.jeebooks 2 (b  a) n (b  a) A2 = a + n  1 ,......, An = a + n  1 n  Ar = nA where A is the single A.M. between a & b. r1 Geometric Progression: a, ar, ar2, ar3, ar4,...... is a G.P. with a as the first term & r as common ratio. (i) nth term = a rn1  (ii) a rn  1 , r 1 Sum of the first n terms i.e. Sn =  r  1  na , r  1 (iii) Sum of an infinite G.P. when r < 1 is given by  a S = 1 r r  1 . Geometric Means (Mean Proportional) (G.M.): If a, b, c > 0 are in G.P., b is the G.M. between a & c, then b² = ac nGeometric Means Between positive number a, b: If a, b are two given numbers & a, G1, G2,....., Gn, b are in G.P.. Then G1, G2, G3,...., Gn are n G.M.s between a & b. G1 = a(b/a)1/n+1, G2 = a(b/a)2/n+1,......, Gn = a(b/a)n/n+1 Harmonic Mean (H.M.): 2ac If a, b, c are in H.P., b is the H.M. between a & c, then b = a  c . 11 1  1  .......  1  H.M. H of a1, a2 , ........ an is given by H = n  a2 an   a1  WWW.JEEBOOKS.IN Page # 25

Relation between G² = AH, A.M.  G.M.  H.M. (only for two numbers) and A.M. = G.M. = H.M. if a = a = a = ...........= a n 123 w.jeebooks Important Results n nn nn     (i) (ar ± br) = ar ± br. (ii) k ar = k ar. r1 r1 r1 r1 r1 n (iii) k = nk; where k is a constant. r1 n n (n 1) (iv) r = 1 + 2 + 3 +...........+ n = r1 2 n n (n 1) (2n 1) (v) r² = 12 + 22 + 32 +...........+ n2 = r1 6 n n2 (n 1)2 (vi) r3 = 13 + 23 + 33 +...........+ n3 = r1 4 BINOMIAL THEOREM 1. Statement of Binomial theorem : If a, b  R and n  N, then (a + b)n = nC0 anb0 + nC1 an–1 b1 + nC2 an–2 b2 +...+ nCr an–r br +...+ nCn a0 bn n = n Cr anrbr r 0 2. Properties of Binomial Theorem : (i) General term : T = nC an–r br r+1 r (ii) Middle term (s) : (a) If n is even, there is only one middle term,  n  2   2  which is th term. (b) If n is odd, there are two middle terms,  n  1  n  1  1  2   2  which are th and th terms. WWW.JEEBOOKS.IN Page # 26

3. Multinomial Theorem : (x + x + x + ........... x )n 1 23 k  n! = r1r2 ...rk n r1! r2!... rk ! w.jeebooks x1r1 . x r2 ...x rk 2 k Here total number of terms in the expansion = Cn+k–1 k–1 4. Application of Binomial Theorem : If ( A  B)n =  + f where  and n are positive integers, n being odd and 0 < f < 1 then ( + f) f = kn where A – B2 = k > 0 and A – B < 1. If n is an even integer, then ( + f) (1 – f) = kn 5. Properties of Binomial Coefficients : (i) nC0 + nC1 + nC2 + ........+ nCn = 2n (ii) nC0 – nC1 + nC2 – nC3 + ............. + (–1)n nCn = 0 (iii) nC + nC + nC + .... = nC + nC + nC + .... = 2n–1 024 135 (iv) nCr + nCr–1 = n+1Cr (v) nCr n  r  1 n Cr1 = r 6. Binomial Theorem For Negative Integer Or Fractional Indices n(n  1) n(n  1)(n  2) (1 + x)n = 1 + nx + 2 ! x2 + 3 ! x3 + .... + n(n  1)(n  2).......(n  r  1) xr + ....,| x | < 1. r! n(n  1)(n  2).........(n  r  1) T= r! xr r+1 WWW.JEEBOOKS.IN Page # 27

PERMUTATION & COMBINNATION 1. Arrangement : number of permutations of n different things taken r at a w.jeebookstime =nP= n (n  1) (n  2)... (n  r + 1) = n! r (n  r )! 2. Circular Permutation : The number of circular permutations of n different things taken all at a time is; (n – 1)! 3. Selection : Number of combinations of n different things taken r at a n! = nPr time = nCr = r! (n  r)! r! 4. The number of permutations of 'n' things, taken all at a time, when 'p' of them are similar & of one type, q of them are similar & of another type, 'r' of them are similar & of a third type & the remaining n  (p + q + r) are all n! different is p ! q ! r ! . 5. Selection of one or more objects (a) Number of ways in which atleast one object be selected out of 'n' distinct objects is nC1 + nC2 + nC3 +...............+ nCn = 2n – 1 (b) Number of ways in which atleast one object may be selected out of 'p' alike objects of one type 'q' alike objects of second type and 'r' alike of third type is (p + 1) (q + 1) (r + 1) – 1 (c) Number of ways in which atleast one object may be selected from 'n' objects where 'p' alike of one type 'q' alike of second type and 'r' alike of third type and rest n – (p + q + r) are different, is (p + 1) (q + 1) (r + 1) 2n – (p + q + r) – 1 6. Multinomial Theorem : Coefficient of xr in expansion of (1  x)n = Cn+r1 (n  N) r 7. Let N = pa. qb. rc...... where p, q, r...... are distinct primes & a, b, c..... are natural numbers then : (a) The total numbers of divisors of N including 1 & N is = (a + 1) (b + 1) (c + 1)........ (b) The sum of these divisors is = (p0 + p1 + p2 +.... + pa) (q0 + q1 + q2 +.... + qb) (r0 + r1 + r2 +.... + rc)........ WWW.JEEBOOKS.IN Page # 28

be resolved as a product of two factors is = 1 (a  1)(b  1)(c  1).... if N is not a perfect square 2 if N is a perfect square w.jeebooks (a 1)....1 1  1)(b  1)(c  2 (d) Number of ways in which a composite number N can be resolved into two factors which are relatively prime (or coprime) to each other is equal to 2n1 where n is the number of different prime factors in N. 8. Dearrangement : Number of ways in which 'n' letters can be put in 'n' corresponding envelopes such that no letter goes to correct envelope is n !  1  1  1  1 .......... ..  (1)n 1  1 1! 2! 3! 4! n!  PROBABILITY 1. Classical (A priori) Definition of Probability : If an experiment results in a total of (m + n) outcomes which are equally likely and mutually exclusive with one another and if ‘m’ outcomes are favorable to an event ‘A’ while ‘n’ are unfavorable, then the probability of occurrence of the event ‘A’ = P(A) = m n(A) =. m  n n(S) We say that odds in favour of ‘A’ are m : n, while odds against ‘A’ are n : m. n P(A) = m  n = 1 – P(A) 2. Addition theorem of probability : P(AB) = P(A) + P(B) – P(AB) De Morgan’s Laws : (a) (A B)c = Ac Bc (b) (A B)c = Ac Bc Distributive Laws : (a) A  (B C) = (A B)  (A C) (b) A  (B C) = (A B)  (A C) (i) P(A or B or C) = P(A) + P(B) + P(C) – P(A B) – P(B C) – P(C A) + P(A B C) (ii) P (at least two of A, B, C occur) = P(B  C) + P(C  A) + P(A  B) – 2P(A B C) (iii) P(exactly two of A, B, C occur) = P(B  C) + P(C  A) + P(A  B) – 3P(A B  C) (iv) P(exactly one of A, B, C occur) = P(A) + P(B) + P(C) – 2P(B C) – 2P(C A) – 2P(A B) + 3P(A B C) WWW.JEEBOOKS.IN Page # 29

P(A  B) 3. Conditional Probability : P(A/B) = P(B) . w.jeebooks4. Binomial Probability Theorem If an experiment is such that the probability of success or failure does not change with trials, then the probability of getting exactly r success in n trials of an experiment is nCr pr qn – r, where ‘p’ is the probability of a success and q is the probability of a failure. Note that p + q = 1. 5. Expectation : If a value Mi is associated with a probability of pi , then the expectation is given by  piMi. n 6. Total Probability Theorem : P(A) = P(Bi ) . P(A / Bi ) i1 7. Bayes’ Theorem : If an event A can occur with one of the n mutually exclusive and exhaustive events B , B , ....., B and the probabilities P(A/B ), P(A/B ) .... P(A/B ) are 12 n 12 n known, then P(B / A) = P(Bi ) . P(A / Bi ) B , B , B ,........,B i 123 n n P(Bi ) . P(A / Bi ) i 1 A = (A  B1)  (A  B2)  (A  B3)  ........  (A  Bn) n P(A) + B P(A  Bi ) = P(A  B) P(A B2) + ....... + P(A  ) = 1 i1 n 8. Binomial Probability Distribution : (i) Mean of any probability distribution of a random variable is given by : µ=  pi xi =  p x = np  pi i i n = number of trials p = probability of success in each probability q = probability of failure (ii) Variance of a random variable is given by, 2 = (xi – µ)2 . pi = pi xi2 – µ2 = npq WWW.JEEBOOKS.IN Page # 30

COMPLEX NUMBER 1. The complex number system w.jeebooks z = a + ib, then a – ib is called congugate of z and is denoted by z . 2. Equality In Complex Number: z1 = z2  Re(z1) = Re(z2) and m (z1) = m (z2). 3. Properties of arguments (i) arg(z1z2) = arg(z1) + arg(z2) + 2m for some integer m. (ii) arg(z1/z2) = arg (z1) – arg(z2) + 2m for some integer m. (iii) arg (z2) = 2arg(z) + 2m for some integer m. (iv) arg(z) = 0  z is a positive real number (v) arg(z) = ± /2  z is purely imaginary and z  0 4. Properties of conjugate (i) |z| = | z | (ii) z z = |z|2(iii) z1  z2 = z1 + z2 (iv) z1  z2 = z1 – z2 (v) z1z2 = z1 z2 (vi)  z1  = z1 (z2  0) z2 z2 (vii) |z1 + z2|2 = (z1 + z2) (z1  z2 ) = |z1|2 + |z2|2 + z1 z2 + z1 z2 (viii) (z1) = z (ix) If w = f(z), then w = f( z ) (x) arg(z) + arg( z ) 5. Rotation theorem If P(z1), Q(z2) and R(z3) are three complex numbers and PQR = , then  z3  z2  = z3  z2 ei z1  z2 z1  z2 6. Demoivre’s Theorem : If n is any integer then (i) (cos  + i sin  )n = cos n + i sin n (ii) (cos 1 + i sin 1) (cos 2 + i sin 2) (cos3 + i sin 2) (cos 3 + i sin 3) .....(cos n + i sin n) = cos (1 + 2 + 3 + ......... n) + i sin (1 + 2 + 3 + ....... + n) WWW.JEEBOOKS.IN Page # 31

7. Cube Root Of Unity : (i) The cube roots of unity are 1, 1  i 3 , 1  i 3 . 22 (ii) If  is one of the imaginary cube roots of unity then 1 +  + ² = 0. In general 1 + r + 2r = 0; where r  I but is not the multiple of 3. w.jeebooks 8. Geometrical Properties: Distance formula : |z1 – z2|. Section formula : z = mz2  nz1 (internal division), z = mz2  nz1 (external mn mn d ivis io n ) (1) amp(z) =  is a ray emanating from the origin inclined at an angle  to the x axis. (2) z  a = z  b is the perpendicular bisector of the line joining a to b. z  z1 (3) If z  z2 = k  1, 0, then locus of z is circle. VECTORS . Position Vector Of A Point: let O be a fixed origin, then the position vector of a point P is the vector  OP . If a and b are position vectors of two points A and B, then,  AB = b  a = pv of B  pv of A.  DISTANCE FORMULA : Distance between the two points A (a) and B (b)  is AB = a  b    r na  mb ab SECTION FORMULA :  . Mid point of AB = . mn 2 WWW.JEEBOOKS.IN Page # 32

. Scalar Product Of Two Vectors:  =  | |  | cos , where  |, |  | a.b |a b |a b   are magnitude of a and b respectively and  is angle between a and b . 1.w.jeebooksi.i = j.j = k.k = 1; i.j = j.k = k.i = 0  projection of     a on b a. b |b|   a2. If = a1i + a2j + a3k & b = b1i + b2j + b3k then a.b = a1b1 + a2b2 + a3b3    a.b  3. The angle  between a&b is given by cos   |a| |b| , 0     4. a . b  0  a  b (a  0 b  0) . Vector Product Of Two Vectors: 1.      If a & b are two vectors &  is the angle between them then axb a b sinn ,     where n is the unit vector perpendicular to both a & b such that a,b&n forms a right handed screw system.  2. Geometrically axb = area of the parallelogram whose two adjacent sides  are represented by a&b . 3. ˆi  ˆi  ˆj  ˆj  kˆ  kˆ   ; ˆi  ˆj  kˆ, ˆj  kˆ  ˆi, kˆ  ˆi  ˆj 0 4.  = a1 ˆi +a2 ˆj + a3 kˆ  If a & b = b1 ˆi + b2 ˆj + b3 kˆ then   ˆi ˆj kˆ a  b  a1 a2 a3 b1 b2 b3    5. a  b  o  a and b are parallel (collinear)   (a  0,b  0) i.e. a  K b , where K is a scalar. WWW.JEEBOOKS.IN Page # 33

   a b a x b 6. Unit vector perpendicular to the plane of & is nˆ    axb  w.jeebooks  If a,b&c are the pv’s of 3 points A, B & C then the vector area of triangle 1    2 axb bxc cxa  ABC = .    The points A, B & C are collinear if a x b  b x c  c x a  0   Area of any quadrilateral whose diagonal vectors are d1 & d2 is given by 1  2 d1xd2  Lagrange's Identity : (a xb )2   2  2 (a.b )2     a b a.aa.b a.bb.b V. Scalar Triple Product:    The scalar triple product of three vectors a , b & c is defined as:    a x b .c  a b c sin  cos  .   Volume of tetrahydron V[abc]  In a scalar triple product the position of dot & cross can be interchanged i.e.       x c) (a x OR [ c a]     a . (b  b). c a b c]  [b  [c a b]      a . (b x c)   a .(c x b) i.e. [a b c]   [a c b ]    a1a2 a3  If a = a i+a j+a k; b = b i+b j+b k &c = c i+c j+c k then [abc] b1b2 b3 . 12 3 12 3 12 3   c1c2 c3  If a , b , c are coplanar [abc]0 .    Volume of tetrahedron OABC with O as origin & A( a ), B( b ) and C( c ) be the vertices = 1    [a b c] 6 WWW.JEEBOOKS.IN Page # 34

 if the pv’s of its vertices   are given by 1  are a , b , c & d [a  b  c  d]. 4 w.jeebooks V. Vector Triple Product:               a x ( b x c ) = (a . c) b  (a . b) c , (a x b) x c = (a . c) b  (b . c)a      (a x b) x c  a x (b x c) , in general 3-DIMENSION 1. Vector representation of a point : Position vector of point P (x, y, z) is x ˆi + y ˆj + z kˆ . 2. Distance formula : (x1  x2 )2  (y1  y2 )2  (z1  z2 )2 , AB = | OB – OA | 3. Distance of P from coordinate axes : PA = y2  z2 , PB = z2  x2 , PC = x2  y2 4. Section Formula : x= mx2  nx1 , y = my2  ny1 , z = mz2  nz1 m n m n mn Mid point : x  x1  x2 , y  y1  y2 , z  z1  z2 2 2 2 5. Direction Cosines And Direction Ratios (i) Direction cosines: Let    be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos , cos cos  are called the direction cosines of the line. The direction cosines are usually denoted by (, m, n). Thus  = cos , m = cos , n = cos . (ii) If , m, n be the direction cosines of a line, then 2 + m2 + n2 = 1 (iii) Direction ratios: Let a, b, c be proportional to the direction cosines , m, n then a, b, c are called the direction ratios. (iv) If , m, n be the direction cosines and a, b, c be the direction ratios of a vector, then    a ,m   b ,n   c a2  b2  c2 a2  b2  c2 a2  b2  c2 WWW.JEEBOOKS.IN Page # 35

, y , z ) and (x , y , z ) then the 11 1 22 2 direction ratios of line PQ are, a = x2  x1, b = y2  y1 & c = z2  z1 and the direction cosines of line PQ are   = x 2  x1 , | PQ | w.jeebooks m = y2  y1 and n = z2  z1 | PQ | | PQ | 6. Angle Between Two Line Segments: cos  = a1a2  b1b2  c1c2 . a12  b12  c12 a 2  b 2  c 2 2 2 2 The line will be perpendicular if a1a2 + b1b2 + c1c2 = 0, parallel if a1 = b1 = c1 a2 b2 c2 7. Projection of a line segment on a line If P(x1, y1, z1) and Q(x2, y2, z2) then the projection of PQ on a line having direction cosines , m, n is (x2  x1)  m(y2  y1)  n(z2  z1) 8. Equation Of A Plane : General form: ax + by + cz + d = 0, where a, b, c are not all zero, a, b, c, d  R. (i) Normal form : x + my + nz = p (ii) Plane through the point (x1, y1, z1) : a (x  x1) + b( y  y1) + c (z  z1) = 0 (iii) Intercept Form: x  y  z  1 a bc   (iv)   a ).  = 0 or  =a.  Vector form: ( r n r .n n (v) Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz +  = 0. Distance between ax + by + cz + d = 0 and 1 ax + by + cz + d2 = 0 is = | d1  d2 | a2  b2  c2 WWW.JEEBOOKS.IN Page # 36

(vi) Equation of a plane passing through a given point & parallel to the given vectors:    r = a + b +  c (parametric form) where  &  are scalars. w.jeebooks   or  . (b   =  . (b   (non parametric form) r c) a c) 9. A Plane & A Point (i) Distance of the point (x, y, z) from the plane ax + by + cz+ d = 0 is given by ax' by' cz'd . a2  b2  c2 (ii)   =d Length of the perpendicular from a point ( a ) to plane r .n  |a.nd | is given by p =  . |n| (iii) Foot (x, y, z) of perpendicular drawn from the point (x , y , z ) to 111 the plane ax + by + cz + d = 0 is given by x'x1  y'y1  z'z1 abc =– (ax1  by1  cz1  d) a2  b2  c2 (iv) To find image of a point w.r.t. a plane: Let P (x1, y1, z1) is a given point and ax + by + cz + d = 0 is given plane Let (x, y, z) is the image point. then x'x1  y'y1  z'z1 =–2 (ax1  by1  cz1  d) abc a2  b2  c2 10. Angle Between Two Planes: cos  = aa'bb'cc' a2  b2  c 2 a'2  b'2  c'2 Planes are perpendicular if aa + bb + cc = 0 and planes are parallel if ab c a' = b' = c' WWW.JEEBOOKS.IN Page # 37

 The angle  between the planes r . n1 = d and r . n2 = d is given by, cos 1 2   = n1 .n2 | n1 |. | n2 | w.jeebooks  Planes are perpendicular if n1 . n2 = 0 & planes are parallel if  =   is a scalar n1  n2 , 11. Angle Bisectors (i) The equations of the planes bisecting the angle between two given planes a x + b y + c z + d = 0 and a x + b y + c z + d = 0 are 1 11 1 2 22 2 a1x  b1y  c1z  d1 = ± a2x  b2y  c2z  d2 a12  b12  c12 a22  b 2  c 2 2 2 (ii) Bisector of acute/obtuse angle: First make both the constant terms positive. Then a1a2 + b1b2 + c1c2 > 0  origin lies on obtuse angle a1a2 + b1b2 + c1c2 < 0  origin lies in acute angle 12. Family of Planes (i) Any plane through the intersection of a1x + b1y + c1z + d1 = 0 & a2x + b2y + c2z + d2 = 0 is a1x + b1y + c1z + d1 +  (a2x + b2y + c2z + d2) = 0 (ii) The equation of plane passing through the intersection of the     planes r . n1 = d1 & r . n2 = d2 is r . (n1 +  n2 ) = d1 d2 where  is arbitrary scalar 13. Volume Of A Tetrahedron: Volume of a tetrahedron with vertices A (x1, y1, z1), B( x2, y2, z2), C (x3, y3, z3) and x1 y1 z1 1 1 x2 y2 z2 1 D (x4, y4, z4) is given by V = 6 x3 y3 z3 1 x4 y4 z4 1 WWW.JEEBOOKS.IN Page # 38

1. Equation Of A Line (i) A straight line is intersection of two planes. it is reprsented by two planes a x + b y + c z + d = 0 and 1 11 1 a2x + b2y + +c2z + d2 =0. w.jeebooks (ii) Symmetric form : x  x1 = y  y1 = z  z1 = r. a bc  (iii) Vector equation:   +b r =a (iv) Reduction of cartesion form of equation of a line to vector form & vice versa x  x1 = y  y1 = z  z1  a b c  r =(x1 ˆi +y1 ˆj + z1 kˆ ) +  (a iˆ + b ˆj + c kˆ ). 2. Angle Between A Plane And A Line: (i) If  is the angle between line x  x1 = y  y1 = z  z1 and the  mn plane ax + by + cz + d = 0, then sin  = a  bm  cn . ( )a2 b2  c2 2  m2  n2   (ii) Vector form: If  is the angle between a line r = ( a +  b ) and     b .n .  .  = d then sin =  | r n | b| | n m n  b xn =0 (iii) Condition for perpendicularity = = ,  b.n =0 ab c (iv) Condition for parallel a + bm + cn = 0 3. Condition For A Line To Lie In A Plane (i) Cartesian form: Line x  x1 = y  y1 = z  z1 would lie in a plane  m n ax + by + cz + d= 0, if ax1 + by1 + cz1 + d = 0 & a + bm + cn = 0.      bin.tnhe=p0la&near. (ii) Vector form: Line r =a + b would lie .n = d if  n= d WWW.JEEBOOKS.IN Page # 39

4. Skew Lines: (i) The straight lines which are not parallel and noncoplanar i.e. nonintersecting are called skew lines. w.jeebooks lines x–  y–  z– & x – '  y – '  z – '  mn ' m' n' ' ' '  0, then lines are skew. If  =  m n ' m' n' (ii) Shortest distance formula for lines    a2 – a1 . b1  b2  b1  b2    r =   and  =   is d a1 +  b1 r a2 +  b2 (iii) Vector Form: For lines  =   and  =   to be skew r a1 +  b1 r a2 +  b2    ( b1 x b2 ). ( a2  a1 )  0 (iv) Shortest distance between parallel lines       (a2  a1)  b   +b & r = a2 + b is d= |b| . r = a1 (v) Condition of coplanarity of two lines    &    is r =a + b r =c + d    [a – c b d]  0 5. Sphere General equation of a sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0. (u, – v, w) is the centre and u 2  v2  w 2  d is the radius of the sphere. WWW.JEEBOOKS.IN Page # 40

SOLUTION OF TRIANGLE 1. Sine Rule: a  b  c . w.jeebooks sin A sinB sin C 2. Cosine Formula: (i) cos A = b2  c2  a2 (ii) cos B = c2  a 2  b2 2bc 2ca a 2  b2  c2 (iii) cos C = 2a b 3. Projection Formula: (i) a = b cosC + c cosB (ii) b = c cosA + a cosC (iii) c = a cosB + b cosA 4. Napier’s Analogy - tangent rule: BC bc A CA ca B (i) tan 2 = b  c cot 2 (ii) tan 2 =c  a cot 2 AB ab C (iii) tan = cot 2 ab 2 5. Trigonometric Functions of Half Angles: A (s  b) (s  c) B (s  c) (s  a) (i) sin = ; sin = ; bc 2 2 ca C (s  a) (s  b) sin = 2 ab (ii) A s (s  a) B s (s  b) ; C = s (s  c) ; cos = cos = 2 ca cos 2 bc 2 ab (iii) A (s  b) (s  c)  where s = a  b  c is semi tan = = 2 s (s  a) s (s  a) 2 perimetre of triangle. 2 2 (iv) sin A = s(s  a)(s  b)(s  c) = bc bc WWW.JEEBOOKS.IN Page # 41

6. Area of Triangle () : 1 11  = 2 ab sin C = 2 bc sin A = 2 ca sin B = s (s  a) (s  b) (s  c) 7. m - n Rule: If BD : DC = m : n, then (m + n) cot   m cot   n cot   n cot B  m cot C w.jeebooks 8. Radius of Circumcirlce : R = a  b  c = abc 2sinA 2sinB 2sinC 4 9. Radius of The Incircle :  ABC (i) r = (ii) r = (s  a) tan = (s  b) tan = (s  c) tan s 222 AB C (iii) r = 4R sin sin sin 22 2 10. Radius of The Ex- Circles :   (i) r = ;r = ;r = 1 sa 2 sb 3 sc ABC Page # 42 (ii) r1 = s tan 2 ; r2 = s tan 2 ; r3 = s tan 2 ABC (iii) r = 4 R sin . cos . cos 1 222 WWW.JEEBOOKS.IN

11. Length of Angle Bisectors, Medians & Altitudes : (i) Length of an angle bisector from the angle A = a = 2 bc cos A ; b c 2 1w.jeebooks2b2  2c2  a2 (ii) Length of median from the angle A = ma = 2 2 & (iii) Length of altitude from the angle A = Aa = a 12. Orthocentre and Pedal Triangle: The triangle KLM which is formed by joining the feet of the altitudes is called the Pedal Triangle. (i) Its angles are  2A,  2B and  2C. (ii) Its sides are a cosA = R sin 2A, b cosB = R sin 2B and c cosC = R sin 2C (iii) Circumradii of the triangles PBC, PCA, PAB and ABC are equal. WWW.JEEBOOKS.IN Page # 43

13. The triangle formed by joining the three excentres 1, 2 and 3 of  ABC is called the excentral or excentric triangle. (i)  ABC is the pedal triangle of the  1 2 3. w.jeebooks (ii) Its angles are   A B &  C. , 2 22 2 2 2 AB C (iii) Its sides are 4 R cos , 4 R cos & 4 R cos . 22 2 A BC (iv) 1 = 4 R sin 2 ; 2 = 4 R sin 2 ; 3 = 4 R sin 2 . (v) Incentre  of  ABC is the orthocentre of the excentral  1 2 3. 14. Distance Between Special Points : (i) Distance between circumcentre and orthocentre OH2 = R2 (1 – 8 cosA cos B cos C) (ii) Distance between circumcentre and incentre A sin B C O2 = R2 (1 – 8 sin sin ) = R2 – 2Rr 222 (iii) Distance between circumcentre and centroid 1 OG2 = R2 – (a2 + b2 + c2) 9 INVERSE TRIGONOMETRIC FUNCTIONS 1. Principal Values & Domains of Inverse Trigonometric/Circular Functions: Function Domain Range (i) y = sin1 x where 1x1  y  22 (ii) y = cos1 x where 1x1 0y (iii) y = tan1 x where xR y  22 (iv) y = cosec1 x where x   1 or x  1   y   ,y0 (v) y = sec1 x where x  1 or x  1 22 (vi) y = cot1 x where xR  0  y  ; y  2 0 < y<  WWW.JEEBOOKS.IN Page # 44

P-2 (i) sin1 (sin x) = x,    x   22 P-3 (ii) cos1 (cos x) = x; 0  x   P-5 w.jeebooks(iii)tan1 (tan x) = x;  2.  x I-1 22 (iv) cot1 (cot x) = x; 0 < x <   (v) sec1 (sec x) = x; 0  x  , x  2 (vi) cosec1 (cosec x) = x; x  0,    x   22 (i) sin1 (x) =  sin1 x, 1  x  1 (ii) tan1 (x) =  tan1 x, x  R (iii) cos1 (x) =   cos1 x, 1  x  1 (iv) cot1 (x) =   cot1 x, x  R (i) sin1 x + cos1 x =  , 1  x  1 2 (ii) tan1 x + cot1 x =  , x  R 2 (iii) cosec1 x + sec1 x =  , x  1 2 Identities of Addition and Substraction: (i) sin1 x + sin1 y = sin1 x 1  y2  y 1  x2  , x  0, y  0 & (x2 + y2)  1  =   sin1 x 1  y2  y 1  x2  , x  0, y  0 & x2 + y2 > 1  (ii) cos1 x + cos1 y = cos1 x y  1  x2 1  y2  , x  0, y  0  (iii) tan1 x + tan1 y = tan1 xy , x > 0, y > 0 & xy < 1 1  xy =  + tan1 xy , x > 0, y > 0 & xy > 1 1 xy  Page # 45 = , x > 0, y > 0 & xy = 1 2 WWW.JEEBOOKS.IN

I - 2 (i) sin1 x  sin1 y = sin1 x 1  y2  y 1  x2  , x  0, y  0 (ii)  (iii) cos1 x  cos1 y = cos1 x y  1  x2 1  y2  x  0, y  0, x  y  w.jeebooks , xy tan1 x  tan1y = tan1 1 xy , x  0, y  0   2 sin1 x if | x |  1 2 sin1  2 x 1  x2   I - 3 (i)  =    2 sin1 x if x  1 2  if x   1    2sin1 x  2 (ii) cos1 (2 x2  1) =  2cos1 x if 0  x  1  2   2 cos 1x if 1  x  0  2 tan1x if | x | 1  (iii) tan1 2x =    2 tan1x if x  1 1 x2     2 tan1x if x 1  2 tan1x if | x | 1 2x  (iv) sin1 1  x2    2 tan1x if x  1  = if x  1    2 tan1x (v) cos1 1 x2 =  2 tan1x if x  0 1 x2   2 tan 1x if x  0 If tan1 x + tan1 y + tan1 z = tan1  x  y  z  xyz  if, x > 0, y > 0, z > 1  xy  yz  zx  0 & (xy + yz + zx) < 1 NOTE: (i) If tan1 x + tan1 y + tan1 z =  then x + y + z = xyz  (ii) If tan1 x + tan1 y + tan1 z = then xy + yz + zx = 1 2 (iii) tan1 1 + tan1 2 + tan1 3 =  WWW.JEEBOOKS.IN Page # 46

STATISTICS 1. Arithmetic Mean / or Mean w.jeebooks If x1, x2, x3 ,.......xn are n values of variate xi then their A.M. x is defined as n x xi = x1  x2  x3  .......  xn = n i1 n If x , x , x , .... x are values of veriate with frequencies f , f , f ,.........f then 123 n 123 n their A.M. is given by n f1x1  f2x2  f3x3  ......fnfn n x = f1  f2  f3  ......  fn = fi x i fi i1 , where N = i1 N 2. Properties of Arithmetic Mean : (i) Sum of deviation of variate from their A.M. is always zero that is  xi  x = 0. (ii) Sum of square of deviation of variate from their A.M. is minimum that is  xi  x2 is minimum (iii) If x is mean of variate xi then A.M. of (xi + ) = x +  A.M. of i . xi = . x A.M. of (axi + b) = a x + b 3. Median The median of a series is values of middle term of series when the values are written is ascending order or descending order. Therefore median, divide on arranged series in two equal parts For ungrouped distribution : If n be number of variates in a series then   n  1th term,when n is odd   2    Median = Mean of  n th and  n  2th term when n is even  2 2  WWW.JEEBOOKS.IN Page # 47

4. Mode If a frequency distribution the mode is the value of that variate which have the maximum frequency. Mode for w.jeebooksFor ungrouped distribution : The value of variate which has maximum frequency. For ungrouped frequency distribution : The value of that variate which have maximum frequency. Relationship between mean, median and mode. (i) In symmetric distribution, mean = mode = median (ii) In skew (moderately asymmetrical) distribution, median divides mean and mode internally in 1 : 2 ratio.  median = 2Mean  Mode 3 5. Range differenceof extremevalues L  S sum of extremevalues = L  S where L = largest value and S = smallest value 6. Mean deviation : n | xi  A | Mean deviation = i1 n n  fi | xi  A | Mean deviation = i1 (for frequency distribution) N 7. Variance : Standard deviation = + variance formula x2 = xi  x2 n WWW.JEEBOOKS.IN Page # 48

 n 2     x2 =  xi  n   n x i2  i1  x 2 i i1 w.jeebooks – n  = i1 – x2 n  n  d2 = di2 –  di  2 , where di = xi – a , where a = assumed mean n  n  (ii) coefficient of S.D. =     x  coefficient of variation =    × 100 (in percentage)  x  Properties of variance : (i) var(xi + ) = var(xi) (ii) var(.xi) = 2(var xi) (iii) var(a x + b) = a2(var x ) ii where , a, b are constant. MATHEMATICAL REASONING Let p and q are statements p q pq pvq p q qp p  q q p TT T T T T T T TF F T F T F F FT F F T F F F FF F F T T T T Tautology : This is a statement which is true for all truth values of its components. It is denoted by t. Consider truth table of p v ~ p p ~ p pv~p TF T FT T WWW.JEEBOOKS.IN Page # 49

This is statement which is false for all truth values of its compo- nents. It is denoted by f or c. Consider truth table of p ^ ~ p p ~ p p~p TF F FT F w.jeebooks (i) Statements p  q pq pq pq Negation (~ p)  (~ q) (~ p)  (~ q) p  (~ q) p  –q Let p  q Then (ii) (Contrapositive of p  q)is(~ q  ~ p) SETS AND RELATION Laws of Algebra of sets (Properties of sets): (i) Commutative law : (A  B) = B  A ; A  B = B  A (ii) Associative law:(A  B)  C=A  (B  C) ; (A  B)C = A  (B  C) (iii) Distributive law : A (B  C) = (A  B)  (A  C) ; A  (B  C) = (A  B)  (A  C) (iv) De-morgan law : (A  B)' = A'  B' ; (A  B)' = A'  B' (v) Identity law : A  U = A ; A   = A (vi) Complement law : A  A' = U, A  A' = , (A')' = A (vii) Idempotent law : A  A = A, A  A = A Some important results on number of elements in sets : If A, B, C are finite sets and U be the finite universal set then (i) n(A  B) = n(A) + n(B) – n(A  B) (ii) n(A – B) = n(A) – n(A  B) (iii) n(A  B  C) = n(A) + n(B) + n(C) – n(A  B) – n(B  C) – n (A  C) + n(A  B  C) (iv) Number of elements in exactly two of the sets A, B, C = n(A  B) + n(B  C) + n(C  A) – 3n(A  B  C) (v) Number of elements in exactly one of the sets A, B, C = n(A) + n(B) + n(C) – 2n(A  B) – 2n(B  C) – 2n(A  C) + 3n(A  B  C) WWW.JEEBOOKS.IN Page # 50


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