JEE mathematics

Sponsor Area

Question
CBSEENMA11015581

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals

  • 1 half left parenthesis 1 minus square root of 5 right parenthesis
  • 1 half square root of 5
  • square root of 5
  • 1 half left parenthesis square root of 5 minus 1 right parenthesis

Solution

D.

1 half left parenthesis square root of 5 minus 1 right parenthesis Given space ar to the power of straight n minus 1 end exponent space equals ar to the power of straight n space plus space ar to the power of straight n plus 1 end exponent
rightwards double arrow space 1 space equals space straight r plus space straight r squared
therefore space straight r space equals space fraction numerator square root of 5 minus 1 over denominator 2 end fraction

Sponsor Area

Question
CBSEENMA11015582

If space sin to the power of negative 1 end exponent space open parentheses straight x over 5 close parentheses space plus space cosec to the power of negative 1 end exponent space open parentheses 5 over 4 close parentheses space equals space straight pi over 2 then the value of x
  • 1

  • 3

  • 4

  • 5

Solution

B.

3

sin to the power of negative 1 end exponent space straight x over 5 space plus space sin to the power of negative 1 end exponent space 4 over 5 space equals space straight pi over 2
rightwards double arrow space sin to the power of negative 1 end exponent space straight x over 5 space equals space cos to the power of negative 1 end exponent space 4 over 5
rightwards double arrow space sin to the power of negative 1 end exponent space straight x over 5
space equals space sin to the power of negative 1 end exponent space 3 over 5
therefore space straight x space equals space 3

Question
CBSEENMA11015583

In the binomial expansion of (a - b)n, n ≥ 5, the sum of 5th and 6th terms is zero, then
a/b equals

  • 5/n −4

  • 6 /n −5

  • n -5 /6

  • n -4 /5

Solution

D.

n -4 /5

straight C presuperscript straight n subscript 4 space straight a to the power of straight n minus 4 end exponent space left parenthesis negative straight b right parenthesis to the power of 4 space plus straight C presuperscript straight n subscript 5 straight a to the power of straight n minus 5 end exponent space left parenthesis negative straight b right parenthesis to the power of 5 space equals space 0
rightwards double arrow space open parentheses straight a over straight b close parentheses space equals space fraction numerator straight n minus 5 plus 1 over denominator 5 end fraction

Question
CBSEENMA11015584

The set S: {1, 2, 3, …, 12} is to be partitioned into three sets A, B, C of equal size. Thus, A ∪ B ∪ C = S, A ∩ B = B ∩ C = A ∩ C = φ. The number of ways to partition S is-

  • 12!/3!(4!)3

  • 12!/3!(3!)4

  • 12!/(4!)3

  • 12!/(3!)4

Solution

C.

12!/(4!)3

Number of ways
straight C presuperscript 12 subscript 4 space straight x space straight C presuperscript 8 subscript 4 space straight x space straight C presuperscript 4 subscript 4 space equals space fraction numerator 12 factorial over denominator left parenthesis 4 factorial right parenthesis cubed end fraction

Question
CBSEENMA11015585

A body weighing 13 kg is suspended by two strings 5 m and 12 m long, their other ends being fastened to the extremities of a rod 13 m long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are 

  • 12 kg and 13 kg

  • 5 kg and 5 kg

  • 5 kg and 12 kg

  • 5 kg and 13 kg

Solution

C.

5 kg and 12 kg


straight T subscript 2 space cos space open parentheses straight pi over 2 minus straight theta close parentheses space equals space straight T subscript 1 space cosθ
rightwards double arrow space straight T subscript 1 space cos space straight theta space equals space straight T subscript 2 space sin space straight theta
straight T subscript 1 space sin space straight theta space plus space straight T subscript 2 space cos space straight theta space equals space 13
because space OC space equals space CA space space equals CB
rightwards double arrow space angle space AOC space equals space angle OAC space and space angle COB space equals space angle OBC
therefore space sin space straight theta space equals space sin space straight A space equals space 5 over 13 space and space cos space straight theta space equals 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 space and space cos space straight theta space equals space 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 straight T subscript 2
straight T subscript 2 space open parentheses 5 over 12.5 over 13.12 over 13 close parentheses space equals space 13
straight T subscript 2 space open parentheses fraction numerator 169 over denominator 12.13 end fraction close parentheses space equals space 13
straight T subscript 2 space equals space 12 space kgs
rightwards double arrow space straight T subscript 1 space equals space 5 space kgs