JEE mathematics

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Question
CBSEENMA11015445

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

  • 5

Solution

A.

3

Given, 
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Question
CBSEENMA11015446

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

  • 58th

Solution

D.

58th

Clearly, number of words start with A = 4!/ 2! = 12
Number of words start with L = 4! = 24
A number of words start with M = 4!/2! = 12
Number of words start with SA = 3!/2! = 3
Number of words starts with SL = 3! = 6
Note that, next word will be 'SMALL'
Hence, the position of word 'SMALL' is 58th.

Question
CBSEENMA11015447

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

  • 729

Solution

D.

729

Clearly, number of terms in the expansion of 

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Question
CBSEENMA11015448

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

  • 7/4

Solution

B.

4/3

Let a be the first term and d be a common difference. Then, we have a+d, a+4d, a+8d in GP,
ie. (a +4d)2 = (a+d)(a+8d)
⇒ a2 +16d2 +8ad = a2 +8ad +ad +8d2
⇒ 8d2 = ad
8d =a  [∴ d≠0]
Now, common ratio,
fraction numerator straight a plus 4 straight d over denominator straight a plus straight d end fraction space equals space fraction numerator 8 straight d space plus 4 straight d over denominator 8 straight d space plus straight d end fraction space equals space fraction numerator 12 straight d over denominator 9 straight d end fraction space equals space 4 over 3

Question
CBSEENMA11015449

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

  • 99

Solution

B.

101

Let S10 be the sum of first ten terms of the series. Then, we have

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