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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 122 markseasy
Matrices
What is the transpose of a matrix? Write two properties of the transpose.
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The transpose of a matrix A, denoted A' (or A^T), is the matrix obtained by interchanging its rows and columns. Two properties are: (1) (A')' = A, that is, the transpose of the transpose is the original matrix; and (2) (AB)' = B' A', the transpose of a product equals the product of the transposes in reverse order. Also (A + B)' = A' + B' and (kA)' = k A'.
(AB)' = B' A'
Marking-scheme points
- ✓Transpose: interchange rows and columns
- ✓(A')' = A
- ✓(AB)' = B' A'
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Matrices
Define a symmetric and a skew-symmetric matrix.
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A square matrix A is symmetric if A' = A, that is, aij = aji for all i and j (it is unchanged on taking its transpose). A square matrix A is skew-symmetric if A' = -A, that is, aij = -aji for all i and j; this forces all its diagonal elements to be zero. Every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.
symmetric: A' = A; skew-symmetric: A' = -A
Marking-scheme points
- ✓Symmetric: A' = A (aij = aji)
- ✓Skew-symmetric: A' = -A (aij = -aji), diagonal elements zero
- ✓Every square matrix = symmetric part + skew-symmetric part
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Matrices
If a matrix has 12 elements, what are the possible orders it can have?
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If a matrix has 12 elements, then the order (number of rows x number of columns) must be a pair of factors whose product is 12. The possible orders are: 1 x 12, 12 x 1, 2 x 6, 6 x 2, 3 x 4 and 4 x 3. So there are 6 possible orders. (In general, the number of possible orders equals the number of ways of writing the total number of elements as a product of two natural numbers.)
Marking-scheme points
- ✓Order (rows x columns) must multiply to give 12
- ✓Factor pairs of 12
- ✓Orders: 1x12, 12x1, 2x6, 6x2, 3x4, 4x3 (six orders)
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Matrices
If A = [[1, 2],[3, 4]] and B = [[2, 1],[0, 5]], find 2A - B.
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First find 2A = [[2, 4],[6, 8]]. Then subtract B element by element: 2A - B = [[2 - 2, 4 - 1],[6 - 0, 8 - 5]] = [[0, 3],[6, 3]].
Marking-scheme points
- ✓2A = [[2, 4],[6, 8]]
- ✓Subtract B element by element
- ✓2A - B = [[0, 3],[6, 3]]
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Determinants
Evaluate the determinant of the matrix [[1, 2, 3],[0, 1, 4],[5, 6, 0]].
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Expand along the first row: |A| = 1 x (1 x 0 - 4 x 6) - 2 x (0 x 0 - 4 x 5) + 3 x (0 x 6 - 1 x 5) = 1 x (0 - 24) - 2 x (0 - 20) + 3 x (0 - 5) = -24 + 40 - 15 = 1. So the value of the determinant is 1.
Marking-scheme points
- ✓Expand along the first row
- ✓= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)
- ✓= -24 + 40 - 15 = 1
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Determinants
State any three properties of determinants.
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(1) The value of a determinant remains unchanged if its rows and columns are interchanged (that is, |A'| = |A|). (2) If any two rows (or two columns) of a determinant are interchanged, the sign of the determinant changes. (3) If any two rows (or columns) of a determinant are identical (or proportional), the value of the determinant is zero. Also, if each element of a row (or column) is multiplied by a constant k, the determinant is multiplied by k.
Marking-scheme points
- ✓Interchanging rows and columns does not change the value
- ✓Interchanging two rows/columns changes the sign
- ✓Two identical/proportional rows or columns give value zero
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Determinants
Define the minor and the cofactor of an element of a determinant.
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The minor of an element aij of a determinant is the value of the determinant obtained by deleting the ith row and the jth column in which that element lies; it is denoted by Mij. The cofactor of the element aij is Aij = (-1)^(i + j) x Mij, that is, the minor with a sign attached according to the position of the element. Cofactors are used to expand a determinant and to find the adjoint of a matrix.
Aij = (-1)^(i+j) Mij
Marking-scheme points
- ✓Minor Mij: determinant after deleting row i and column j
- ✓Cofactor Aij = (-1)^(i+j) Mij
- ✓Used in expansion of determinants and finding the adjoint
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Determinants
What is a singular and a non-singular matrix? State the condition for a matrix to be invertible.
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A square matrix A is called singular if its determinant is zero (|A| = 0), and non-singular if its determinant is non-zero (|A| is not equal to 0). The condition for a square matrix to be invertible (to have an inverse) is that it must be non-singular, that is, |A| must not be zero, because the inverse A inverse = (1/|A|) adj A is defined only when |A| is not zero.
A inverse exists if and only if |A| is not 0
Marking-scheme points
- ✓Singular: |A| = 0; non-singular: |A| not equal to 0
- ✓A matrix is invertible only if it is non-singular
- ✓Because A inverse = (1/|A|) adj A needs |A| not zero
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Determinants
If A is a 3 x 3 matrix with |A| = 5, find the values of |adj A| and |2A|.
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For an n x n matrix, |adj A| = |A|^(n - 1). Here n = 3, so |adj A| = |A|^2 = 5^2 = 25. Also, for a scalar k, |kA| = k^n |A|. Here |2A| = 2^3 x |A| = 8 x 5 = 40. So |adj A| = 25 and |2A| = 40.
|adj A| = |A|^(n-1); |kA| = k^n |A|
Marking-scheme points
- ✓|adj A| = |A|^(n-1) = 5^2 = 25
- ✓|kA| = k^n |A|
- ✓|2A| = 2^3 x 5 = 40
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Continuity and Differentiability
State the conditions for a function f(x) to be continuous at a point x = a.
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A function f(x) is continuous at a point x = a if the following three conditions are all satisfied: (1) f(a) is defined (the function has a value at a); (2) the limit of f(x) as x approaches a exists (that is, the left-hand limit equals the right-hand limit); and (3) this limit is equal to the value of the function, that is, limit of f(x) as x approaches a = f(a).
limit(x->a) f(x) = f(a)
Marking-scheme points
- ✓f(a) must be defined
- ✓The limit as x approaches a must exist (LHL = RHL)
- ✓The limit must equal f(a)
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Continuity and Differentiability
What is the relationship between continuity and differentiability of a function?
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If a function is differentiable at a point, then it is necessarily continuous at that point. However, the converse is not always true: a function that is continuous at a point need not be differentiable there. For example, f(x) = |x| is continuous at x = 0 but not differentiable at x = 0 (it has a sharp corner). So differentiability implies continuity, but continuity does not imply differentiability.
Marking-scheme points
- ✓Differentiable at a point => continuous at that point
- ✓Continuous does not always imply differentiable
- ✓Example: |x| is continuous but not differentiable at x = 0
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Continuity and Differentiability
Differentiate sin(x^2) with respect to x.
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Let y = sin(x^2). Using the chain rule, dy/dx = cos(x^2) x d/dx(x^2) = cos(x^2) x 2x = 2x cos(x^2). The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) x g'(x).
d/dx f(g(x)) = f'(g(x)) g'(x)
Marking-scheme points
- ✓Apply the chain rule to y = sin(x^2)
- ✓dy/dx = cos(x^2) x 2x
- ✓dy/dx = 2x cos(x^2)
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Continuity and Differentiability
Write the derivatives of sin inverse (x) and tan inverse (x) with respect to x.
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The derivative of sin inverse (x) with respect to x is 1/sqrt(1 - x^2), valid for -1 < x < 1. The derivative of tan inverse (x) with respect to x is 1/(1 + x^2), valid for all real x. These standard results are obtained by implicit differentiation of y = sin inverse (x) and y = tan inverse (x).
d/dx(sin inverse x) = 1/sqrt(1 - x^2)
Marking-scheme points
- ✓d/dx (sin inverse x) = 1/sqrt(1 - x^2)
- ✓d/dx (tan inverse x) = 1/(1 + x^2)
- ✓Obtained by implicit differentiation
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Continuity and Differentiability
If x = a cos t and y = a sin t, find dy/dx.
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Differentiate each with respect to the parameter t: dx/dt = -a sin t and dy/dt = a cos t. Then dy/dx = (dy/dt)/(dx/dt) = (a cos t)/(-a sin t) = -cos t/sin t = -cot t. So dy/dx = -cot t.
dy/dx = (dy/dt)/(dx/dt)
Marking-scheme points
- ✓dx/dt = -a sin t, dy/dt = a cos t
- ✓dy/dx = (dy/dt)/(dx/dt)
- ✓dy/dx = -cot t
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Continuity and Differentiability
State Rolle's theorem and the Lagrange mean value theorem.
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Rolle's theorem: if a function f is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) such that f'(c) = 0. Lagrange's mean value theorem: if a function f is continuous on [a, b] and differentiable on (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
f'(c) = (f(b) - f(a))/(b - a)
Marking-scheme points
- ✓Rolle's: continuous on [a,b], differentiable on (a,b), f(a) = f(b) => f'(c) = 0
- ✓Mean value theorem: f'(c) = (f(b) - f(a))/(b - a)
- ✓Both guarantee a point c in the open interval
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Application of Derivatives
What is meant by the derivative as a rate of change? Write the rate of change of the area of a circle with respect to its radius.
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The derivative dy/dx represents the instantaneous rate of change of y with respect to x. When a quantity changes with time, its derivative with respect to time gives the rate of change. The area of a circle is A = pi r^2, so the rate of change of area with respect to the radius is dA/dr = 2 pi r. If the radius changes with time, dA/dt = 2 pi r (dr/dt).
dA/dr = 2 pi r
Marking-scheme points
- ✓Derivative = instantaneous rate of change
- ✓Area A = pi r^2 -> dA/dr = 2 pi r
- ✓With time: dA/dt = 2 pi r (dr/dt)
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Application of Derivatives
State the condition for a function to be increasing or decreasing on an interval.
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A differentiable function f is increasing on an interval if its derivative f'(x) is greater than or equal to 0 (strictly increasing if f'(x) > 0) for all x in that interval. It is decreasing if its derivative f'(x) is less than or equal to 0 (strictly decreasing if f'(x) < 0) for all x in that interval. So the sign of the first derivative tells us whether the function rises or falls.
f'(x) > 0 increasing; f'(x) < 0 decreasing
Marking-scheme points
- ✓Increasing: f'(x) >= 0 (strictly increasing if f'(x) > 0)
- ✓Decreasing: f'(x) <= 0 (strictly decreasing if f'(x) < 0)
- ✓Sign of the first derivative decides the behaviour
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Application of Derivatives
How do you find the slope of the tangent and the normal to a curve at a given point?
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The slope of the tangent to a curve y = f(x) at a point is the value of the derivative dy/dx at that point. The normal is the line perpendicular to the tangent at that point, so its slope is the negative reciprocal of the tangent's slope, that is, slope of normal = -1/(dy/dx). Using the point-slope form with these slopes gives the equations of the tangent and the normal.
slope of normal = -1/(slope of tangent)
Marking-scheme points
- ✓Slope of tangent = dy/dx at the point
- ✓Slope of normal = -1/(dy/dx)
- ✓Use point-slope form for the equations
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Application of Derivatives
State the second derivative test for local maxima and minima.
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Let f be a function and c be a point where f'(c) = 0 (a critical point). By the second derivative test: if f''(c) < 0, then f has a local maximum at c; if f''(c) > 0, then f has a local minimum at c; and if f''(c) = 0, the test fails and we must use the first derivative test (or higher derivatives) to decide.
Marking-scheme points
- ✓Applies at a critical point where f'(c) = 0
- ✓f''(c) < 0 -> local maximum; f''(c) > 0 -> local minimum
- ✓f''(c) = 0 -> test fails, use first derivative test
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Integrals
What is integration? Write the power rule for the integral of x^n.
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Integration is the reverse (inverse) process of differentiation; it is used to find a function whose derivative is given. The indefinite integral of a function includes an arbitrary constant of integration C. The power rule for integration is: the integral of x^n dx = x^(n + 1)/(n + 1) + C, valid for n not equal to -1. For n = -1, the integral of (1/x) dx = ln|x| + C.
integral of x^n dx = x^(n+1)/(n+1) + C
Marking-scheme points
- ✓Integration is the inverse of differentiation
- ✓integral of x^n dx = x^(n+1)/(n+1) + C (n not -1)
- ✓integral of (1/x) dx = ln|x| + C
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