Statement − 1: For every natural number n ≥ 2 Statement −

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 Multiple Choice QuestionsLong Answer Type

21.

Using principle of mathematical induction, prove that

cos space straight alpha space space cos space 2 straight alpha space space cos space 4 straight alpha space............... cos left parenthesis 2 to the power of straight n minus 1 end exponent straight alpha right parenthesis space equals space fraction numerator sin space 2 to the power of straight n straight alpha over denominator 2 to the power of straight n sinα end fraction for all straight n space element of space straight N

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 Multiple Choice QuestionsShort Answer Type

22.

Using principle of mathematical induction, show that straight n less than 2 to the power of straight n space space for space all space straight n space element of space straight N

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23.

Using principle of mathematical induction, show that  3 to the power of straight n greater than 2 to the power of straight n for all straight n space element of space straight N


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 Multiple Choice QuestionsMultiple Choice Questions

24.

If the number of terms in the expansion of open parentheses 1 minus 2 over straight x space plus 4 over straight x squared close parentheses to the power of straight n comma straight x not equal to 0 comma is 28, then the sum of the coefficients of all the terms in this expansion is

  • 64

  • 2187

  • 243

  • 243

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25.

Statement − 1: For every natural number n ≥ 2 fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction space greater than space square root of straight n

Statement −2: For every natural number n ≥ 2,straight n greater or equal than 2 comma space square root of straight n left parenthesis straight n plus 1 right parenthesis space end root space less than space straight n plus 1

  • Statement −1 is false, Statement −2 is true

  • Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.


C.

Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

straight P space left parenthesis straight n right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction
straight P space left parenthesis 2 right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space greater than space square root of 2
Let space us space assume space that space straight P space left parenthesis straight k right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space...... fraction numerator 1 over denominator square root of straight k end fraction space plus fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space greater than square root of straight k plus 1 end root
has space to space be space true.
straight L. straight H. straight S greater than thin space square root of straight k space plus space fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space equals space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root space plus 1 over denominator square root of straight k plus 1 end root end fraction
since space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space greater than straight k space space left parenthesis for all space straight k greater or equal than 0 right parenthesis
therefore space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root plus 1 over denominator square root of straight k plus 1 end root end fraction space greater than space fraction numerator straight k plus 1 over denominator square root of straight k plus 1 end root end fraction space equals space square root of straight k plus 1 end root
Let space straight p left parenthesis straight n right parenthesis space space equals space square root of straight n space left parenthesis straight n plus 1 right parenthesis end root space less than space straight n plus 1
State space minus 1 space is space correct.
straight P space left parenthesis 2 right parenthesis space space equals space square root of 2 space straight x space 3 end root space less than space 3
If space straight P space left parenthesis straight k right parenthesis space space equals space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space less than space left parenthesis straight k plus 1 right parenthesis space is space true
Now space space straight P space left parenthesis straight k plus 1 right parenthesis space equals space square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space less than space straight k plus 2 space has space to space be space true
square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space thin space left parenthesis straight k plus 2 right parenthesis
Hence Statement −2 is not a correct explanation of Statement −1. 
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26.

If A = open square brackets table row 1 0 row 1 1 end table close square brackets and I = open square brackets table row 1 0 row 0 1 end table close square brackets , then which one of the following holds for all n ≥ 1, by the principle of mathematical induction

  • An = nA – (n – 1)I

  • An = 2n-1A – (n – 1)I

  • An = nA + (n – 1)I

  • An = nA + (n – 1)I

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27.

Let S(K) = 1 +3+5+..... (2K-1) = 3+K2. Then which of the following is true?

  • S(1) is correct

  • Principle of mathematical induction can be used to prove the formula

  • S(K) ≠S(K+1)

  • S(K) ≠S(K+1)

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28.

Maximum sum of coefficient in the expansion of (1 – x sinθ + x2 )n is

  • 1

  • 2n

  • 3n

  • 3n

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29.

For positive integer n, n3 + 2n is always divisible by

  • 3

  • 7

  • 5

  • 6


30.

The acceleration of a particle starting from rest moving in a straight line with uniform acceleration is 8 m/s2. The time taken by the particle to move the second metre is

  • (√2-1)/2 S

  • (√2+1)/2 S

  • (1 + √2)S

  • (√2-1)S


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