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

Get solved practice paper for CBSE class 11 Maths in PDF format. This paper is important for coming CBSE class 11 Maths exam 2017.

Feb 9, 2017 16:00 IST
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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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