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The questions your board exam loves to ask
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 112 markseasy
Straight Lines
Find the equation of the line with slope 2 that passes through the point (1, -3).
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Using the point-slope form y - y1 = m(x - x1) with m = 2 and (x1, y1) = (1, -3): y - (-3) = 2(x - 1), so y + 3 = 2x - 2, giving y = 2x - 5 or 2x - y - 5 = 0.
y - y1 = m(x - x1)
Marking-scheme points
- ✓Point-slope form: y - y1 = m(x - x1)
- ✓y + 3 = 2(x - 1)
- ✓y = 2x - 5
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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
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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
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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
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Straight Lines
Find the distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 5 = 0.
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For parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, the distance is d = |C1 - C2|/sqrt(A^2 + B^2). Here A = 3, B = -4, C1 = 7, C2 = 5. d = |7 - 5|/sqrt(9 + 16) = 2/5.
d = |C1 - C2|/sqrt(A^2 + B^2)
Marking-scheme points
- ✓d = |C1 - C2|/sqrt(A^2 + B^2)
- ✓= |7 - 5|/sqrt25
- ✓d = 2/5 units
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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
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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
Still unsure? Ask the AI tutor →MathsClass 115 markshard
Straight Lines
Find the image (reflection) of the point (3, 8) in the line x + 3y - 7 = 0.
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For the image (h, k) of point (x1, y1) in line Ax + By + C = 0, we use (h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2). Here A = 1, B = 3, C = -7, (x1, y1) = (3, 8). A x1 + B y1 + C = 3 + 24 - 7 = 20, and A^2 + B^2 = 1 + 9 = 10, so the common ratio = -2(20)/10 = -4. Then h = 3 + 1(-4) = -1 and k = 8 + 3(-4) = 8 - 12 = -4. Hence the image is (-1, -4).
(h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2)
Marking-scheme points
- ✓(h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2)
- ✓Ratio = -2(20)/10 = -4
- ✓Image = (-1, -4)
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Conic Sections
Find the centre and radius of the circle x^2 + y^2 - 4x + 6y - 12 = 0.
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Compare with x^2 + y^2 + 2gx + 2fy + c = 0: 2g = -4 so g = -2, 2f = 6 so f = 3, c = -12. Centre = (-g, -f) = (2, -3). Radius = sqrt(g^2 + f^2 - c) = sqrt(4 + 9 + 12) = sqrt25 = 5.
centre (-g, -f), radius sqrt(g^2 + f^2 - c)
Marking-scheme points
- ✓Compare to x^2 + y^2 + 2gx + 2fy + c = 0
- ✓Centre = (-g, -f) = (2, -3)
- ✓Radius = sqrt(g^2 + f^2 - c) = 5
Still unsure? Ask the AI tutor →MathsClass 112 markseasy
Conic Sections
Find the equation of the circle with centre (2, -3) and radius 5.
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The equation of a circle with centre (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. Here (h, k) = (2, -3) and r = 5, so (x - 2)^2 + (y + 3)^2 = 25. Expanding: x^2 + y^2 - 4x + 6y - 12 = 0.
(x - h)^2 + (y - k)^2 = r^2
Marking-scheme points
- ✓(x - h)^2 + (y - k)^2 = r^2
- ✓(x - 2)^2 + (y + 3)^2 = 25
- ✓Expanded: x^2 + y^2 - 4x + 6y - 12 = 0
Still unsure? Ask the AI tutor →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
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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
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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
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Conic Sections
Find the equation of the parabola with focus (0, 3) and directrix y = -3.
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The focus is on the positive y-axis and the directrix is y = -3, so the parabola opens upwards with vertex at the origin and form x^2 = 4ay. Here a = 3 (distance from vertex to focus), so x^2 = 4(3)y = 12y.
x^2 = 4ay
Marking-scheme points
- ✓Focus on y-axis, directrix y = -3 -> opens up, x^2 = 4ay
- ✓a = 3
- ✓x^2 = 12y
Still unsure? Ask the AI tutor →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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Conic Sections
Find the eccentricity of an ellipse whose latus rectum is half of its major axis.
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Latus rectum = 2b^2/a and major axis = 2a. Given 2b^2/a = (1/2)(2a) = a, so 2b^2 = a^2. Since b^2 = a^2(1 - e^2), we get 2a^2(1 - e^2) = a^2, so 2(1 - e^2) = 1, giving 1 - e^2 = 1/2, e^2 = 1/2 and e = 1/sqrt2.
b^2 = a^2(1 - e^2)
Marking-scheme points
- ✓2b^2/a = a -> 2b^2 = a^2
- ✓Use b^2 = a^2(1 - e^2)
- ✓e^2 = 1/2 -> e = 1/sqrt2
Still unsure? Ask the AI tutor →MathsClass 112 markseasy
Introduction to Three Dimensional Geometry
Find the distance between the points (2, 3, 5) and (4, 3, 1).
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The distance between (x1, y1, z1) and (x2, y2, z2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]. Here = sqrt[(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2] = sqrt[4 + 0 + 16] = sqrt20 = 2 sqrt5.
d = sqrt[(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2]
Marking-scheme points
- ✓Distance = sqrt[(dx)^2 + (dy)^2 + (dz)^2]
- ✓= sqrt[4 + 0 + 16] = sqrt20
- ✓= 2 sqrt5 units
Still unsure? Ask the AI tutor →MathsClass 112 markseasy
Introduction to Three Dimensional Geometry
Find the distance of the point (1, -2, 3) from the origin.
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The distance of a point (x, y, z) from the origin is sqrt(x^2 + y^2 + z^2). Here = sqrt(1^2 + (-2)^2 + 3^2) = sqrt(1 + 4 + 9) = sqrt14 units.
d = sqrt(x^2 + y^2 + z^2)
Marking-scheme points
- ✓Distance from origin = sqrt(x^2 + y^2 + z^2)
- ✓= sqrt(1 + 4 + 9)
- ✓= sqrt14 units
Still unsure? Ask the AI tutor →MathsClass 113 marksmedium
Introduction to Three Dimensional Geometry
Find the coordinates of the point which divides the line segment joining (1, -2, 3) and (3, 4, -5) internally in the ratio 2 : 3.
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By the section formula for internal division in ratio m : n, the coordinates are ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n), (m z2 + n z1)/(m + n)) with m = 2, n = 3. x = (2(3) + 3(1))/5 = 9/5; y = (2(4) + 3(-2))/5 = 2/5; z = (2(-5) + 3(3))/5 = -1/5. The point is (9/5, 2/5, -1/5).
((m x2 + n x1)/(m+n), ...)
Marking-scheme points
- ✓Use internal section formula with m = 2, n = 3
- ✓x = 9/5, y = 2/5, z = -1/5
- ✓Point = (9/5, 2/5, -1/5)
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Introduction to Three Dimensional Geometry
Find the centroid of the triangle whose vertices are (1, 2, 3), (3, -1, 5) and (4, 0, -2).
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The centroid of a triangle with vertices (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) is ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3). x = (1+3+4)/3 = 8/3; y = (2-1+0)/3 = 1/3; z = (3+5-2)/3 = 6/3 = 2. Centroid = (8/3, 1/3, 2).
G = ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3)
Marking-scheme points
- ✓Centroid = average of the three vertices
- ✓x = 8/3, y = 1/3, z = 2
- ✓Centroid = (8/3, 1/3, 2)
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