Mathematics and Pedagogy is a Maths Part-II Graduation (Senior Teacher) topic in the RAS/RPSC syllabus. This page gathers exam-style Mathematics and Pedagogy multiple-choice questions with correct answers and explanations, so aspirants can test recall and revise frequently examined concepts.
Practice 142 Mathematics and Pedagogy multiple-choice questions with detailed answers and explanations. Ideal for RAS/RPSC exam preparation.
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Q1. In an assignment problem, if all cᵢⱼ ≥ 0, then xᵢⱼ = Xᵢⱼ will be an optimal solution where
Explanation
In a minimization assignment problem with all cᵢⱼ ≥ 0, the objective value cannot be negative. If a feasible assignment X gives ΣᵢΣⱼ cᵢⱼXᵢⱼ = 0, it has attained the least possible objective value. Therefore xᵢⱼ = Xᵢⱼ is optimal, so option D is correct.
Q2. A particle is thrown vertically upward with a velocity of 490 cm/s. It will return to this position after –
Explanation
For vertical projection, the time to return to the same level is T = 2u/g. With u = 490 cm/s and g = 980 cm/s², T = 2 × 490 / 980 = 1 second. Therefore option A is correct.
Q3. The equation of plane parallel to x-axis is
Explanation
A plane parallel to the x-axis must contain a direction parallel to (1, 0, 0). Therefore its normal vector must be perpendicular to the x-axis, so the coefficient of x in the plane equation must be zero. The equation then has the form by + cz + d = 0, which is option C.
Q4. If the primal problem has an unbounded solution, then dual has -
Explanation
By linear-programming duality, if the primal objective is unbounded, the dual cannot have an ordinary finite optimum. In the standard feasible-primal case, the dual is infeasible; among the given choices, option D is the only one that includes the no-solution outcome. Options A, B, and C assert definite forms that do not follow.
Q5. The maximum number of normals which can be drawn common to y² = 4ax and x² = 4by is
Explanation
For y² = 4ax, the normal with parameter t is y = -tx + a(t³ + 2t). For x² = 4by, using parameter s, the normal is y = -x/s + b(s² + 2). A common normal must have the same slope, so s = 1/t; equating intercepts gives a fifth-degree equation in t. Therefore at most 5 common normals are possible, so option D is correct.
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