Download algebra and mathamatics and more Lecture notes Mathematics in PDF only on Docsity!
.
1
Class – X Mathematics Formulae & some basic concepts
Real Number
Euclid’s Division Algorithm(lemma) : Given positive integers a’ andb’, there exists unique integers q and r such that a = b.q + r, where 0 ≤ r < b ( where a = dividend, b = divisor, q = quotient, and r = remainder.
Polynomials
In step1 : Factorize the given polynomials, a) Either by splitting the terms, (OR) b) Using these identities : (i) (a + b) 2 = a 2 + 2ab + b^2 (ii) (a – b)^2 = a^2 – 2ab + b 2 (iii) a^2 – b^2 = (a + b)(a – b) (iv) a 4 – b^4 = (a^2 ) 2 – (b^2 ) 2 .= (a 2 + b^2 ) (a^2 – b^2 ) = (a^2 + b^2 ) (a – b ) (a + b ) (v) (a + b) 3 = a 3 + b^3 + 3ab (a +b) (vi) a^3 + b^3 = (a + b)( a^2 + ab + b 2 ) (vii) (a – b) 3 = a^3 – b^3 – 3ab (a – b ) (viii) a^3 – b^3 = (a – b)( a 2 + ab + b^2 ) (ix) ( a + b + c) 2 = a^2 + b^2 + c 2 + 2ab + 2bc + 2ac (x) a^3 + b^3 + c^3 – 3abc = ( a + b + c )(a^2 + b^2 + c^2 – ab – bc – ac ) Trial & Error method. In step2 : Take the product of ‘Common terms’ as their HCF. In step3 : Take the product of All the terms , Omit, the HCF value which gives you the value of LCM. Product of LCM × HCF = Product of the two polynomials. Note: If cubical expression is given, it may be factorized by using ‘Trial & Error” method. Remainder theorem
If (x – 2 ) is a factor of the given expression, then take x-2 = 0,therefore x = 2, then substitute this value in p(x) = 5x² + 3x – 6 as p(2) : 5(2)² + 3(2) – 6 = 0 ( Here taking =0 is very important. If not taken answer can’t be found.) If (x-2) leaves a remainder of 4 p(2) : 5(2)² + 3(2) – 6 = 4 ( Here taking =4 is very important. If not taken answer can’t be found.)
Linear Equation in two variables
If pair of linear equation is : a 1 + b 1 y + c 1 = 0 & a 2 x + b 2 y + c 2 = 0 Then nature of roots/zeroes/solutions : (i) If a 1 /a 2 ≠ b 1 /b 2 → system has unique solution, is consistent OR graph is two intersecting lines. (ii) If a 1 /a 2 = b 1 /b 2 ≠ c 1 / c 2 → system has no solution, is inconsistent OR graph are parallel lines. (iii)If a 1 /a 2 = b 1 /b 2 = c 1 / c 2 → system has infinite solution, is consistent OR graph are coincident lines. Quadratic Equations
Note: To find the value of ‘x’ you may adopt either ‘splitting the middle term’ or ‘formula method’,
x = – b ± √D ( where D = b 2 – 4ac ) Hence 2a Sum of the roots = – b/a & Product of roots = c/a If roots of an equation are given, then : Quadratic Equation : x² – (sum of the roots).x + (product of the roots) = 0 If Discriminant > 0, then the roots are Real & unequal or unique, lines are intersecting. Discriminant = 0, then the roots are Real & equal, lines are coincident. Discriminant < 0, then the roots are Imaginary (not real), parallel lines
Downloaded from: http://www.cbseportal.com
.
If speed of boat = x km/hr and that of stream = y km/hr then speed in upstream = (x–y) km/hr and speed in downstream = (x+y) km/hr.
Ratio & Proportion
Duplicate ratio of a : b is a^2 : b^2 ( Incase of Sub-duplicate ratio you have to take ‘Square root’) Triplicate ratio of a : b is a^3 : b^3 ( Incase of Sub-triplicate ratio you have to take ‘Cube root’) Proportion a : b = c : d, Continued Proportion a : b = b : c, (Middle value is repeated) 1 st^2 nd^3 rd^4 th^ proportionals 1 st^2 nd^2 nd^3 rd^ proportionals Product of ‘Means’(Middle values) = Product of ‘Extremes’(Either end values)
If is given, then Componendo & Dividendo is Note : “ Where to take “K” method ?” You may adopt it in the following situations. If a/b = c/d = e/f are given, then you may assume as a/b = c/d = e/f = k Therefore a = b.k, c = d.k, e = f.k, then substitute the values of ‘a’ ‘b’ and ‘c’ in the given problem. Incase of continued proportion : a/b = b/c = k hence, a = bk, b = ck therefore putting the value of b we can get a = ck² & b = ck.(putting these values equation can be solved) Similarity
If two triangles are similar then, ratio of their sides are equal. i.e if Δ ABC ~ Δ PQR then AB = BC = AC PQ QR PR. If Δ ABC ~ Δ PQR then Area of Δ ABC = Side^2 = AB^2 = BC^2 = AC^2 Area of Δ PQR Side^2 PQ^2 QR^2 PR^2
Distance & Section Formulae
Distance = / (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2. ( The same formula is to be used to find the length of line
segment, sides of a triangle, square, rectangle, parallelogram etc.,) To prove co-linearity of the given three points A,B, and C, You have to find length of AB, BC, AC then use the condition AB + BC = AC. OR use this condition to solve the question easily : Area of triangle formed by these points : ½.[x 1 (y2 – y 3 ) (^) + x 2 (y (^) 3 - y 1 ) (^) + x 3 (y1 – y 2 )] = 0
Section formula: point (x, y) = Mid point =
If line is trisected then take m:n ratio as 1:2 and find co-ordinates of point p(x,y). Equation of a line
If two points are given, then Slope (m) = y 2 – y 1 x 2 – x (^1) If a point, and slope are given, then Slope (m) = y – y 1 x – x (^1) If two lines are ‘Parallel’ to each other then their slopes are equal i.e m 1 = m 2 If two lines are ‘Perpendicular’ to each other then product of their slopes is – 1. i.e m 1 × m 2 = – 1 Depending upon the question You may have to use equation od straight line as a) y = mx + c, where ‘c’ is the y-intercept. OR b) (y (^) – y 1 ) = m.(x (^) – x 1 )
.
While solving the problems based on combination of solids it would be better if you take common. T.S.A. of combined solid = C.S.A of solid 1 + C.S.A of solid 2 + C.S.A of solid 3 If a solid is melted and, recast into number of other small solids, then Volume of the larger solid = No of small solids x Volume of the smaller solid For Ex: A cylinder is melted and cast into smaller spheres. Find the number of spheres Volume of Cylinder = No of sphere x Volume of sphere. If an ‘Ice cream cone with hemispherical top’ is given then you have to take a) Total Volume = Volume of Cone + Volume of Hemisphere b) Surface area = CSA of Cone + CSA of hemisphere (usually Surface area will not be asked) Trigonometric Identities
Wherever ‘Square’ appears think of using the identities (i) Sin^2 θ + Cos^2 θ = 1 (ii) Sec^2 θ – Tan^2 θ = 1 (iii) Coseec^2 θ – Cot^2 θ = 1 Try to convert all the values of the given problem in terms of Sin θ and Cos θ Cosec θ may be written as 1/Sin θ Sin θ. Cosec θ = 1 Sec θ may be written as 1 / Cos θ Cos θ. Sec θ = 1 Cot θ may be written as 1 / Tan θ Tan θ. Cot θ = 1 Tan θ may be written as Sin θ / Cos θ Cot θ = Cos θ / Sin θ Wherever fractional parts appears then think taking their ‘LCM’ Think of using ( a + b )^2 , ( a – b )^2 , ( a + b )^3 , ( a – b )^3 formulae etc., Rationalize the denominator [ If a + b, (or) a – b format is given in the denominator] You may separate the denominator For Ex : Sin θ + Cos θ as Sin θ + Cos θ = 1 + Cot θ Sin θ Sin θ Sin θ
If you are not able to solve the LHS part completely, Do the problem to such an extent you can solve, then start working with RHS, and finally you will end up at a step where LHS = RHS Sin ( 90 – θ ) = Cos θ : Cos ( 90 – θ ) = Sin θ. Sec( 90 – θ ) = Cosec θ : Cosec ( 90 – θ ) = Sec θ Tan ( 90 – θ ) = Cot θ : Cot ( 90 – θ ) = Tan θ Values of Trigonometric Identities :
0° 30° 45° 60° 90° Sin θ 0 1/2 1/√2 √3/2 1 Cos θ 1 √3/2 1/√2 1/2 0 Tan θ^0 1/√3^1 √3^ ∞
Graphical Representation
Don’t forget to write the scale on x-axis, and on y-axis. (^) To find the ‘Lower quartile’ take N / 4 [Here N is ∑ f] then take the corresponding point on X-axis To find the ‘Upper quartile’ take 3N / 4, then take the corresponding point on X-axis To find the ‘Median’ take N / 2, then take the corresponding point on X-axis
Measures of Central Tendency
For un-grouped data
Arithmetic Mean = Sum of observations No of observations Mode = The most frequently occurred value of the raw data.
.
To find the Median first of all arrange the data in ‘Ascending’ or ‘Descending’ order, then Median = (N+1)/2 term value of the given data, in case of the data is having odd no of observations. Median = [(N/2) + (N+1)/2)] / 2 term value of the given data, in case of the data is having even number of observations. For grouped data Arithmetic Mean = ∑ fx (Direct method) ∑ f Arithmetic Mean = a + ∑ fd (short cut method) ∑ f
Arithmetic Mean = a + ∑ fu × C (where C is class interval) (step-deviation method) ∑ f Probability
Probability of an event : P(event) = Number of favorable outcomes Total number of outcomes If probability of happening an event is x then probability of not happening that event is (1-x). For e.g. If probability of winning a game is 0.4 then probability of loosing it is (1-0.4) = 0. If probability of finding a defective bulb is 3/7 then probability of finding a non-defective bulb is (1-3/7) = 4/7. In a deck of playing cards, there are four types of cards : ♠ (Spades in Black colour) having A, 2,3,4,5,6,7,8,9,10,J,K, and Q total 13 cards ♣ (Clubs in Black colour) having A, 2,3,4,5,6,7,8,9,10,J,K, and Q total 13 cards ♥ (Hearts in Red colour) having A, 2,3,4,5,6,7,8,9,10,J,K, and Q total 13 cards ♦ (Diamond in Red colour) having A, 2,3,4,5,6,7,8,9,10,J,K, and Q total 13 cards 52 cards Jack, King and Queen are known as ‘Face Cards’ , As these cards are having some pictures on it. Always remember Ace is not a face card as it doesn’t carry any face on it. If one coin is tossed the total number of outcomes are 2 either a Head or a Tail. If two coins are tossed the total number of outcomes are 2 ×2 = 4 If three coins are tossed the total number of outcomes are 2 ×2×2 = 8 Similarly for Dice, In a single roll total number of outcomes are 6 If two Dices are rolled, total number of outcomes are 6×6 = 36
Look out for my Sample Papers and Guess Papers and be the best of the rest…….
