If the number of terms in the expansion of  is 28, then the su

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The sum of all real values of x satisfying the equation
left parenthesis straight x squared minus 5 straight x space plus 5 right parenthesis to the power of straight x squared plus 4 straight x minus 60 space equals space 1 end exponent is:

  • 3

  • -4

  • 6

  • 6

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2.

If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:

  • 46th

  • 59th

  • 52nd

  • 52nd

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3.

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


D.

243

Clearly, number of terms in the expansion of 


open parentheses 1 minus 2 over straight x plus 4 over straight x squared close parentheses space is space fraction numerator left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction space or space straight C presuperscript straight n plus 2 end presuperscript subscript 2
left square bracket assuming space 1 over straight x space and space 1 over straight x squared space distinct right square bracket
therefore comma space fraction numerator left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction space equals space 28
rightwards double arrow space left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis space equals space 56 space
space equals space left parenthesis 6 plus 1 right parenthesis left parenthesis 6 plus 2 right parenthesis space rightwards double arrow straight n space equals space 6
Hence comma space sum space of space coefficients equals space left parenthesis 1 minus 2 plus 4 right parenthesis to the power of 6 space equals space 3 to the power of 6 space equals space 729

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4.

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

  • 8/5

  • 4/3

  • 1

  • 1

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5.

If the sum of the first ten terms of the series,

open parentheses 1 3 over 5 close parentheses squared space plus open parentheses 2 2 over 5 close parentheses squared space plus open parentheses 3 1 fifth close parentheses squared space plus space 4 squared space plus space open parentheses 4 4 over 5 close parentheses squared space plus.....is 16/5 m, the m is equal to

  • 102

  • 101

  • 100

  • 100

347 Views

6.

Let straight p space equals space limit as straight x rightwards arrow 0 to the power of plus of left parenthesis 1 plus tan squared square root of straight x right parenthesis to the power of 1 divided by 2 straight x end exponent comma space then log p is equal to 

  • 2

  • 1

  • 1/2

  • 1/2

371 Views

7.

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60o. Then the time taken (in minutes) by him, from B to reach the pillar, is:

  • 6

  • 10

  • 20

  • 20

182 Views

8.

If 
straight f left parenthesis straight x right parenthesis space plus space 2 straight f open parentheses 1 over straight x close parentheses space equals space 3 straight x comma
straight x space not equal to space 0 space and space straight S space equals space open curly brackets straight x space straight epsilon space straight R colon space straight f space left parenthesis straight x right parenthesis space equals space straight f left parenthesis negative straight x right parenthesis close curly brackets semicolon space then space straight S

  • is an empty set

  • contains exactly one element.

  • contains exactly two elements.

  • contains exactly two elements.

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9.

A value of θ for which fraction numerator 2 space plus space 3 straight i space sin space straight theta over denominator 1 minus 2 straight i space sin space straight theta end fraction is purely imaginary is

  • π/3

  • π/6

  • sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 4 end fraction close parentheses
  • sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 4 end fraction close parentheses
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10.

If A = open square brackets table row cell 5 straight a end cell cell negative straight b end cell row 3 2 end table close square brackets and A adj A = AAT, then 5a +b is equal to

  • -1

  • 5

  • 4

  • 4

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