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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 112 markseasy

Introduction to Three Dimensional Geometry

Find the midpoint of the line segment joining the points (2, 3, 4) and (6, 7, 8).

Reveal model answer + marking points

The midpoint of the segment joining (x1,y1,z1) and (x2,y2,z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2). = ((2+6)/2, (3+7)/2, (4+8)/2) = (4, 5, 6).

M = ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)

Marking-scheme points

  • Midpoint = average of endpoints
  • = ((2+6)/2, (3+7)/2, (4+8)/2)
  • = (4, 5, 6)
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MathsClass 112 markseasy

Limits and Derivatives

Evaluate the limit: lim as x approaches 2 of (x^2 - 4)/(x - 2).

Reveal model answer + marking points

Direct substitution gives 0/0, so factorise the numerator: x^2 - 4 = (x - 2)(x + 2). Then (x^2 - 4)/(x - 2) = x + 2 for x not equal to 2. Taking the limit as x approaches 2 gives 2 + 2 = 4.

a^2 - b^2 = (a - b)(a + b)

Marking-scheme points

  • 0/0 form -> factorise
  • (x^2 - 4)/(x - 2) = x + 2
  • Limit = 4
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MathsClass 112 markseasy

Limits and Derivatives

State and use the standard limit to evaluate lim as x approaches 0 of (sin x)/x.

Reveal model answer + marking points

The standard limit is lim as x approaches 0 of (sin x)/x = 1 (with x in radians). This is a fundamental result used throughout limits and derivatives of trigonometric functions.

lim x->0 (sin x)/x = 1

Marking-scheme points

  • Standard result (x in radians)
  • lim x->0 (sin x)/x = 1
  • Used for trigonometric derivatives
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MathsClass 113 marksmedium

Limits and Derivatives

Evaluate lim as x approaches 0 of (sin 3x)/(sin 5x).

Reveal model answer + marking points

Write (sin 3x)/(sin 5x) = [(sin 3x)/(3x) x 3x] / [(sin 5x)/(5x) x 5x]. As x approaches 0, (sin 3x)/(3x) -> 1 and (sin 5x)/(5x) -> 1, so the limit = (1 x 3x)/(1 x 5x) = 3/5.

lim x->0 sin(kx)/(kx) = 1

Marking-scheme points

  • Multiply and divide to make sin(kx)/(kx) forms
  • Each such ratio tends to 1
  • Limit = 3x/5x = 3/5
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MathsClass 113 marksmedium

Limits and Derivatives

Evaluate lim as x approaches 3 of (x^2 - 9)/(x^2 - x - 6).

Reveal model answer + marking points

Direct substitution gives 0/0. Factorise: x^2 - 9 = (x - 3)(x + 3) and x^2 - x - 6 = (x - 3)(x + 2). Cancelling (x - 3): the expression becomes (x + 3)/(x + 2). Taking x approaching 3 gives (3 + 3)/(3 + 2) = 6/5.

Marking-scheme points

  • 0/0 form -> factorise both
  • Cancel (x - 3): get (x + 3)/(x + 2)
  • Limit = 6/5
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MathsClass 113 marksmedium

Limits and Derivatives

Evaluate lim as x approaches 0 of (sqrt(1 + x) - 1)/x.

Reveal model answer + marking points

Direct substitution gives 0/0. Rationalise by multiplying numerator and denominator by (sqrt(1 + x) + 1): [(sqrt(1+x) - 1)(sqrt(1+x) + 1)]/[x(sqrt(1+x) + 1)] = (1 + x - 1)/[x(sqrt(1+x) + 1)] = x/[x(sqrt(1+x) + 1)] = 1/(sqrt(1+x) + 1). As x approaches 0 this equals 1/(1 + 1) = 1/2.

(sqrt a - sqrt b)(sqrt a + sqrt b) = a - b

Marking-scheme points

  • 0/0 form -> rationalise numerator
  • Numerator becomes x; cancel x
  • Limit = 1/(1 + 1) = 1/2
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MathsClass 112 marksmedium

Limits and Derivatives

Find the derivative of f(x) = x^2 from first principles.

Reveal model answer + marking points

By definition f'(x) = lim as h approaches 0 of [f(x + h) - f(x)]/h = lim [ (x + h)^2 - x^2 ]/h = lim [x^2 + 2xh + h^2 - x^2]/h = lim [2xh + h^2]/h = lim (2x + h) = 2x. So f'(x) = 2x.

f'(x) = lim h->0 [f(x+h) - f(x)]/h

Marking-scheme points

  • f'(x) = lim h->0 [f(x+h) - f(x)]/h
  • (x+h)^2 - x^2 = 2xh + h^2
  • Limit = 2x
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MathsClass 112 markseasy

Limits and Derivatives

Differentiate y = x^3 + 3x^2 - 5x + 2 with respect to x.

Reveal model answer + marking points

Using the power rule d/dx(x^n) = n x^(n-1) term by term: dy/dx = 3x^2 + 3(2x) - 5 + 0 = 3x^2 + 6x - 5.

d/dx(x^n) = n x^(n-1)

Marking-scheme points

  • Apply power rule to each term
  • Derivative of constant = 0
  • dy/dx = 3x^2 + 6x - 5
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MathsClass 113 marksmedium

Limits and Derivatives

Differentiate y = (x^2 + 1)(x - 3) using the product rule.

Reveal model answer + marking points

Let u = x^2 + 1 and v = x - 3, so u' = 2x and v' = 1. By the product rule dy/dx = u'v + uv' = (2x)(x - 3) + (x^2 + 1)(1) = 2x^2 - 6x + x^2 + 1 = 3x^2 - 6x + 1.

(uv)' = u'v + uv'

Marking-scheme points

  • Product rule: (uv)' = u'v + uv'
  • u' = 2x, v' = 1
  • dy/dx = 3x^2 - 6x + 1
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MathsClass 113 marksmedium

Limits and Derivatives

Differentiate y = (x + 1)/(x - 1) using the quotient rule.

Reveal model answer + marking points

Let u = x + 1 and v = x - 1, so u' = 1 and v' = 1. By the quotient rule dy/dx = (u'v - uv')/v^2 = [1(x - 1) - (x + 1)(1)]/(x - 1)^2 = (x - 1 - x - 1)/(x - 1)^2 = -2/(x - 1)^2.

(u/v)' = (u'v - uv')/v^2

Marking-scheme points

  • Quotient rule: (u/v)' = (u'v - uv')/v^2
  • Numerator = (x - 1) - (x + 1) = -2
  • dy/dx = -2/(x - 1)^2
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MathsClass 112 marksmedium

Limits and Derivatives

Find the derivative of y = sin x cos x.

Reveal model answer + marking points

Using the product rule with u = sin x (u' = cos x) and v = cos x (v' = -sin x): dy/dx = cos x cos x + sin x (-sin x) = cos^2 x - sin^2 x = cos 2x. (Alternatively, y = (1/2) sin 2x, whose derivative is cos 2x.)

cos^2 x - sin^2 x = cos 2x

Marking-scheme points

  • Product rule: cos x cos x + sin x(-sin x)
  • = cos^2 x - sin^2 x
  • = cos 2x
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MathsClass 113 markshard

Limits and Derivatives

Find the derivative of sin x from first principles.

Reveal model answer + marking points

f'(x) = lim h->0 [sin(x + h) - sin x]/h. Expand sin(x + h) = sin x cos h + cos x sin h, so the numerator = sin x cos h + cos x sin h - sin x = sin x(cos h - 1) + cos x sin h. Thus f'(x) = sin x . lim h->0 (cos h - 1)/h + cos x . lim h->0 (sin h)/h. Since lim (cos h - 1)/h = 0 and lim (sin h)/h = 1, we get f'(x) = sin x . 0 + cos x . 1 = cos x.

d/dx(sin x) = cos x

Marking-scheme points

  • Use sin(x+h) = sinx cosh + cosx sinh
  • lim (cos h - 1)/h = 0 and lim (sin h)/h = 1
  • Derivative = cos x
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MathsClass 112 markseasy

Statistics

Find the mean of the first 10 natural numbers.

Reveal model answer + marking points

The first 10 natural numbers are 1, 2, ..., 10. Their sum = n(n + 1)/2 = 10 x 11/2 = 55. Mean = sum/number of observations = 55/10 = 5.5.

mean = (sum of observations)/n

Marking-scheme points

  • Sum of first 10 natural numbers = 55
  • Mean = sum/n
  • = 55/10 = 5.5
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MathsClass 113 marksmedium

Statistics

Find the mean deviation about the mean for the data 6, 7, 10, 12, 13, 4, 8, 12.

Reveal model answer + marking points

There are 8 observations. Mean = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9. Absolute deviations from the mean: |6-9|=3, |7-9|=2, |10-9|=1, |12-9|=3, |13-9|=4, |4-9|=5, |8-9|=1, |12-9|=3. Sum of absolute deviations = 3+2+1+3+4+5+1+3 = 22. Mean deviation = 22/8 = 2.75.

MD = (sum of |xi - mean|)/n

Marking-scheme points

  • Mean = 72/8 = 9
  • Sum of |xi - mean| = 22
  • Mean deviation = 22/8 = 2.75
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MathsClass 113 marksmedium

Statistics

Find the variance and standard deviation of the data 2, 4, 6, 8, 10.

Reveal model answer + marking points

Mean = (2+4+6+8+10)/5 = 30/5 = 6. Squared deviations: (2-6)^2=16, (4-6)^2=4, (6-6)^2=0, (8-6)^2=4, (10-6)^2=16; their sum = 40. Variance = 40/5 = 8. Standard deviation = sqrt(variance) = sqrt8 = 2 sqrt2 (approximately 2.83).

variance = (sum of (xi - mean)^2)/n; SD = sqrt(variance)

Marking-scheme points

  • Mean = 6; sum of squared deviations = 40
  • Variance = 40/5 = 8
  • SD = sqrt8 = 2 sqrt2 (about 2.83)
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MathsClass 112 markseasy

Statistics

Define variance and standard deviation of a set of observations.

Reveal model answer + marking points

Variance is the arithmetic mean of the squares of the deviations of the observations from their mean; for n observations, variance = (1/n) sum of (xi - mean)^2. Standard deviation is the positive square root of the variance; it has the same units as the observations and measures the spread or dispersion of the data about the mean.

SD = sqrt(variance)

Marking-scheme points

  • Variance = mean of squared deviations from the mean
  • SD = positive square root of variance
  • Both measure dispersion; SD has same units as data
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MathsClass 113 marksmedium

Statistics

The mean of a data set is 6 and its standard deviation is 2 sqrt2. Find the coefficient of variation.

Reveal model answer + marking points

Coefficient of variation (CV) = (standard deviation/mean) x 100. Here SD = 2 sqrt2 which is approximately 2.83, and mean = 6. CV = (2.83/6) x 100, which is approximately 47.1 percent. A higher CV indicates greater relative variability.

CV = (SD/mean) x 100

Marking-scheme points

  • CV = (SD/mean) x 100
  • SD = 2 sqrt2 = 2.83, mean = 6
  • CV = (2.83/6) x 100 = about 47.1 percent
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MathsClass 112 markseasy

Probability

A fair die is thrown once. Find the probability of getting a prime number.

Reveal model answer + marking points

The sample space is {1, 2, 3, 4, 5, 6}, so total outcomes = 6. The prime numbers on a die are 2, 3 and 5, which is 3 favourable outcomes. Probability = favourable/total = 3/6 = 1/2.

P(E) = favourable outcomes / total outcomes

Marking-scheme points

  • Total outcomes = 6
  • Primes on die: 2, 3, 5 (3 outcomes)
  • P = 3/6 = 1/2
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MathsClass 113 marksmedium

Probability

Two fair dice are thrown together. Find the probability that the sum of the numbers on them is 8.

Reveal model answer + marking points

When two dice are thrown, the total number of outcomes = 6 x 6 = 36. The outcomes giving a sum of 8 are (2,6), (3,5), (4,4), (5,3) and (6,2), which is 5 favourable outcomes. Probability = 5/36.

P(E) = favourable outcomes / total outcomes

Marking-scheme points

  • Total outcomes = 36
  • Sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) = 5 outcomes
  • P = 5/36
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MathsClass 113 marksmedium

Probability

One card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is a king or a heart.

Reveal model answer + marking points

Let A = the card is a king and B = the card is a heart. P(A) = 4/52, P(B) = 13/52, and P(A and B) = 1/52 (the king of hearts). By the addition theorem P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

P(A or B) = P(A) + P(B) - P(A and B)

Marking-scheme points

  • P(king) = 4/52, P(heart) = 13/52
  • P(king and heart) = 1/52
  • P(king or heart) = 16/52 = 4/13
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