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

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

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

392 Views

Advertisement

2.

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. 
125 Views

Advertisement
3.

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

131 Views

4.

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)

158 Views

Advertisement
5.

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

  • 1

  • 2n

  • 3n

  • 3n

126 Views

6.

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

  • 3

  • 7

  • 5

  • 6


7.

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


 Multiple Choice QuestionsShort Answer Type

8.

Prove by induction that for n  N, n2 + n is an even integer (n  1)


Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

9.

A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by x = t - 6t2 + t3. Its acceleration will be zero at

  • t = 1 unit time

  • t = 2 unit time

  • t = 3 unit time

  • t = 4 unit time


10.

For each n N, 23n - 1is divisible by

  • 7

  • 8

  • 6

  • 16


Advertisement