Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

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31.

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is-

  • tan to the power of negative 1 end exponent space straight b over ac
  • 45°

  • tan to the power of negative 1 end exponent fraction numerator bc over denominator straight a left parenthesis straight c minus straight a right parenthesis end fraction
  • tan to the power of negative 1 end exponent fraction numerator bc over denominator straight a left parenthesis straight c minus straight a right parenthesis end fraction


C.

tan to the power of negative 1 end exponent fraction numerator bc over denominator straight a left parenthesis straight c minus straight a right parenthesis end fraction


straight a space equals space left parenthesis straight u space cos space straight alpha right parenthesis straight t space and space straight b space equals space left parenthesis straight u space sinα right parenthesis space straight t space minus 1 over 22 gt squared
straight b space equals space straight a space tan space straight alpha space minus space 1 half space straight g fraction numerator straight a squared over denominator straight u squared space cos squared straight alpha end fraction
also comma space straight c space equals space fraction numerator straight u squared space sin space 2 straight alpha over denominator straight g end fraction
straight b space equals space straight a space tan space straight alpha space minus space fraction numerator straight a squared straight g over denominator 2 end fraction space open parentheses fraction numerator sin space 2 straight alpha over denominator cg end fraction close parentheses sec squared space straight alpha
straight b space equals space straight a space tan space straight alpha space minus space fraction numerator straight a squared over denominator 2 straight c end fraction 2 space space tanα
rightwards double arrow space open parentheses straight a minus straight a squared over straight c close parentheses tanα space equals space straight b
tan space straight alpha space equals space fraction numerator bc over denominator straight a left parenthesis straight c minus straight a right parenthesis end fraction
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32.

If (2, 3, 5) is one end of a diameter of the sphere x2+ y2+ z2 − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are

  • (4, 9, –3)

  • (4, –3, 3)

  • (4, 3, 5)

  • (4, 3, 5)

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33. Let space straight a with minus on top space equals space straight i with hat on top space plus straight j with hat on top space plus straight k comma space straight b with bar on top space equals space straight i with hat on top space minus straight j with hat on top space plus 2 straight k with hat on top space and space straight c with bar on top space equals space straight x straight i with hat on top space plus space left parenthesis straight x minus 2 right parenthesis straight j with hat on top minus straight k with hat on top.
If space the space vector space straight c with bar on top space lies space in space the space plane space of space straight a with bar on top space and space straight b with bar on top comma space then space straight x space equals
  • 0

  • -1

  • -2

  • -2

103 Views

34.

Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right-angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which ‘k’ can take is given by  

  • {1, 3}

  • {0, 2}

  • {–1, 3}

  • {–1, 3}

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35.

Let P = (−1, 0), Q = (0, 0) and R = ( 3, 3 √3) be three points. The equation of the bisector of the angle PQR

  • square root of 3 straight x space plus straight y space equals space 0
  • straight x space plus space fraction numerator square root of 3 over denominator 2 end fraction straight y space equals space 0
  • fraction numerator square root of 3 over denominator 2 end fraction space straight x space plus straight y space equals 0
  • fraction numerator square root of 3 over denominator 2 end fraction space straight x space plus straight y space equals 0
223 Views

36.

If one of the lines of my2+ (1 − m2)xy − mx2 = 0 is a bisector of the angle between the lines xy = 0, then m is

  • −1/2

  • -2

  • 1

  • 1

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37. let space straight F left parenthesis straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis space plus space straight f space open parentheses 1 over straight x close parentheses comma space where space straight f left parenthesis straight x right parenthesis space equals space integral subscript 1 superscript straight x fraction numerator log space straight t over denominator 1 plus straight t end fraction dt. Then F(e) equals
  • 1/2

  • 1

  • 2

  • 2

90 Views

38.

The solution for x of the equation integral subscript square root of 2 end subscript superscript straight x fraction numerator dt over denominator straight t square root of straight t squared minus 1 end root end fraction space equals straight pi over 2 space is

  • 2

  • π

  • square root of 3 divided by 2
  • square root of 3 divided by 2
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39. integral fraction numerator dx over denominator cos space straight x space plus space square root of 3 space sin space straight x end fraction equals
  • 1 half space log space tan space open parentheses straight x over 2 plus straight pi over 12 close parentheses space plus straight C
  • 1 half space log space tan space open parentheses straight x over 2 minus straight pi over 12 close parentheses plus straight c
  • log space tan space open parentheses straight x over 2 minus straight pi over 12 close parentheses plus straight c
  • log space tan space open parentheses straight x over 2 minus straight pi over 12 close parentheses plus straight c
137 Views

40.

The area enclosed between the curves y2 = x and y = |x| is

  • 2/3

  • 1/3

  • 1/6

  • 1/6

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