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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.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

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MathsClass 123 marksmedium

Determinants

Using determinants, find the area of the triangle whose vertices are (1, 0), (6, 0) and (4, 3).

Reveal model answer + marking points▾

The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substituting: area = (1/2)|1(0 - 3) + 6(3 - 0) + 4(0 - 0)| = (1/2)|-3 + 18 + 0| = (1/2)(15) = 7.5 square units.

Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

Marking-scheme points

  • ✓Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
  • ✓= (1/2)|-3 + 18 + 0|
  • ✓= 7.5 square units
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MathsClass 123 marksmedium

Determinants

Find the inverse of the matrix A = [[2, 3],[1, 4]] using the adjoint method.

Reveal model answer + marking points▾

First find the determinant: |A| = (2)(4) - (3)(1) = 8 - 3 = 5, which is non-zero, so the inverse exists. For a 2 x 2 matrix [[a, b],[c, d]], the adjoint is [[d, -b],[-c, a]], so adj A = [[4, -3],[-1, 2]]. Then the inverse is A inverse = (1/|A|) adj A = (1/5)[[4, -3],[-1, 2]] = [[4/5, -3/5],[-1/5, 2/5]].

A inverse = (1/|A|) adj A

Marking-scheme points

  • ✓|A| = 8 - 3 = 5 (non-zero, so inverse exists)
  • ✓adj A = [[4, -3],[-1, 2]]
  • ✓A inverse = (1/5)[[4, -3],[-1, 2]]
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MathsClass 123 markshard

Determinants

Solve the system of equations 2x + 3y = 8 and x + 2y = 5 using determinants (Cramer's rule).

Reveal model answer + marking points▾

Write D = determinant of coefficients = |[[2, 3],[1, 2]]| = (2)(2) - (3)(1) = 4 - 3 = 1. Dx (replace first column by constants) = |[[8, 3],[5, 2]]| = (8)(2) - (3)(5) = 16 - 15 = 1. Dy (replace second column by constants) = |[[2, 8],[1, 5]]| = (2)(5) - (8)(1) = 10 - 8 = 2. By Cramer's rule, x = Dx/D = 1/1 = 1 and y = Dy/D = 2/1 = 2. So x = 1, y = 2.

x = Dx/D, y = Dy/D

Marking-scheme points

  • ✓D = 1, Dx = 1, Dy = 2
  • ✓Cramer's rule: x = Dx/D, y = Dy/D
  • ✓x = 1, y = 2
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