Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

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61.

If a = 2i^ +k^, b = i^ + j^ + k^, c = 4i^ - 3j^ + 7k^, then the vector r satisfying r x b = c x b and r · a = 0 is

  •  i^ + 8j^ +2k^

  •  i^ - 8j^ +2k^

  •  i^ - 8j^ -2k^

  •  - i^ - 8j^ +2k^


D.

 - i^ - 8j^ +2k^

Given that, a =2i^ + k^, b = i^ +j^ + k^ and c = 4i^ - 3j^ + 7k^ r × b - c × b = 0      r - c × b = 0  r - c  b r - c = λb      i  r . a = 0 c + λb . a = 0 4i^ - 3j^ + 7k^ + λi^ + λj^ + λk^ . 2i^ + k^ = 0 4 + λ . i^ + - 3 +λj^ + 7 + λk^ . 2i^ + k^ = 0 4 + λ 2 + 7 + λ . 1 = 0 8 + 2λ + 7 + λ = 0 3λ = - 15 λ = - 5Put the value of λ in Eq (1), we getr = 4i^ - 3j^ + 7k^ - 5i^ + j^ + k^= 4i^ - 3j^ +7k^ - 5i^ - 5j^ - 5k^= - i^ - 8j^ +2k^


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62.

If a, b and c are three vectors such that a = 1, b = 2, c = 3 and a . b = b . c = c . a = 0, then a b c = ? is equal to

  • 2

  • 3

  • 4

  • 5


63.

If a × b b × c c × a = λa b c2, then λ is  equal to 

  • 0

  • 1

  • 2

  • 3


64.

The cartesion equation of the plane passing through the point (3, - 2, - 1) and parallel to the vectors b = i^ - 2j^ + 4k^ and c = 3i^ + 2j^ - 5k^ is

  • 2x - 17y - 8z + 63 = 0 

  • 3x + 17y + 8z + 36 = 0

  • 2x + 17y + 8z + 36 = 0

  • 3x - 16y + 8z - 63 = 0


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65.

The probability distribution of a random variable is given below

X = x 0 1 2 3 4 5 6 7
P(X = x) 0 k 2k 2k 3k k2 2k2 7k2 + k

Then P(0 ) < X < 5) =?

  • 110

  • 310

  • 810


66.

If (2, - 1, 2) and (K, 3, 5) are the triads of direction ratios of two lines and the angle between them is 45°, then the value of K is

  • 2

  • 3

  • 4

  • 6


67.

The length of perpendicular from the origin to the plane which makes intercepts 13, 14 and 15 respectively on the coordinate axes is

  • 152

  • 110

  • 52

  • 5


68.

The volume of sphere is increasing at the rate of 1200 cu cm/s. The rate of increase in its surface area when the radius is 10 cm is

  • 120 sqcm/s

  • 240 sqcm/s

  • 200 sqcm/s

  • 100 sqcm/s


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69.

The slope of the tangent to the curve

y = 0x11 + t3dt at the point, where x = 1 is 

  • 14

  • 13

  • 12

  • 1


70.

dxx - 1x2 - 1 is equal to

  • - x - 1x + 1 + C

  • x - 1x2 + 1 + C

  • - x + 1x - 1 + C

  • x2 + 1x - 1 + C


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