Tips and Formulas Basic Maths

BASIC MATHS & VECTOR BASIC MATHEMATICS 1. Quadratic Equation x Roots of ax2 + bx + c = 0 are Sum of roots = x1 + x2 = – 2.  b  b 2  4 ac 2a b c ; Product of roots = x1 x 2  a a Binomial Theorem (1  x) n  1  nx  (1  x)n  1  nx  n  n  1 2 n  n  1 2 x2  n(n  1)(n  2) 3 x  ...... 6 x2  n(n  1)(n  2) 3 x  ..... 6 If x < < 1 then (1  x ) n  1  nx & (1  x) n  1  nx Logarithm (i) log mn = log m+log n 4. Componendo and Dividendo Rule If 5. p a pq ab  then  q b p q ab Arithmetic progression (AP) a, a + d, a +2d, ..., a + (n–1)d Sum of n terms Sn = Note : here d = common difference n [2a+(n–1)d] 2 (i) 1 +2 + 3 + 4 + 5 ............ n  2 (ii) 12 + 22 + 32 + ...... + n  6. n(n  1) 2 n( n  1)(2n  1) 6 2  Geometrical Progression (GP) a, ar, ar2,..., arn–1 here r = common ratio Sum of n terms S n  7. m =logm-log n (iii) log mn = n log m (iv) loge m = 2.303 log10m n (ii) log  a 1r n 1r  2 a +b a 3.  Sum of  term S   b a 1r Trigonometry 2 rad = 360°  1 rad = 57.3° (i) sin θ= perpendicular hypotenuse (iv) cosec   hypotenuse perpendiculaer (ii) cos θ= base hypotenuse (iii) tan θ= (v) sec   hypotenuse base (vi) cot   perpendicular base base perpendicular a (vii) sin   (x) cosec  2 a b (viii) cos   2 1 sin  (xi) sec   a b 2 1 cos  (xvii) cos( A  B)  cos A cos B msin A sin B tan A ± tan B 1 mtan A tan B (xviii) sin 2A = 2sinAcosA (xx) tan 2A= 2tan A 1  tan 2 A sine law A A c cosine law b2  c 2  a2 , 2bc cos B = c 2  a 2  b2 , 2ca cos C = tan    sin   tan  B C B a C cos   1 Differentiation (i) y = xn  dy = nxn-1 dx (iv) y = cosx  (ii)y = ln x  dy = –sinx dx 1 x (iii)y = sinx  dy  e x  dx (vi) y  uv  dy = dx x   (v) y  e df  g  x   dg ( x)  (vii) y  f  g  x    dg ( x) dx dy = cosx dx dy dv du u v dx dx dx du dv v u dy u dy (viii)y = k = constant  =0 (ix) y    dx 2 dx dx v dx v Integration n (i)  x dx  x n 1  C , n  1 n 1 (iv)  cos xdx  sin x  C 13. a2  b2  c2 2ab b For small  sin    12. 1 tan  (xvi) sin( A  B )  sin A cos B  cos A sin B cos A = 11. (xii) cot   (xv) 1  cot 2   c osec2  sin A sin B sin C   a b c 10. a b (xiv) 1  tan 2   sec2  (xix) cos2A = cos2A – sin2A = 1-2sin2A = 2cos2A –1 9. (ix) tan   2 2 (xiii) sin   cos   1 (xvii) tan (A±B)= 8. b 2 1 (ii)  dx  l nx  C x x  (v)  e dx  Maxima & Minima of a function y = f(x) (i)For maximum value dy d2y 0&  ve dx dx 2 (ii)For minimum value dy d2y 0&  ve dx dx 2 1 x  e C  (iii)  sin xdx   cos x  C 2 (vi)  (x  ) dx  (x  ) n 1 C  (n  1) BASIC MATHS & VECTOR 14. Average of a varying quantity x2  x1 If y = f(x) then  y  y x2 ydx  x2  dx  ydx x1 x2  x1 x1 15. Formulae for determination of area (i) Area of a square = (side)2 (i)i Area of a rectangle = length × breadth 1 ×base×height 2 1 (iv) Area of a trapezoid = ×(distance between parallel sides) × (sum of parallel sides) 2 (v) Area enclosed by a circle =  r2 (r = radius) 2 (vi) Surface area of a sphere =  r (r = radius) (vii) Area of parallelogram = base × height (iii) Area of a triangle = (viii)Area of curved surface of cylinder = 2 r l (r = radius and  = length) (ix) Area of whole surface of cylinder = 2r (r + ) (x) Area of ellipse =  ab ( = length) (a & b are semi major axis respectively) (xi) Surface area of a cube = 6(side)2 where  r l   r r 2  h 2  lateral area (xii) Total surface area of a cone = r2 + r, 16. Formulae for determination of volume (i) Volume of rectangle slab = length × breadth × height (ii) Volume of a cubee = (side)3 (iii) Volume of a sphere = 4 3 r 3 (r = radius) (iv) Volume of cylinder =  r 2 l (v) Volume of a cone = 17. 1 2 r h 3 19. (r = radius and h =height) To convert an angle from degree to radian, we have to multiply it by degree, we have to multiply it by 18. (r = radius and  = length) 180   180 and to convert an angle from radian to . dy dy and by help of integration, if is given, we can find y.. dx dx The maximum and minimum values of function A cos + B sin are By help of differentiation, if y is given, we can find A2  B 2 and – A2  B 2 respectively.. (i) (a+b)2 = a2 + b2 + 2ab (ii) (a–b)2 = a2 + b2 – 2ab (iii) (a+b)(a–b) = a2 – b2 (iv ) (a+b)3 = a3 + b3 + 3ab (a+b) (v) (a–b)3 = a3 – b3 – 3ab (a–b)