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Algebra Formulas for Class 11: Algebra is about using letters to represent numbers. It is about stating the relationship between different numbers. Algebra is used in Maths when we do not know the exact number(s) or values in a calculation. In algebra, we use letters to represent unknown values or values that can change. Algebra can be used in business when predicting sales, growth, and profit.
We have compiled important Algebra formulas for Class 11. These formulas and Algebraic Identities will help all Class 11 students in their studies as well as final exams. Class 11 Algebra chapters contain concepts of progressions, quadratic equations, linear inequalities, binomial theorem, and permutations & combinations, and mathematical induction. We have provided all the formulas and Algebraic Identities for Class 11 that are important for these chapters on this page. CBSE Class 11 students can download these formulas as a PDF from this page.
Students who are looking for the complete list of Maths formulas for Class 11 PDF for Algebra can refer to this article. Class 11 Maths contain 6 chapters related to Algebra. You can check the formula chapter-wise in the table below:
| Complex Numbers and Quadratic Equations ★ √-1 = i or i2 = -1 which means i can be assumed as the solution of the equation x2 + 1 = 0. i is called Iota in complex numbers. ★ i2 = -1 i3 = i2 * i = -i i4 = i2 * i2 = 1 ★ y = ax2+ bx + c is a quadratic equation, where a ≠ 0, b and c can be any real (or complex) number. Here, a, b and c are called as the coefficients. ★ Quadratic equation has two roots: (α, β) = [–b ± √(b2 – 4ac)]/2ac, where α and β are the roots of the equation. (i) If b2 − 4ac > 0, then the quadratic equation has two distinct real roots. (ii) If b2 − 4ac < 0, then the quadratic equation has two imaginary roots. (iii) If b2 − 4ac = 0, then the quadratic equation has two equal real roots. |
| Linear Inequalities ★ The graph of a linear inequality in one variable is a number line. Use an open circle for < and > and a closed circle for ≤ and ≥. ★ Addition Property of Inequality: (i) If x > y, then (x + z) > (y + z) (ii) If x < y, then (x + z) < (y + z) ★ Subtraction Property of Inequality: (i) If x > y, then (x − z) > (y − z) (ii) If x < y, then (x − z) < (y − z) ★ Multiplication Property of Inequality: (i) If x > y and z > 0, then xz > yz (ii) If x< y and z > 0, then xz < yz (iii) If x > y and z < 0, then xz < yz (iv) If x < y and z < 0, then xz > yz ★ Division Property of Inequality: (i) If x > y and z > 0, then x/z > y/z (ii) If x < y and z > 0, then x/z < y/z (iii) If x > y and z < 0, then x/z < y/z (iv) If x < y and z < 0, then x/z > y/z |
| Permutations and Combinations ★ The number of permutations of n objects taken r at a time is determined by the following formula: P(n, r) = n!/(n − r)! ★ The number of combinations of n objects taken r at a time is determined by the following formula: C(n, r) = n!/(n − r)!r! |
| Binomial Theorem ★ The Binomial Theorem states that: (a + b)n = an + (nC1)an-1b + (nC2)an-2b2 + … + (nCn-1)abn-1 + bn where n is a positive integer |
| Sequence and Series ★ Arithmetic Progression: (i) Sequence: a, a+d, a+2d, ……, a + (n – 1)d, …. (ii) Common Difference: d = (a2 – a1), where a2 and a1 are successive term and preceding term respectively. (iii) General Term (nth term): an = a + (n – 1)d (iv) nth Term from the last term: an = l – (n – 1)d (v) Sum of first n terms: Sn = n/2[2a + (n – 1)d] ★ Geometric Progression: (i) Sequence: a, ar, ar2, …., ar(n-1), … (ii) Common Ratio: r = ar(n-1)/ar(n-2), where ar(n-1) and ar(n-2) are successive term and preceding term respectively. (iii) General Term (nth term): an = ar(n-1) (iv) nth Term from the last term: an = 1/r(n-1) (v) Sum of first n terms: Sn = a(1 – rn)/(1 – r) if r < 1 Sn = a(rn -1)/(r – 1) if r > 1 * Here, a = first term, d = common difference, r = common ratio, n = position of term, l = last term |
| Principle of Mathematical Induction ★ Mathematical induction is a technique for proving a statement, a theorem, or a formula, that is asserted about every natural number. It has only 2 steps: Step 1: Show it is true for the first one. Step 2: Show that if anyone is true then the next one is true. Then all are true. |
Check out other important Maths articles for Class 11:
| NCERT Solutions for Class 11 Maths | Class 11 Maths Syllabus |
| NCERT Books for Class 11 Maths | Maths Formulas for Class 11 PDF Download |
We have provided the direct link to download Algebra formulas for Class 11 PDF.
We have provided some important Class 11 Algebra questions with solutions related to linear inequalities, quadratic formulas, binomial theorem, mathematical induction, and progressions:
| Question 1: Solve (√2 + 1)5 + (√2 − 1)5 using binomial theorem. |
| Solution: We have (x + y)5 + (x – y)5 = 2[5C0 x5 + 5C2 x3 y2 + 5C4 xy4] = 2(x5 + 10 x3 y2 + 5xy4) Now (√2 + 1)5 + (√2 − 1)5 = 2[(√2)5 + 10(√2)3(1)2 + 5(√2)(1)4] = 58√2 |
| Question 2: Solve the following quadratic equation: (x2 + 2x − 35) = 0 using the quadratic formula. |
| Solution: Firstly, we have to identify what a, b, and c: a = 1, b = 2, c = −35 Next we need to substitute these into the formula: (α, β) = [–b ± √(b2 – 4ac)]/2ac = [-2 ± √(22 – 4x1x-35)]/2x1x-35 Solving it, we get: (α, β) = (-2 ± √144)/2 = (-2 ± 12)/2 = 5, -7 Hence, the roots of the equation are 5 and -7. |
| Question 3: The sum of the three numbers in A.P is 21 and the product of the first and third number of the sequence is 45. What are the three numbers? |
| Solution: Let the numbers are be (a – d), a, and (a + d) Then (a – d) + a + (a + d) = 21 (Given) 3a = 21 a = 7 and (a – d)(a + d) = 45 a2 – d2 = 45 72 – d2 = 45 d2 = 4 d = ± 2 Hence, the numbers are 5, 7 and 9 when d = 2 and 9, 7 and 5 when d = -2. In both the cases numbers are the same. |
| Question 4: If a rubber ball consistently bounces back 2/3 of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped? |
| Solution: Each time the ball is dropped and it bounces back, it reaches 2/3 of the height it was dropped from. After the first bounce, the ball will reach 2/3 of the height from which it was dropped – let us call it the original height. After the second bounce, the ball will reach 2/3 of the height it would have reached after the first bounce. So, at the end of the second bounce, the ball would have reached 2/3 x 2/3 of the original height = 4/9th of the original height. After the third bounce, the ball will reach 2/3 of the height it would have reached after the second bounce. So, at the end of the third bounce, the ball would have reached 2/3 x 2/3 x 2/3 = 8/27th of the original height. After the fourth and last bounce, the ball will reach 2/3 of the height it would have reached after the third bounce. So, at the end of the last bounce, the ball would have reached 2/3 x 2/3 x 2/3 x 2/3 of the original height = 16/81 of the original height. |
| Question 5: In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together? |
| Solution: In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters = 7!/2! = 2520. Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways. ∴ Required number of ways = (2520 x 20) = 50400 |
Check Algebra Formulas for other classes as well:
Here we have provided some of the practice questions on the formula of Algebraic Identities for Class 11 for you to practice:
| Q1: How many 6-digit numbers can be formed using the digits 0, 1, 2, 5, 7, and 9 which are divisible by 11 and no digit is repeated. Q2: Find the smallest number that must be added to 289279 to make it divisible by 8. Q3: A grandfather is 10 times older than his granddaughter. He is 54 years older than her. Find their present ages. Q4: Prove: (2n + 7) < (n + 3)2, for n∈N. Q5: Find the point(s) of intersection of the parabola with equation y = x2 – 5x + 4 and the line with equation y = 2x – 2. Q6: Find the constant k so that the quadratic equation 2x2 + 5x – k = 0 has two real solutions. Q7: Simplify i231 where i is the imaginary unit and is defined as: i = √(-1). Q8: Find all zeros of the polynomial P(x) = x3 – 3x2 – 10x + 24 knowing that x = 2 is a zero of the polynomial. Q9: If x is an integer, what is the greatest value of x which satisfies 5 < 2x + 2 < 9? Q10: Factor the expression 6x2 – 13x + 5. Q11: Find the constant k so that the system of the two equations: 2x + ky = 2 and 5x – 3y = 7 has no solutions. Q12: Find the constant k so that : -x2 – (k + 7)x – 8 = -(x – 2)(x – 4). Q13: Find the squares of the quadratic function f given by f(x) = 2x2 – 6x + 4. Q14: What is the remainder when f(x) = (x – 2)54 is divided by x – 1? Q15: Simplify | – x2 + 4x – 4 | |
Also, Check
Here we have provided some of the frequently asked questions related to formula of algebra Class 11.
A: Algebra unit for Class 11 contains the following chapters:
(i) Chapter 1: Principle of Mathematical Induction
(ii) Chapter 2: Complex Numbers and Quadratic Equations
(iii) Chapter 3: Linear Inequalities
(iv) Chapter 4: Permutations and Combinations
(v) Chapter 5: Binomial Theorem
(vi) Chapter 6: Sequence and Series
A: You can solve important algebra questions for Class 11 here on this page.
A: We have provided the direct Algebra formulas Class 11 PDF download link on this page. Download for without any signup.
A: An Algebraic Equation or identity is made of two or more algebraic expressions separated by an equal sign. The LHS (Left Hand Side) of the equation is always equal to the RHS (Right Hand Side) for any value of variables in it.
A: An Algebraic Expression is a combination of constants, variables, and algebraic operations (+, -, ×, ÷). We can derive the Algebraic Expression for a given situation or condition by using these combinations.
Now you are provided with all the necessary information regarding Class 11 Algebra formulas. We hope that you have downloaded the Maths Formulas for Class 11 PDF available on this page. Students can also make use of NCERT Solutions for Maths provided by Embibe for their exam preparation.
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