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800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

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MathsClass 113 marksmedium

Complex Numbers and Quadratic Equations

Express (3 + 4i)/(1 - 2i) in the form a + i b.

Reveal model answer + marking points▾

Multiply numerator and denominator by the conjugate of the denominator (1 + 2i): [(3 + 4i)(1 + 2i)]/[(1 - 2i)(1 + 2i)]. Numerator = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i. Denominator = 1^2 + 2^2 = 5. So the result = (-5 + 10i)/5 = -1 + 2i.

multiply by conjugate of denominator

Marking-scheme points

  • ✓Multiply by conjugate (1 + 2i)
  • ✓Numerator = -5 + 10i; denominator = 5
  • ✓Result = -1 + 2i
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MathsClass 113 marksmedium

Complex Numbers and Quadratic Equations

Solve the equation x^2 - 2x + 5 = 0.

Reveal model answer + marking points▾

Here a = 1, b = -2, c = 5. Discriminant = b^2 - 4ac = 4 - 20 = -16. So x = [2 +/- sqrt(-16)]/2 = [2 +/- 4i]/2 = 1 +/- 2i. The roots are 1 + 2i and 1 - 2i.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Marking-scheme points

  • ✓Discriminant = 4 - 20 = -16
  • ✓sqrt(-16) = 4i
  • ✓x = (2 +/- 4i)/2 = 1 +/- 2i
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MathsClass 113 markshard

Complex Numbers and Quadratic Equations

Find the square root of the complex number -5 + 12i.

Reveal model answer + marking points▾

Let sqrt(-5 + 12i) = a + i b. Squaring: a^2 - b^2 + 2ab i = -5 + 12i. So a^2 - b^2 = -5 and 2ab = 12 (ab = 6). Also |z| = sqrt((-5)^2 + 12^2) = sqrt(25 + 144) = 13, so a^2 + b^2 = 13. Adding a^2 - b^2 = -5 and a^2 + b^2 = 13 gives 2a^2 = 8, a^2 = 4, a = +/-2; then b^2 = 9, b = +/-3. Since ab = 6 is positive, a and b have the same sign. Hence sqrt(-5 + 12i) = +/- (2 + 3i).

(a + ib)^2 = a^2 - b^2 + 2ab i

Marking-scheme points

  • ✓Let sqrt = a + ib; a^2 - b^2 = -5, 2ab = 12
  • ✓a^2 + b^2 = |z| = 13
  • ✓Solve: a = +/-2, b = +/-3 -> +/- (2 + 3i)
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MathsClass 113 marksmedium

Linear Inequalities

Solve 5x - 3 >= 3x - 5 and show the solution set on a number line.

Reveal model answer + marking points▾

5x - 3 >= 3x - 5. Subtract 3x from both sides: 2x - 3 >= -5. Add 3: 2x >= -2. Divide by 2: x >= -1. The solution set is {x : x >= -1} = [-1, infinity). On the number line, a closed (filled) circle at -1 with shading to the right.

Marking-scheme points

  • ✓Bring x-terms together: 2x - 3 >= -5
  • ✓2x >= -2 -> x >= -1
  • ✓Solution [-1, infinity), closed circle at -1
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MathsClass 113 marksmedium

Linear Inequalities

Solve the system of inequalities 2x - 1 < 5 and 3x + 2 > -4, and find the common solution.

Reveal model answer + marking points▾

First inequality: 2x - 1 < 5 gives 2x < 6, so x < 3. Second inequality: 3x + 2 > -4 gives 3x > -6, so x > -2. The common solution is the intersection: -2 < x < 3, i.e. the interval (-2, 3).

Marking-scheme points

  • ✓First: x < 3
  • ✓Second: x > -2
  • ✓Common solution: -2 < x < 3 = (-2, 3)
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MathsClass 113 marksmedium

Linear Inequalities

The longest side of a triangle is three times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Reveal model answer + marking points▾

Let the shortest side = x cm. Then the longest side = 3x and the third side = 3x - 2. Perimeter = x + 3x + (3x - 2) = 7x - 2. Given perimeter is at least 61: 7x - 2 >= 61, so 7x >= 63, giving x >= 9. Hence the minimum length of the shortest side is 9 cm.

perimeter = sum of sides

Marking-scheme points

  • ✓Sides: x, 3x, 3x - 2; perimeter = 7x - 2
  • ✓7x - 2 >= 61 -> 7x >= 63
  • ✓x >= 9, so minimum shortest side = 9 cm
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MathsClass 113 marksmedium

Permutations and Combinations

How many distinct arrangements can be made using all the letters of the word 'MATHEMATICS'?

Reveal model answer + marking points▾

MATHEMATICS has 11 letters in which M appears 2 times, A appears 2 times and T appears 2 times; the rest (H, E, I, C, S) are distinct. Number of distinct arrangements = 11! / (2! x 2! x 2!) = 39916800 / 8 = 4989600.

arrangements = n!/(p! q! r!)

Marking-scheme points

  • ✓11 letters with M x2, A x2, T x2
  • ✓Arrangements = 11!/(2! 2! 2!)
  • ✓= 39916800/8 = 4989600
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MathsClass 113 marksmedium

Permutations and Combinations

Find the value of n if nC2 = 15.

Reveal model answer + marking points▾

nC2 = n(n-1)/2 = 15, so n(n-1) = 30. We need two consecutive integers whose product is 30: 6 x 5 = 30, so n = 6. (n = -5 is rejected since n must be a positive integer.)

nC2 = n(n-1)/2

Marking-scheme points

  • ✓nC2 = n(n-1)/2
  • ✓n(n-1) = 30
  • ✓n = 6 (positive integer)
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MathsClass 113 marksmedium

Permutations and Combinations

How many 4-digit numbers can be formed using the digits 0 to 9 if no digit is repeated?

Reveal model answer + marking points▾

The thousands place cannot be 0, so it can be filled in 9 ways (digits 1 to 9). After using one digit, the hundreds place can be filled in 9 ways (including 0 now), the tens place in 8 ways and the units place in 7 ways. Total = 9 x 9 x 8 x 7 = 4536.

Marking-scheme points

  • ✓Thousands place: 9 ways (cannot be 0)
  • ✓Then 9, 8, 7 ways for remaining places
  • ✓9 x 9 x 8 x 7 = 4536
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MathsClass 113 marksmedium

Permutations and Combinations

From 7 men and 5 women, in how many ways can a committee of 5 members be formed consisting of exactly 3 men and 2 women?

Reveal model answer + marking points▾

Choose 3 men from 7 in 7C3 ways and 2 women from 5 in 5C2 ways. 7C3 = (7 x 6 x 5)/(3 x 2 x 1) = 35 and 5C2 = (5 x 4)/(2 x 1) = 10. By the multiplication principle, total ways = 35 x 10 = 350.

total = 7C3 x 5C2

Marking-scheme points

  • ✓Choose men: 7C3 = 35
  • ✓Choose women: 5C2 = 10
  • ✓Total = 35 x 10 = 350
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MathsClass 113 marksmedium

Binomial Theorem

Expand (2x - 3)^4 using the binomial theorem.

Reveal model answer + marking points▾

(2x - 3)^4 = 4C0 (2x)^4 + 4C1 (2x)^3(-3) + 4C2 (2x)^2(-3)^2 + 4C3 (2x)(-3)^3 + 4C4 (-3)^4 = 16x^4 + 4(8x^3)(-3) + 6(4x^2)(9) + 4(2x)(-27) + 81 = 16x^4 - 96x^3 + 216x^2 - 216x + 81.

(a + b)^4 expansion

Marking-scheme points

  • ✓Use T(r+1) = 4Cr (2x)^(4-r)(-3)^r
  • ✓Compute each of the 5 terms
  • ✓= 16x^4 - 96x^3 + 216x^2 - 216x + 81
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MathsClass 113 marksmedium

Binomial Theorem

Find the coefficient of x^5 in the expansion of (x + 3)^8.

Reveal model answer + marking points▾

The general term is T(r+1) = 8Cr x^(8-r) 3^r. For the term in x^5, we need 8 - r = 5, so r = 3. The coefficient = 8C3 x 3^3 = 56 x 27 = 1512.

T(r+1) = nCr a^(n-r) b^r

Marking-scheme points

  • ✓General term = 8Cr x^(8-r) 3^r
  • ✓For x^5: 8 - r = 5 -> r = 3
  • ✓Coefficient = 8C3 x 3^3 = 56 x 27 = 1512
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MathsClass 113 markshard

Binomial Theorem

Find the term independent of x in the expansion of (x^2 - 1/x)^6.

Reveal model answer + marking points▾

General term T(r+1) = 6Cr (x^2)^(6-r) (-1/x)^r = 6Cr (-1)^r x^(12 - 2r) x^(-r) = 6Cr (-1)^r x^(12 - 3r). For the term independent of x, the power of x must be zero: 12 - 3r = 0, so r = 4. The term = 6C4 (-1)^4 = 15 x 1 = 15.

set power of x to 0

Marking-scheme points

  • ✓General term = 6Cr (-1)^r x^(12 - 3r)
  • ✓Independent of x: 12 - 3r = 0 -> r = 4
  • ✓Term = 6C4 = 15
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MathsClass 113 marksmedium

Binomial Theorem

Using the binomial theorem, evaluate (sqrt2 + 1)^4 + (sqrt2 - 1)^4.

Reveal model answer + marking points▾

For (a + b)^4 + (a - b)^4, the odd-power terms cancel and we get 2[a^4 + 6 a^2 b^2 + b^4]. With a = sqrt2 and b = 1: a^4 = (sqrt2)^4 = 4, a^2 b^2 = 2 x 1 = 2 so 6 a^2 b^2 = 12, and b^4 = 1. Sum = 2[4 + 12 + 1] = 2 x 17 = 34.

(a+b)^4 + (a-b)^4 = 2[a^4 + 6a^2 b^2 + b^4]

Marking-scheme points

  • ✓(a+b)^4 + (a-b)^4 = 2[a^4 + 6a^2 b^2 + b^4]
  • ✓a = sqrt2, b = 1: 4 + 12 + 1 = 17
  • ✓Answer = 2 x 17 = 34
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MathsClass 113 marksmedium

Binomial Theorem

Find the coefficient of x^6 y^3 in the expansion of (x + 2y)^9.

Reveal model answer + marking points▾

General term T(r+1) = 9Cr x^(9-r) (2y)^r = 9Cr 2^r x^(9-r) y^r. For x^6 y^3 we need 9 - r = 6 and r = 3, which is consistent (r = 3). Coefficient = 9C3 x 2^3 = 84 x 8 = 672.

T(r+1) = nCr a^(n-r) b^r

Marking-scheme points

  • ✓General term = 9Cr 2^r x^(9-r) y^r
  • ✓For x^6 y^3: r = 3
  • ✓Coefficient = 9C3 x 2^3 = 84 x 8 = 672
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MathsClass 113 marksmedium

Sequences and Series

Find the sum of the first 20 terms of the arithmetic progression 3, 7, 11, 15, ...

Reveal model answer + marking points▾

Here a = 3, d = 4, n = 20. The sum of n terms is S_n = (n/2)[2a + (n - 1)d]. So S_20 = (20/2)[2 x 3 + 19 x 4] = 10[6 + 76] = 10 x 82 = 820.

S_n = (n/2)[2a + (n - 1)d]

Marking-scheme points

  • ✓a = 3, d = 4, n = 20
  • ✓S_n = (n/2)[2a + (n - 1)d]
  • ✓S_20 = 10 x 82 = 820
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MathsClass 113 markshard

Sequences and Series

The sum of three numbers in AP is 24 and their product is 440. Find the numbers.

Reveal model answer + marking points▾

Let the three numbers be a - d, a, a + d. Their sum = 3a = 24, so a = 8. Their product = (a - d)(a)(a + d) = a(a^2 - d^2) = 440. So 8(64 - d^2) = 440, giving 64 - d^2 = 55, so d^2 = 9 and d = 3 (or -3). The numbers are 5, 8, 11.

AP terms: a - d, a, a + d

Marking-scheme points

  • ✓Take terms as a - d, a, a + d; sum 3a = 24 -> a = 8
  • ✓Product 8(64 - d^2) = 440 -> d^2 = 9
  • ✓Numbers are 5, 8, 11
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MathsClass 113 marksmedium

Sequences and Series

Find the sum of the first 6 terms of the geometric progression 5, 15, 45, ...

Reveal model answer + marking points▾

Here a = 5, r = 3, n = 6. Since r > 1, S_n = a(r^n - 1)/(r - 1). So S_6 = 5(3^6 - 1)/(3 - 1) = 5(729 - 1)/2 = 5 x 728/2 = 5 x 364 = 1820.

S_n = a(r^n - 1)/(r - 1), r > 1

Marking-scheme points

  • ✓a = 5, r = 3, n = 6
  • ✓S_n = a(r^n - 1)/(r - 1)
  • ✓S_6 = 5 x 728/2 = 1820
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MathsClass 113 marksmedium

Sequences and Series

Find the sum to infinity of the geometric progression 1 + 1/3 + 1/9 + 1/27 + ...

Reveal model answer + marking points▾

Here a = 1 and r = 1/3. Since |r| < 1, the sum to infinity exists and is S_infinity = a/(1 - r) = 1/(1 - 1/3) = 1/(2/3) = 3/2.

S_infinity = a/(1 - r), |r| < 1

Marking-scheme points

  • ✓a = 1, r = 1/3, |r| < 1 so sum exists
  • ✓S_infinity = a/(1 - r)
  • ✓= 1/(2/3) = 3/2
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MathsClass 113 marksmedium

Sequences and Series

Find the sum of the squares of the first 10 natural numbers, that is 1^2 + 2^2 + ... + 10^2.

Reveal model answer + marking points▾

The sum of squares of the first n natural numbers is given by n(n + 1)(2n + 1)/6. For n = 10: sum = 10 x 11 x 21 / 6 = 2310/6 = 385.

sum of squares = n(n + 1)(2n + 1)/6

Marking-scheme points

  • ✓Formula: sum = n(n + 1)(2n + 1)/6
  • ✓n = 10: 10 x 11 x 21
  • ✓= 2310/6 = 385
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