Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equationd V(t)/dt = - k(T - t), where k > 0 is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is

  • straight T squared minus 1 over straight k
  • straight I space minus KT squared over 2
  • straight I space minus space kT squared over 2
  • straight I space minus space kT squared over 2
290 Views

22.

For x ∈, (0, 5π/2) define f(x). Then f (x) = integral subscript 0 superscript straight x space square root of straight t space sin space straight t space dt has

  • local maximum at π and 2π.

  • local minimum at π and 2π

  • local minimum at π and the local maximum at 2π.

  • local minimum at π and the local maximum at 2π.

95 Views

23.

The area of the region enclosed by the curves y = x, x = e, y =1/x and the positive x-axis is

  • 1/2 square units

  • 1 square units

  • 3/2 square units

  • 3/2 square units

139 Views

24.

The line L1: y = x = 0 and L2: 2x + y = 0 intersect the line L3: y + 2 = 0 at P and Q respectively. The bisectorof the acute angle between L1 and L2 intersects L3 at R.
Statement-1: The ratio PR: RQ equals 2√2:√5
Statement-2: In any triangle, the bisector of an angle divides the triangle into two similar triangles.

  • Statement-1 is true, Statement-2 is true ; Statement-2 is 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

  • Statement-1 is true, Statement-2 is false

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25.

If the angle between the line x =fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 3 over denominator straight lambda end fraction and the plane x + 2y + 3z = 4 is cos-1 square root of 5 over 14 end root then λ equal

  • 2/3

  • 3/2

  • 2/5

  • 2/5


A.

2/3

fraction numerator straight x minus 0 over denominator 1 end fraction space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 3 over denominator straight lambda end fraction space... space left parenthesis 1 right parenthesis
x + 2y + 3z = 4 ....... (2)

Angle between the line and plane is
cos space left parenthesis 90 minus space straight theta right parenthesis space equals space fraction numerator straight a subscript 1 straight a subscript 2 space plus straight b subscript 1 straight b subscript 2 space plus space straight c subscript 1 straight c subscript 2 over denominator square root of straight a subscript 1 superscript 2 plus straight b subscript 1 superscript 2 plus straight c subscript 1 superscript 2 end root space square root of straight a subscript 2 superscript 2 plus straight b subscript 2 superscript 2 plus straight c subscript 2 superscript 2 end root end fraction
rightwards double arrow space sin space straight theta space equals space fraction numerator 1 space plus space 4 space plus space 3 straight lambda over denominator square root of 14 space straight x space square root of 5 space plus space straight lambda squared end root end fraction space equals space fraction numerator 5 space plus space 3 straight lambda over denominator square root of 14 space straight x space end root square root of 5 space plus space straight lambda squared end root end fraction space... space left parenthesis 3 right parenthesis

But given that angle between line and plane is

rightwards double arrow space sin space straight theta space equals space fraction numerator 3 over denominator square root of 14 end fraction
therefore space from space left parenthesis 3 right parenthesis
fraction numerator 3 over denominator square root of 14 end fraction space equals space fraction numerator space 5 space plus space 3 straight lambda over denominator square root of 14 space end root space straight x square root of space 5 space plus straight lambda squared end root end fraction
rightwards double arrow space 9 space left parenthesis 5 space plus space straight lambda squared right parenthesis space equals space 25 space plus space 9 straight lambda squared space plus space 30 space straight lambda
30 space straight lambda space equals space 20
straight lambda space equals space 2 divided by 3
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26.

Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3
Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3 .

  • 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.

  • Statement-1 is true, Statement-2 is false.

149 Views

27. If space straight a with rightwards arrow on top space equals space fraction numerator 1 over denominator square root of 10 end fraction space left parenthesis 3 straight i with hat on top space plus straight k with hat on top right parenthesis space and space straight b with rightwards arrow on top space equals space 1 over 7 left parenthesis 2 straight i with hat on top space plus space 3 straight j with hat on top space minus 6 straight k with hat on top right parenthesis comma space then space the space value space of space left parenthesis 2 straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis. left square bracket left parenthesis straight a with rightwards arrow on top space straight x space straight b with rightwards arrow on top right parenthesis space straight x left parenthesis straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top right parenthesis right square bracket space is
  • -5

  • -3

  • 5

  • 5

134 Views

28.

If C and D are two events such that C ⊂ D and P(D) ≠0, then the correct statement among the following is:

  • P(C|D) = P(C)

  • P(C|D)  ≥ P(C)

  • P(C|D) < P(C)

  • P(C|D) < P(C)

233 Views

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29.

The vector straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are not perpendicular  and straight c with rightwards arrow on top space and space straight d with rightwards arrow on top are two vectors satisfying: straight b with rightwards arrow on top space straight x straight c with rightwards arrow on top space equals space straight b with rightwards arrow on top space straight x straight d with rightwards arrow on top space and space straight a with rightwards arrow on top. straight b with rightwards arrow on top space equals space 0. The vector straight d with rightwards arrow on top is equal to

  • straight b with rightwards arrow on top space minus space open parentheses fraction numerator straight b. straight c over denominator straight a with rightwards arrow on top. straight d with rightwards arrow on top end fraction close parentheses straight c with rightwards arrow on top
  • straight c with rightwards arrow on top space plus open parentheses fraction numerator straight a with rightwards arrow on top. straight c with rightwards arrow on top over denominator straight a with rightwards arrow on top. straight b with rightwards arrow on top end fraction close parentheses straight b with rightwards harpoon with barb upwards on top
  • straight b with rightwards arrow on top space plus space open parentheses fraction numerator straight b with rightwards arrow on top. straight c with rightwards arrow on top over denominator straight a with rightwards arrow on top. straight b with rightwards arrow on top end fraction close parentheses straight c with rightwards arrow on top
  • straight b with rightwards arrow on top space plus space open parentheses fraction numerator straight b with rightwards arrow on top. straight c with rightwards arrow on top over denominator straight a with rightwards arrow on top. straight b with rightwards arrow on top end fraction close parentheses straight c with rightwards arrow on top
121 Views

30.

Statement-1 : The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line: straight x over 1 space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 3 end fraction
Statement-2: The line:straight x over 1 space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 3 end fraction  bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).

  • 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.

  • Statement-1 is true, Statement-2 is false.

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