[Released by CBSE, New Delhi in July 2010 for First Term (SA-I) to be held during the academic year 2010-11 and onward] Time : 3 to 3
1 hours 2
M.M. : 80
General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 10 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each. (iii) Question numbers 1 to 10 in Section A are multiple choice questions, where you are to select one correct option out of the given four. (iv) There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculators is not permitted. SECTION – A (Question numbers 1 to 10 are of one mark each.) 1. Euclid’s Division Lemma states that for any two positive integers a and b, there exist unique integres q and r such that a = bq + r, where r must satisfy : (a) 1 < r < b (b) 0 < r < b (c) 0 < r < b (d) 0 < r < b Sol. (c) 0 < r < b 2. In the figure, the graph of a polynomial p(x) is shown. The number of zeroes of p(x) is : (a) 4 (b) 1 (c) 2 (d) 3 Sol. (b) Since the graph cuts the x-axis at only one point, hence p(x) has only one zero.
3. In the figure, if DE || BC, then x equals : (a) 6 cm (b) 8 cm (c) 10 cm (d) 12.5 cm AD DE = Sol. (c) We have [By BPT] AB BC 2 4 ⇒ = ⇒ x = 10 cm 5 x
CBSE Sample Question Paper (Solved) (Term - I)
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4. If sin 3θ = cos (θ – 6°), where (3θ) and (θ – 6°) are both acute angles, then the value of θ is : (a) 18° (b) 24° (c) 36° (d) 30° Sol. (b) sin 3θ = cos{90° – (96° – θ)} = sin (96° – θ) 96° = 24° ⇒ 3θ = 96° – θ ⇒ θ = 4 1 cosec 2 θ − sec 2 θ 5. Given that tan θ = , the value of is : 3 cosec 2 θ + sec 2 θ