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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.

800 board questionsModel answersMarking-scheme pointsEvery chapterCBSE · ISC · State boards

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MathsClass 122 marksmedium

Differential Equations

Find the order and degree of the differential equation (d2y/dx2)^3 + (dy/dx) + y = 0.

Reveal model answer + marking points

The highest order derivative present in the equation is d2y/dx2 (the second derivative), so the order of the differential equation is 2. The power (exponent) of this highest order derivative is 3, and the equation is a polynomial in the derivatives, so the degree is 3. Hence the order is 2 and the degree is 3.

Marking-scheme points

  • Highest derivative is d2y/dx2 -> order 2
  • Its power is 3 -> degree 3
  • Order = 2, degree = 3
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MathsClass 122 markseasy

Differential Equations

Distinguish between the general solution and a particular solution of a differential equation.

Reveal model answer + marking points

The general solution of a differential equation is the solution that contains as many arbitrary constants as the order of the differential equation; it represents a whole family of curves. A particular solution is obtained from the general solution by giving definite values to these arbitrary constants using given initial or boundary conditions; it represents one specific member of the family.

Marking-scheme points

  • General solution: contains arbitrary constants (= order of the equation)
  • Represents a family of curves
  • Particular solution: constants fixed using given conditions
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MathsClass 122 marksmedium

Differential Equations

What is a homogeneous differential equation? What substitution is used to solve it?

Reveal model answer + marking points

A differential equation of the form dy/dx = f(x, y) is called homogeneous if the function f(x, y) can be written as a function of the ratio y/x only (that is, f depends on x and y only through y/x). To solve it, we use the substitution y = v x (so v = y/x), which gives dy/dx = v + x (dv/dx). This substitution reduces the equation to a variable separable form in v and x, which can then be integrated.

y = v x, dy/dx = v + x(dv/dx)

Marking-scheme points

  • Homogeneous: dy/dx can be written as a function of y/x
  • Substitute y = v x, so dy/dx = v + x(dv/dx)
  • Reduces to a variable separable equation
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MathsClass 122 markseasy

Vector Algebra

Distinguish between a scalar and a vector quantity. What is a position vector?

Reveal model answer + marking points

A scalar quantity is one that has only magnitude and no direction (for example, mass, length, temperature). A vector quantity is one that has both magnitude and direction and obeys the laws of vector addition (for example, displacement, velocity, force). The position vector of a point P with coordinates (x, y, z) with respect to the origin O is the vector OP = x i + y j + z k, and its magnitude is sqrt(x^2 + y^2 + z^2).

|OP| = sqrt(x^2 + y^2 + z^2)

Marking-scheme points

  • Scalar: magnitude only (mass, temperature)
  • Vector: magnitude and direction (force, velocity)
  • Position vector of (x, y, z) = x i + y j + z k
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MathsClass 122 markseasy

Vector Algebra

Find the magnitude of the vector a = 2 i + 3 j + 6 k and a unit vector in its direction.

Reveal model answer + marking points

The magnitude of a = 2 i + 3 j + 6 k is |a| = sqrt(2^2 + 3^2 + 6^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7. A unit vector in the direction of a is obtained by dividing a by its magnitude: a hat = a/|a| = (1/7)(2 i + 3 j + 6 k) = (2/7) i + (3/7) j + (6/7) k.

a hat = a/|a|

Marking-scheme points

  • |a| = sqrt(4 + 9 + 36) = 7
  • Unit vector = a/|a|
  • a hat = (2 i + 3 j + 6 k)/7
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MathsClass 122 markseasy

Vector Algebra

Define a unit vector, a zero vector and collinear vectors.

Reveal model answer + marking points

A unit vector is a vector whose magnitude is 1; it is used to specify direction. A zero (null) vector is a vector whose magnitude is zero and whose direction is indeterminate; it is denoted by 0. Collinear vectors are two or more vectors that are parallel to the same line, irrespective of their magnitudes and directions; such vectors can be written as scalar multiples of one another.

Marking-scheme points

  • Unit vector: magnitude 1 (gives direction)
  • Zero vector: magnitude 0, direction indeterminate
  • Collinear vectors: parallel to the same line (scalar multiples)
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MathsClass 122 marksmedium

Vector Algebra

Write the formula for the scalar projection of vector a on vector b. What is the geometrical meaning of the dot product being zero?

Reveal model answer + marking points

The scalar projection (component) of vector a on vector b is given by (a . b)/|b|. It represents the length of the shadow of a along the direction of b. If the dot product a . b = 0 (and neither vector is zero), then cos theta = 0, so theta = 90 degrees; this means the two vectors are perpendicular (orthogonal) to each other.

projection of a on b = (a . b)/|b|

Marking-scheme points

  • Scalar projection of a on b = (a . b)/|b|
  • It is the component of a along b
  • a . b = 0 means the vectors are perpendicular
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MathsClass 122 marksmedium

Vector Algebra

Find the value of lambda for which the vectors a = 2 i + lambda j + k and b = i - 2 j + 3 k are perpendicular.

Reveal model answer + marking points

Two vectors are perpendicular if their dot product is zero. So a . b = (2)(1) + (lambda)(-2) + (1)(3) = 2 - 2 lambda + 3 = 5 - 2 lambda. Setting this equal to zero: 5 - 2 lambda = 0, so 2 lambda = 5 and lambda = 5/2.

a . b = 0 for perpendicular vectors

Marking-scheme points

  • Perpendicular vectors: a . b = 0
  • a . b = 2 - 2 lambda + 3 = 5 - 2 lambda
  • 5 - 2 lambda = 0 gives lambda = 5/2
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MathsClass 122 markseasy

Three Dimensional Geometry

What are direction cosines and direction ratios of a line? State the relation between the direction cosines.

Reveal model answer + marking points

The direction cosines of a line are the cosines of the angles that the line makes with the positive directions of the x-, y- and z-axes; they are usually denoted by l, m and n. The direction ratios are any set of numbers proportional to the direction cosines. The direction cosines always satisfy the relation l^2 + m^2 + n^2 = 1.

l^2 + m^2 + n^2 = 1

Marking-scheme points

  • Direction cosines l, m, n = cosines of angles with the axes
  • Direction ratios are proportional to direction cosines
  • l^2 + m^2 + n^2 = 1
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MathsClass 122 marksmedium

Three Dimensional Geometry

Find the direction cosines of a line whose direction ratios are 1, 2, 2.

Reveal model answer + marking points

The direction cosines are obtained by dividing the direction ratios by the square root of the sum of their squares. Here the magnitude = sqrt(1^2 + 2^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3. So the direction cosines are l = 1/3, m = 2/3 and n = 2/3. We can check l^2 + m^2 + n^2 = 1/9 + 4/9 + 4/9 = 9/9 = 1.

l = a/sqrt(a^2+b^2+c^2), etc.

Marking-scheme points

  • Magnitude of direction ratios = sqrt(1 + 4 + 4) = 3
  • Divide each ratio by 3
  • Direction cosines: 1/3, 2/3, 2/3
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MathsClass 122 marksmedium

Three Dimensional Geometry

Write the general Cartesian equation of a plane and the equation of a plane in intercept form.

Reveal model answer + marking points

The general Cartesian equation of a plane is Ax + By + Cz + D = 0, where A, B, C are the direction ratios of the normal to the plane and are not all zero. The equation of a plane that makes intercepts a, b and c on the x-, y- and z-axes respectively is x/a + y/b + z/c = 1 (the intercept form). A plane is uniquely determined by three non-collinear points.

Ax + By + Cz + D = 0

Marking-scheme points

  • General equation: Ax + By + Cz + D = 0
  • (A, B, C) are direction ratios of the normal
  • Intercept form: x/a + y/b + z/c = 1
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MathsClass 122 marksmedium

Three Dimensional Geometry

How is the angle between two planes found? State the condition for two planes to be perpendicular.

Reveal model answer + marking points

The angle between two planes is equal to the angle between their normals. If the planes are A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0, then cos theta = |A1 A2 + B1 B2 + C1 C2| / (sqrt(A1^2 + B1^2 + C1^2) x sqrt(A2^2 + B2^2 + C2^2)). The two planes are perpendicular if A1 A2 + B1 B2 + C1 C2 = 0, and parallel if A1/A2 = B1/B2 = C1/C2.

cos theta = |A1A2 + B1B2 + C1C2|/(|n1||n2|)

Marking-scheme points

  • Angle between planes = angle between their normals
  • cos theta = |A1A2 + B1B2 + C1C2|/(product of magnitudes)
  • Perpendicular: A1A2 + B1B2 + C1C2 = 0
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MathsClass 122 marksmedium

Three Dimensional Geometry

Find the distance between the parallel planes 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0.

Reveal model answer + marking points

For two parallel planes Ax + By + Cz + D1 = 0 and Ax + By + Cz + D2 = 0, the distance between them is |D1 - D2|/sqrt(A^2 + B^2 + C^2). Here A = 2, B = -2, C = 1, D1 = 3, D2 = 9. Distance = |3 - 9|/sqrt(2^2 + (-2)^2 + 1^2) = 6/sqrt(4 + 4 + 1) = 6/sqrt(9) = 6/3 = 2 units.

d = |D1 - D2|/sqrt(A^2 + B^2 + C^2)

Marking-scheme points

  • Distance = |D1 - D2|/sqrt(A^2 + B^2 + C^2)
  • = |3 - 9|/sqrt(9) = 6/3
  • Distance = 2 units
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MathsClass 122 markseasy

Linear Programming

Define a linear programming problem, the objective function and constraints.

Reveal model answer + marking points

A linear programming problem (LPP) is a problem of maximising or minimising a linear function of variables subject to certain linear conditions. The objective function is the linear function Z = ax + by that is to be optimised (maximised or minimised). The constraints are the linear inequalities (or equations) in the variables that restrict the values the variables can take; together with the non-negativity restrictions (x >= 0, y >= 0) they define the region of possible solutions.

Z = ax + by (objective function)

Marking-scheme points

  • LPP: optimise a linear function under linear constraints
  • Objective function: Z = ax + by to be maximised/minimised
  • Constraints: linear inequalities restricting the variables
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MathsClass 122 markseasy

Linear Programming

Define feasible region, feasible solution and optimal solution in a linear programming problem.

Reveal model answer + marking points

The feasible region is the common region determined by all the constraints (including the non-negativity restrictions) of a linear programming problem; it is the set of all points that satisfy every constraint. A feasible solution is any point that lies in the feasible region. An optimal solution is a feasible solution that gives the maximum or minimum value of the objective function, whichever is required.

Marking-scheme points

  • Feasible region: common region satisfying all constraints
  • Feasible solution: any point in the feasible region
  • Optimal solution: feasible point that optimises the objective function
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MathsClass 122 marksmedium

Linear Programming

Distinguish between a bounded and an unbounded feasible region.

Reveal model answer + marking points

A feasible region is said to be bounded if it can be enclosed within a circle (or a rectangle) of finite radius, that is, it does not extend to infinity. It is unbounded if it extends indefinitely (to infinity) in some direction and cannot be enclosed. For a bounded feasible region, the objective function always has both a maximum and a minimum value at the corner points. For an unbounded region, a maximum or a minimum may not exist and must be checked separately.

Marking-scheme points

  • Bounded: can be enclosed in a circle of finite radius
  • Unbounded: extends to infinity
  • Bounded region always gives both maximum and minimum
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MathsClass 122 markseasy

Probability

Define conditional probability. Write its formula.

Reveal model answer + marking points

Conditional probability is the probability of occurrence of an event A given that another event B has already occurred; it is written as P(A | B). Its formula is P(A | B) = P(A intersection B)/P(B), provided P(B) is not zero. Here P(A intersection B) is the probability that both A and B occur, and P(B) is the probability of B.

P(A | B) = P(A intersection B)/P(B)

Marking-scheme points

  • Probability of A given that B has occurred
  • P(A | B) = P(A intersection B)/P(B)
  • Defined only when P(B) is not zero
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MathsClass 122 marksmedium

Probability

When are two events said to be independent? State the multiplication rule for independent events.

Reveal model answer + marking points

Two events A and B are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other, that is, P(A | B) = P(A) and P(B | A) = P(B). For independent events, the multiplication rule takes the simple form P(A intersection B) = P(A) x P(B). This is used as the test for independence: if P(A intersection B) = P(A) P(B), the events are independent.

P(A intersection B) = P(A) x P(B)

Marking-scheme points

  • Independent: occurrence of one does not affect the other
  • P(A | B) = P(A) and P(B | A) = P(B)
  • Multiplication rule: P(A intersection B) = P(A) P(B)
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MathsClass 122 markseasy

Probability

What is a random variable and a probability distribution?

Reveal model answer + marking points

A random variable is a real-valued function whose value is determined by the outcome of a random experiment; it assigns a number to each outcome of the sample space (for example, the number of heads when two coins are tossed). A probability distribution of a random variable is a table (or description) that lists all the possible values of the random variable together with their corresponding probabilities; the sum of all these probabilities is always 1.

sum of all p(x) = 1

Marking-scheme points

  • Random variable: assigns a real number to each outcome
  • Probability distribution: values with their probabilities
  • Sum of all probabilities = 1
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