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Vocabulary and conceptual flashcards generated from the CBSE Class 12 Mathematics Question Paper Code 65/5/3.
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Q.P. Code 65/5/3 Exam Structure
A Class 12 Mathematics examination comprising 38 compulsory questions divided into five sections (A, B, C, D, and E) with a maximum of 80 marks and 3 hours time allowed.
Section A Specification
Comprises 20 questions carrying 1 mark each, consisting of 18 multiple choice questions (MCQs, Questions 1 to 18) and 2 Assertion-Reason based questions (Questions 19 and 20).
Section B Specification
Comprises 5 Very Short Answer (VSA) type questions carrying 2 marks each (Questions 21 to 25).
Section C Specification
Comprises 6 Short Answer (SA) type questions carrying 3 marks each (Questions 26 to 31).
Section D Specification
Comprises 4 Long Answer (LA) type questions carrying 5 marks each (Questions 32 to 35).
Section E Specification
Comprises 3 case study based questions carrying 4 marks each (Questions 36 to 38).
Principal Value of cot−1(31)
The principal value of cot−1(31) is 3π.
Determinant of a 3×3 Diagonal Matrix
For a 3×3 diagonal matrix A=[aij] with a11=1, a22=5, and a33=−2, the determinant ∣A∣ equals −10.
Determinant Property ∣A∣=kn∣B∣
If A=kB, where A and B are square matrices of order n and k is a scalar, then ∣A∣=kn∣B∣.
Greatest Integer Function Continuity at x=2
For f(x)=[x], where f is the greatest integer function, f is neither continuous nor differentiable at x=2.
Order and Degree Sum of (dxdy)2=dx2d2y
The differential equation (dxdy)2=dx2d2y has order 2 and degree 1, giving a sum of order and degree equal to 3.
Perpendicular Vector Calculation
The problem of finding a vector of magnitude 5 that is perpendicular to both 3i^−2j^+k^ and 4i^+3j^−2k^.
Case Study 1: Student Roll Number Function
A function f:A→N defined on a set A of 30 students in class XII, where f(x) is given by the roll number of student x.
Case Study 2: Seed Germination Probabilities
A setup involving 10 brinjal seeds, 12 cabbage seeds, and 8 radish seeds with germination probabilities of 25%, 35%, and 40% respectively.
Case Study 3: Cuboidal Box Optimization
Optimization problem for a closed wooden cuboidal box with square base (length = breadth = xm) and height = ym to achieve minimum surface area S for a fixed volume V.