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Binomial Theorem

Sponsor Area

Question
CBSEENMA11015575

WiredFaculty
  • 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 false. 

Solution

B.

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

WiredFaculty
Hence Statement −2 is also true and is a correct explanation of Statement −1.

Sponsor Area

Question
CBSEENMA11015669

If space straight S subscript straight n space equals space sum from straight r equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight r end fraction space and space straight t subscript straight n space equals space sum from straight r equals space 0 to straight n of space straight r over straight C subscript straight r space then comma straight t subscript straight n over straight S subscript straight n space is
  • 1 half straight n
  • 1 half straight n minus 1
  • n-1

  • n

Solution

A.

1 half straight n WiredFaculty

Question
CBSEENMA11015493

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

  • 17/35

  • 13/35

  • 11/35

  • 10/35

Solution

C.

11/35

WiredFaculty

Question
CBSEENMA11015613

For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is

  • (20, 45)

  • (35, 20)

  • (45, 35)

  • (35, 45)

Solution

D.

(35, 45)

WiredFaculty

Question
CBSEENMA11015472

If (10)9 +2 (11)2(10)7 + .....+10 (11)9 = K(10)9, then k is equal to

  • 121/10

  • 441/100

  • 100

  • 110

Solution

C.

100

1