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CBSE Class 11 Mathematics Solved Practice Paper 2017: Set ‒ II

Feb 9, 2017 16:00 IST

    CBSE Class 11 Mathematics Solved Practice Paper is available for download in PDF format. This solved paper is based on latest CBSE syllabus. With the help solutions provided in this paper, student can easily understand the concept involved in the problems.

    Inside CBSE Class 11 Mathematics Solved Practice Paper

    In this solved practice paper, there are 26 questions. These questions are divided into three sections; section ‒ A, section ‒ B and section ‒ C.

    Section ‒ A contains 6 questions, section ‒ B contains 13 questions and section ‒ C contains 7 questions. Overall choices in this question paper are not provided. However, you will find internal choices in some questions.  

    CBSE Class 11 Mathematics Syllabus 2017

    General Instructions to solved this paper are given below

    All questions are compulsory.

    Please check that this Question Paper contains 26 Questions.

    Questions 1-6 in Section A are very short-answer type questions carrying 1 mark each.

    Questions 7-19 in Section B are long-answer I type questions carrying 4 marks each.

    Questions 20-26 in Section C are long-answer II type questions carrying 6 marks each.

    Some randomly selected solved questions are given below

    Question: Prove that: sec2x + cosec2x ≥ 4

    Solution:

    sec2x + cosec2x = (1 + tan2x) + (1 + cot2x)

    ⇒ sec2x + cosec2x = 2 + tan2x + cot2x

    ⇒ sec2x + cosec2x = 2 + tan2x + cot2x ‒ 2 tan x cot x + 2 tan x cot x

    ⇒ sec2x + cosec2x = 2 + (tan x ‒ cot x)2 + 2

    ⇒ sec2x + cosec2x = 4 + (tan x ‒ cot x)2 ≥ 4

    ∵ (tan x ‒ cot x)2 ≥ 0

    ∴ 4 + (tan x ‒ cot x)2 ≥ 4

    NCERT Solutions

    Question: Calculate the value of a3 + 7a2a + 16, when a = 1 + 2 i

    Solution:

    Given,

    a = 1 + 2 i

    a – 1 = 2i

    ⇒ (a – 1)2 = 4 i2

    ⇒ a2 – 2 a + 5 =0

    Now, a3 + 7a2a + 16 = a (a2 ‒ 2a + 5) + 9 (a2 ‒ 2a + 5) + (12 a ‒ 29)

    a3 + 7a2a + 16 = a (0) + 9 (0) + 12 a ‒ 29 [∵ a2 ‒ 2a + 5 = 0]

    a3 + 7a2a + 16 = 12 (1 + 2i) ‒ 29 [∵ a = 1 + 2i]

    a3 + 7a2a + 16 = ‒17 + 24i

    Therefore, the value of the given polynomial, when a = 1 + 2i is ‒17 + 24i.

    Question: sin8x ‒ cos8x = (sin2x ‒ cos2x) (1 ‒ 2 sin2x cos2x)

    Solution:

    LHS = (sin8x ‒ cos8x)

            = (sin4x)2 ‒ (cos4x)2

            = (sin4x ‒ cos4x) (sin4x +cos4x)

            = (sin2x ‒ cos2x) (sin2x + cos2x) (sin4x +cos4x)

            = (sin2x ‒ cos2x) (sin2x + cos2x) (sin4x +cos4x)

            = (sin2x ‒ cos2x) (sin4x +cos4x)

            = (sin2x ‒ cos2x) (sin4x +cos4x ‒ 2 sin2x cos2x + 2 sin2x cos2x)

            = (sin2x ‒ cos2x) {(sin2x +cos2x)2 ‒ 2 sin2x cos2x}

            = (sin2x ‒ cos2x) {1 ‒ 2 sin2x cos2x}

            = RHS

    Download the completely solved practice paper form this link

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