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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.
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?
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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
MathsClass 113 marksmedium
Binomial Theorem
Expand (2x - 3)^4 using the binomial theorem.
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(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
MathsClass 113 marksmedium
Binomial Theorem
Find the coefficient of x^5 in the expansion of (x + 3)^8.
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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
MathsClass 113 markshard
Binomial Theorem
Find the term independent of x in the expansion of (x^2 - 1/x)^6.
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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
MathsClass 113 marksmedium
Binomial Theorem
Using the binomial theorem, evaluate (sqrt2 + 1)^4 + (sqrt2 - 1)^4.
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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
MathsClass 113 marksmedium
Binomial Theorem
Find the coefficient of x^6 y^3 in the expansion of (x + 2y)^9.
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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
MathsClass 113 marksmedium
Sequences and Series
Find the sum of the first 20 terms of the arithmetic progression 3, 7, 11, 15, ...
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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
MathsClass 113 markshard
Sequences and Series
The sum of three numbers in AP is 24 and their product is 440. Find the numbers.
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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
MathsClass 113 marksmedium
Sequences and Series
Find the sum of the first 6 terms of the geometric progression 5, 15, 45, ...
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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
MathsClass 113 marksmedium
Sequences and Series
Find the sum to infinity of the geometric progression 1 + 1/3 + 1/9 + 1/27 + ...
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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
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.
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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
MathsClass 113 marksmedium
Straight Lines
Find the equation of the line passing through the points (1, 2) and (3, 6).
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Slope m = (6 - 2)/(3 - 1) = 4/2 = 2. Using point-slope form with point (1, 2): y - 2 = 2(x - 1), so y - 2 = 2x - 2, giving y = 2x, or 2x - y = 0.
y - y1 = m(x - x1)
Marking-scheme points
- ✓Slope = (6 - 2)/(3 - 1) = 2
- ✓y - 2 = 2(x - 1)
- ✓Equation: 2x - y = 0
MathsClass 113 marksmedium
Straight Lines
Find the acute angle between the lines x + sqrt(3) y = 1 and sqrt(3) x + y = 1.
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Slope of first line m1 = -1/sqrt3; slope of second line m2 = -sqrt3. The angle between them satisfies tan(theta) = |(m1 - m2)/(1 + m1 m2)| = |(-1/sqrt3 + sqrt3)/(1 + (-1/sqrt3)(-sqrt3))| = |((-1 + 3)/sqrt3)/(1 + 1)| = |(2/sqrt3)/2| = 1/sqrt3. So theta = 30 degrees.
tan theta = |(m1 - m2)/(1 + m1 m2)|
Marking-scheme points
- ✓m1 = -1/sqrt3, m2 = -sqrt3
- ✓tan theta = |(m1 - m2)/(1 + m1 m2)| = 1/sqrt3
- ✓theta = 30 degrees
MathsClass 113 marksmedium
Straight Lines
Find the distance of the point (3, -5) from the line 3x - 4y - 26 = 0.
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The distance of point (x1, y1) from line Ax + By + C = 0 is d = |A x1 + B y1 + C|/sqrt(A^2 + B^2). Here A = 3, B = -4, C = -26 and (x1, y1) = (3, -5). d = |3(3) - 4(-5) - 26|/sqrt(9 + 16) = |9 + 20 - 26|/sqrt25 = |3|/5 = 3/5.
d = |A x1 + B y1 + C|/sqrt(A^2 + B^2)
Marking-scheme points
- ✓d = |A x1 + B y1 + C|/sqrt(A^2 + B^2)
- ✓Numerator = |9 + 20 - 26| = 3; denominator = 5
- ✓d = 3/5 units
MathsClass 113 marksmedium
Straight Lines
Find the equation of the line which makes intercepts -3 and 2 on the x-axis and y-axis respectively.
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The intercept form of a line is x/a + y/b = 1, where a and b are the x- and y-intercepts. Here a = -3 and b = 2, so x/(-3) + y/2 = 1. Multiplying throughout by 6: -2x + 3y = 6, i.e. 2x - 3y + 6 = 0.
x/a + y/b = 1
Marking-scheme points
- ✓Intercept form: x/a + y/b = 1
- ✓x/(-3) + y/2 = 1
- ✓Simplify to 2x - 3y + 6 = 0
MathsClass 113 marksmedium
Straight Lines
Find the equation of the perpendicular bisector of the line segment joining the points (1, 1) and (3, 5).
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Midpoint of the segment = ((1 + 3)/2, (1 + 5)/2) = (2, 3). Slope of the segment = (5 - 1)/(3 - 1) = 2, so the slope of the perpendicular bisector = -1/2. Using point-slope form through (2, 3): y - 3 = -1/2 (x - 2), so 2(y - 3) = -(x - 2), 2y - 6 = -x + 2, giving x + 2y - 8 = 0.
product of perpendicular slopes = -1
Marking-scheme points
- ✓Midpoint = (2, 3); segment slope = 2
- ✓Perpendicular slope = -1/2
- ✓Equation: x + 2y - 8 = 0
MathsClass 113 marksmedium
Conic Sections
For the parabola y^2 = 12x, find the coordinates of the focus, the equation of the directrix, the axis and the length of the latus rectum.
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Compare y^2 = 12x with y^2 = 4ax: 4a = 12, so a = 3. The parabola opens to the right. Focus = (a, 0) = (3, 0). Directrix: x = -a, i.e. x = -3. Axis: the x-axis, i.e. y = 0. Length of latus rectum = 4a = 12.
y^2 = 4ax: focus (a,0), directrix x = -a
Marking-scheme points
- ✓y^2 = 4ax with 4a = 12 -> a = 3
- ✓Focus (3, 0), directrix x = -3
- ✓Axis y = 0, latus rectum = 4a = 12
MathsClass 113 marksmedium
Conic Sections
For the ellipse x^2/25 + y^2/9 = 1, find the vertices, foci, eccentricity and length of the latus rectum.
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Here a^2 = 25 (a = 5) and b^2 = 9 (b = 3), with a > b so the major axis is along the x-axis. c = sqrt(a^2 - b^2) = sqrt(25 - 9) = 4. Vertices = (+/-5, 0); foci = (+/-4, 0); eccentricity e = c/a = 4/5; length of latus rectum = 2b^2/a = 2(9)/5 = 18/5.
e = c/a; latus rectum = 2b^2/a
Marking-scheme points
- ✓a = 5, b = 3, c = sqrt(a^2 - b^2) = 4
- ✓Vertices (+/-5, 0), foci (+/-4, 0)
- ✓e = 4/5, latus rectum = 2b^2/a = 18/5
MathsClass 113 marksmedium
Conic Sections
For the hyperbola x^2/16 - y^2/9 = 1, find the vertices, foci and eccentricity.
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Here a^2 = 16 (a = 4) and b^2 = 9 (b = 3). For a hyperbola, c = sqrt(a^2 + b^2) = sqrt(16 + 9) = sqrt25 = 5. Vertices = (+/-4, 0); foci = (+/-5, 0); eccentricity e = c/a = 5/4. (Length of latus rectum = 2b^2/a = 9/2.)
hyperbola: c = sqrt(a^2 + b^2), e = c/a
Marking-scheme points
- ✓a = 4, b = 3, c = sqrt(a^2 + b^2) = 5
- ✓Vertices (+/-4, 0), foci (+/-5, 0)
- ✓e = c/a = 5/4
MathsClass 113 marksmedium
Conic Sections
Find the equation of the ellipse with vertices (+/-5, 0) and foci (+/-4, 0).
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The vertices and foci are on the x-axis, so the ellipse is x^2/a^2 + y^2/b^2 = 1 with a > b. From vertices, a = 5; from foci, c = 4. Then b^2 = a^2 - c^2 = 25 - 16 = 9. Hence the equation is x^2/25 + y^2/9 = 1.
b^2 = a^2 - c^2
Marking-scheme points
- ✓a = 5 (vertices), c = 4 (foci)
- ✓b^2 = a^2 - c^2 = 9
- ✓Equation: x^2/25 + y^2/9 = 1
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