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# Maths Formulas PDF Class 10 And 12 Basic/Vedic Topics Wise Sample Questions

## Maths Formulas

Every chapter as well as subject plays a vital role in class 10 and 12 and so does Maths, which is incomplete without Maths Formulas…! For those visitors who are in search of complete Class 10 And 12 Basic/Vedic Topics Wise Important Maths Formulas, we have gathered on this single page. Speaking from the Knowledge point of view, Math Formulas is quite helpful in scoring the best grade in subject Math which will automatically affect your percentage. So Candidates, start your preparation according to the latest Maths Formulas and along with that check the best Sample Questions from here, you also have option of downloading the same in the form of PDF.

At the time of preparation of Math subject, students are advised to make proper chart of all formulas that are stated below and stick it on your wall, revise them regularly for the best results. Practice is what you need in Math subject, so start preparation with the collection of formulas given below. Apart from this, Candidates should know the accuracy and speed are other components that students have to make level of both in Math exam. You must remember that the marks of these classes are very vital for further life. Go through this page of www.recruitmentinboxx.com and save your time in searching it.

### Maths Formulas

Maths Formula for 10th Class

Below is the list of important formulas of class 10, it will be definitely easy for the candidates to collect all according to their chapters such as Algebra, Geometry, Circles  Constructions, Trigonometry, Probability, Coordinate Geometry and Menstruation. Check the marking scheme as well topics from below stipulated table and list of formulas.

 Chapter Topics Marks Algebra Quadratic Equations,Arithmetic Progressions 26 Geometry 12 Circles  Constructions 22 Trigonometry 10 Probability 12 Coordinate Geometry 08 Mensuration 10 Total 100

Maths Algebraic Identities Formula

• (a+b)2=a2+2ab+b2
• (a−b)2=a2−2ab+b2
• (a+b)(a–b)=a2–b2
• (x+a)(x+b)=x2+(a+b)x+ab
• (x+a)(x–b)=x2+(a–b)x–ab
• (x–a)(x+b)=x2+(b–a)x–ab
• (x–a)(x–b)=x2–(a+b)x+ab
• (a+b)3=a3+b3+3ab(a+b)
• (a–b)3=a3–b3–3ab(a–b)
• (a – b)4= a4 – 4a3b + 6a2b2 – 4ab3 + b4)
• a– b4= (a – b)(a + b)(a2 + b2)
• a5 – b5 = (a – b)(a4 + a3b + a2b2+ ab3 + b4)
• If n is a natural number, a– bn= (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)
• If n is even (n = 2k), an+ bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)
• If n is odd (n = 2k + 1), an+ bn = (a + b)(an-1 – an-2b +…- bn-2a + bn-1)
• (a + b + c + …)2 = a2+ b2 + c2 + … + 2(ab + ac + bc + ….

Example:

Question: Which of the following expressions is in the sum-of-products (SOP) form?

1. (A + B)(C + D)
2. (A)B(CD)
3. AB(CD)
4. AB + CD

Combination:

Number of combinations of n different things taken r at a time = ⁿCr = n!r!(n−r)!n!r!(n−r)!

ⁿP₀ = r!∙ ⁿC₀.

ⁿC₀ = ⁿCn = 1

ⁿCr = ⁿCn – r

ⁿCr + ⁿCn – 1 = n+1n+1Cr

If p ≠ q and ⁿCp = ⁿCq then p + q = n.

ⁿCr/ⁿCr – 1 = (n – r + 1)/r.

The total number of combinations of n different things taken any number at a time = ⁿC₁ + ⁿC₂ + ⁿC₃ + …………. + ⁿC₀ = 2ⁿ – 1.

The total number of combinations of (p + q + r + . . . .) things of which p things are alike of a first kind, q things are alike of a second kind r things are alike of a third kind and so on, taken any number at a time is [(p + 1) (q + 1) (r + 1) . . . . ] – 1.

Maths Logarithm Formula

If ax = M then logM = x and conversely.

• log1 = 0.
• logaa = 1.
• alogam = M.
• logaMN = loga M + loga
• loga(M/N) = loga M – loga
• logaMn = n loga
• logaM = logb M x loga
• logba x 1oga b = 1.
• logba = 1/logb
• logM = logbM/loga

Maths Exponential Series

• For all x, ex= 1 + x/1! + x2/2! + x3/3! + …………… + xr/r! + ………….. ∞.
• e = 1 + 1/1! + 1/2! + 1/3! + ………………….. ∞.
• 2 < e < 3; e = 2.718282 (correct to six decimal places).
• a= 1 + (logea) x + [(loge a)2/2!] ∙ x+ [(loge a)3/3!] ∙ x3 + …………….. ∞.

Maths Logarithmic Series

• loge(1 + x) = x – x2/2 + x3/3 – ……………… ∞ (-1 < x ≤ 1).
• loge(1 – x) = – x – x2/ 2 – x3/3 – ………….. ∞ (- 1 ≤ x < 1).
• ½ loge[(1 + x)/(1 – x)] = x + x3/3 + x5/5 + ……………… ∞ (-1 < x < 1).
• loge2 = 1 – 1/2 + 1/3 – 1/4 + ………………… ∞.
• log10m = µ loge m where µ = 1/loge 10 = 0.4342945 and is a positive number.

Maths Laws of Exponents

• (am)(an) = am+n
• (ab)m= ambm
• (am)n= amn

Maths Fractional Exponents

• a= 1
• am/an=am−n
• am=1/ a−m
• a−m=1/am

All Mensuration Formulas In Maths

Basic Formula:

• sin(−θ)=−sinθ
• cos(−θ)=cosθ
• tan(−θ)=−tanθ
• cosec(−θ)=−cosecθ
• sec(−θ)=secθ
• cot(−θ)=−cotθ

Maths Trigonometry Formulas:

Math Formula for 12th Class

Candidates who are going to attempt class 12th board exams must collect all vital formulas of math subject from below

Laws of Indices:

aᵐ ∙ aⁿ = aᵐ + ⁿ

aᵐ/aⁿ = aᵐ – ⁿ

(iii) (aᵐ)ⁿ = aᵐⁿ

(iv) a = 1 (a ≠ 0).

(v) a-ⁿ = 1/aⁿ

(vi) ⁿ√aᵐ = aᵐ/ⁿ

(vii) (ab)ᵐ = aᵐ ∙ bⁿ.

(viii) (a/b)ᵐ = aᵐ/bⁿ

(ix) If aᵐ = bᵐ (m ≠ 0), then a = b.

(x) If aᵐ = aⁿ then m = n.

Surds:

• The surd conjugate of √a + √b (or a + √b) is √a – √b (or a – √b) and conversely.
• If a is rational, √b is a surd and a + √b (or, a – √b) = 0 then a = 0 and b = 0.
• If a and x are rational, √b and √y are surds and a + √b = x + √y then a = x and b = y.

Complex Numbers:

(i) The symbol z = (x, y) = x + iy where x, y are real and i = √-1, is called a complex (or, imaginary) quantity;x is called the real part and y, the imaginary part of the complex number z = x + iy.

(ii) If z = x + iy then z = x – iy and conversely; here, z is the complex conjugate of z.

(iii) If z = x+ iy then

(a) mod. z (or, | z | or, | x + iy | ) = + √(x² + y²) and

(b) amp. z (or, arg. z) = Ф = tan−1−1 y/x (-π < Ф ≤ π).

(iv) The modulus – amplitude form of a complex quantity z is

z = r (cosф + i sinф); here, r = | z | and ф = arg. z (-π < Ф <= π).

(v) | z | = | -z | = z ∙ z = √ (x² + y²).

(vi) If x + iy= 0 then x = 0 and y = 0(x,y are real).

(vii) If x + iy = p + iq then x = p and y = q(x, y, p and q all are real).

(viii) i = √-1, i² = -1, i³ = -i, and i⁴ = 1.

(ix) | z₁ + z₂| ≤ | z₁ | + | z₂ |.

(x) | z₁ z₂ | = | z₁ | ∙ | z₂ |.

(xi) | z₁/z₂| = | z₁ |/| z₂ |.

(xii) (a) arg. (z₁ z₂) = arg. z₁ + arg. z₂ + m

(b) arg. (z₁/z₂) = arg. z₁ – arg. z₂ + m where m = 0 or, 2π or, (- 2π).

(xiii) If ω be the imaginary cube root of unity then ω = ½ (- 1 + √3i) or, ω = ½ (-1 – √3i)

(xiv) ω³ = 1 and 1 + ω + ω² = 0

Do You KnowHow to Prepare For Maths Exam

Variation:

(i) If x varies directly as y, we write x ∝ y or, x = ky where k is a constant of variation.

(ii) If x varies inversely as y, we write x ∝ 1/y or, x = m ∙ (1/y) where m is a constant of variation.

(iii) If x ∝ y when z is constant and x ∝ z when y is constant then x ∝ yz when both y and z vary.

Arithmetical Progression (A.P.):

(i) The general form of an A. P. is a, a + d, a + 2d, a + 3d,…..

where a is the first term and d, the common difference of the A.P.

(ii) The nth term of the above A.P. is t₀ = a + (n – 1)d.

(iii) The sum of first n terns of the above A.P. is s = n/2 (a + l) = (No. of terms/2)[1st term + last term] or, S = ⁿ/₂ [2a + (n – 1) d]

(iv) The arithmetic mean between two given numbers a and b is (a + b)/2.

(v) 1 + 2 + 3 + …… + n = [n(n + 1)]/2.

(vi) 1² + 2² + 3² +……………. + n² = [n(n+ 1)(2n+ 1)]/6.

(vii) 1³ + 2³ + 3³ + . . . . + n³ = [{n(n + 1)}/2 ]².

Geometrical Progression (G.P.):

(i) The general form of a G.P. is a, ar, ar², ar³, . . . . . where a is the first term and r, the common ratio of the G.P.

(ii) The n th term of the above G.P. is t₀ = a.rn−1n−1 .

(iii) The sum of first n terms of the above G.P. is S = a ∙ [(1 – rⁿ)/(1 – r)] when -1 < r < 1

or, S = a ∙ [(rⁿ – 1)/(r – 1) ]when r > 1 or r < -1.

(iv) The geometric mean of two positive numbers a and b is √(ab) or, -√(ab).

(v) a + ar + ar² + ……………. ∞ = a/(1 – r) where (-1 < r < 1).

ax² + bx + c = 0 … (1)

(i) Roots of the equation (1) are x = {-b ± √(b² – 4ac)}/2a.

(ii) If α and β be the roots of the equation (1) then,

sum of its roots = α + β = – b/a = – (coefficient of x)/(coefficient of x² );

and product of its roots = αβ = c/a = (Constant term /(Coefficient of x²).

(iii) The quadratic equation whose roots are α and β is

x² – (α + β)x + αβ = 0

i.e. , x² – (sum of the roots) x + product of the roots = 0.

(iv) The expression (b² – 4ac) is called the discriminant of equation (1).

(v) If a, b, c are real and rational then the roots of equation (1) are

(a) real and distinct when b² – 4ac > 0;

(b) real and equal when b² – 4ac = 0;

(c) imaginary when b² – 4ac < 0;

(d) rational when b²- 4ac is a perfect square and

(e) irrational when b² – 4ac is not a perfect square.

(vi) If α + iβ be one root of equation (1) then its other root will be conjugate complex quantity α – iβ and conversely (a, b, c are real).

(vii) If α + √β be one root of equation (1) then its other root will be conjugate irrational quantity α – √β (a, b, c are rational).

START NOWLogical Reasoning Test

Permutation:

(i) ⌊n (or, n!) = n (n – 1) (n – 2) ∙∙∙∙∙∙∙∙∙ 3∙2∙1.

(ii) 0! = 1.

(iii) Number of permutations of n different things taken r ( ≤ n) at a time ⁿP₀ = n!/(n – 1)! = n (n – 1)(n – 2) ∙∙∙∙∙∙∙∙ (n – r + 1).

(iv) Number of permutations of n different things taken all at a time = ⁿP₀ = n!.

(v) Number of permutations of n things taken all at a time in which p things are alike of a first kind, q things are alike of a second kind, r things are alike of a third kind and the rest are all different, is ⁿ<span style=’font-size: 50%’>!/₀

(vi) Number of permutations of n different things taken r at a time when each thing may be repeated upto r times in any permutation, is nʳ .

Combination:

(i) Number of combinations of n different things taken r at a time = ⁿCr = n!r!(n−r)!n!r!(n−r)!

(ii) ⁿP₀ = r!∙ ⁿC₀.

(iii) ⁿC₀ = ⁿCn = 1.

(iv) ⁿCr = ⁿCn – r.

(v) ⁿCr + ⁿCn – 1 = n+1n+1Crr

(vi) If p ≠ q and ⁿCp = ⁿCq then p + q = n.

(vii) ⁿCr/ⁿCr – 1 = (n – r + 1)/r.

(viii) The total number of combinations of n different things taken any number at a time = ⁿC₁ + ⁿC₂ + ⁿC₃ + …………. + ⁿC₀ = 2ⁿ – 1.

(ix) The total number of combinations of (p + q + r + . . . .) things of which p things are alike of a first kind, q things are alike of a second kind r things are alike of a third kind and so on, taken any number at a time is [(p + 1) (q + 1) (r + 1) . . . . ] – 1.

Permutation and Combination Formula, Aptitude Questions, Problems and Solutions

Binomial Theorem:

(i) Statement of Binomial Theorem : If n is a positive integer then

(a + x)n = an + nC1 an – 1 x + nC2 an – 2 x2 + …………….. + nCr an – r xr + ………….. + xn …….. (1)

(ii) If n is not a positive integer then

(1 + x)n = 1 + nx + [n(n – 1)/2!] x2 + [n(n – 1)(n – 2)/3!] x3 + ………… + [{n(n-1)(n-2)………..(n-r+1)}/r!] xr+ ……………. ∞ (-1 < x < 1) ………….(2)

(iii) The general term of the expansion (1) is (r+ 1)th term

= tr + 1 = nCr an – r xr

(iv) The general term of the expansion (2) is (r + 1) th term

= tr + 1 = [{n(n – 1)(n – 2)….(n – r + l)}/r!] ∙ xr.

(v) There is one middle term is the expansion ( 1 ) when n is even and it is (n/2 + 1)th term ; the expansion ( I ) will have two middle terms when n is odd and they are the {(n – 1)/2 + 1} th and {(n – 1)/2 + 1} th terms.

(vi) (1 – x)-1 = 1 + x + x2 + x3 + ………………….∞.

(vii) (1 + x)-1 = I – x + x2 – x3 + ……………∞.

(viii) (1 – x)-2 = 1 + 2x + 3×2 + 4×3 + . . . . ∞ .

(ix) (1 + x)-2 = 1 – 2x + 3×2 – 4×3 + . . . . ∞ .

Logarithm:

(i) If ax = M then loga M = x and conversely.

(ii) loga 1 = 0.

(iii) loga a = 1.

(iv) a logam = M.

(v) loga MN = loga M + loga N.

(vi) loga (M/N) = loga M – loga N.

(vii) loga Mn = n loga M.

(viii) loga M = logb M x loga b.

(ix) logb a x 1oga b = 1.

(x) logb a = 1/logb a.

(xi) logb M = logb M/loga b.

Must Read ThisHow to Reduce Negative Marking in Exam

Exponential Series:

(i) For all x, ex = 1 + x/1! + x2/2! + x3/3! + …………… + xr/r! + ………….. ∞.

(ii) e = 1 + 1/1! + 1/2! + 1/3! + ………………….. ∞.

(iii) 2 < e < 3; e = 2.718282 (correct to six decimal places).

(iv) ax = 1 + (loge a) x + [(loge a)2/2!] ∙ x2 + [(loge a)3/3!] ∙ x3 + …………….. ∞.

Logarithmic Series:

(i) loge (1 + x) = x – x2/2 + x3/3 – ……………… ∞ (-1 < x ≤ 1).

(ii) loge (1 – x) = – x – x2/ 2 – x3/3 – ………….. ∞ (- 1 ≤ x < 1).

(iii) ½ loge [(1 + x)/(1 – x)] = x + x3/3 + x5/5 + ……………… ∞ (-1 < x < 1).

(iv) loge 2 = 1 – 1/2 + 1/3 – 1/4 + ………………… ∞.

(v) log10 m = µ loge m where µ = 1/loge 10 = 0.4342945 and m is a positive number.

Maths Sample Questions

Well, we all know that math subject needs a lot of knowledge which involve lot of hard work but if you want to do it easier way the Math previous year paper, Math sample paper is the best option. Visitors are advised to do practice for this vital subject from the below settled sample paper, you just have to press the links given below.

Press Here: Maths Sample Paper for class 10

Press Here: Maths Sample Paper for class 12

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