Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directives is x= –4, then the equation of the normal to it at (1,3/2) is

  • x + 2y = 4

  • 2y – x = 2

  • 4x – 2y = 1

  • 4x – 2y = 1

763 Views

12.

If two different numbers are taken from the set {0, 1, 2, 3, ......., 10), then the probability that their sum, as well as absolute difference, are both multiple of 4, is

  • 7/55

  • 6/55

  • 14/55

  • 14/55

259 Views

13.

For three events A, B and C,
P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P(Exactly one of C or A occurs) = 1/4and P(All the three events occur simultaneously) = 1/16.Then the probability that at least one of the events occurs, is

  • 3/16

  • 7/32

  • 7/16

  • 7/16

410 Views

14.

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB.If ∠BPC = β , then tanβ is equal to

  • 4/9

  • 6/7

  • 1/4

  • 1/4

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15.

If S is the set of distinct values of 'b' for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is

  • a singleton

  • an empty set

  • an infinite set

  • an infinite set

297 Views

16.

Let  ω be a complex number such that 2ω +1 = z where z = √-3. if

open vertical bar table row 1 1 1 row 1 cell negative straight omega squared end cell cell straight omega squared end cell row 1 cell straight omega squared end cell cell straight omega to the power of 7 end cell end table close vertical bar space equals space 3 straight k
then k is equal to 

  • 1

  • -z

  • z

  • z

307 Views

17.

If for  x ∈ (0, 1/4)  the derivatives tan to the power of negative 1 end exponent space open parentheses fraction numerator 6 straight x square root of straight x over denominator 1 minus 9 straight x cubed end fraction close parentheses space is space square root of straight x. end root space straight g left parenthesis straight x right parenthesis comma then g(x) is equals to 

  • fraction numerator 3 over denominator 1 plus 9 straight x cubed end fraction
  • fraction numerator 9 over denominator 1 plus 9 straight x cubed end fraction
  • fraction numerator 3 straight x square root of straight x over denominator 1 minus 9 straight x cubed end fraction
  • fraction numerator 3 straight x square root of straight x over denominator 1 minus 9 straight x cubed end fraction
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18.

The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is

  • 4 space left parenthesis square root of 2 space plus 1 right parenthesis
  • 2 left parenthesis square root of 2 space plus 1 right parenthesis
  • 2 left parenthesis square root of 2 space minus 1 right parenthesis
  • 2 left parenthesis square root of 2 space minus 1 right parenthesis


D.

2 left parenthesis square root of 2 space minus 1 right parenthesis


There are two circles satisfying the given
conditions. The circle shown is of least area.
Let radius of circle is 'r'
∴ co-ordinates of centre = (0, 4 – r)
∴ circle touches the line y = x in first quadrant
therefore space open vertical bar fraction numerator 0 minus left parenthesis 4 minus 2 right parenthesis over denominator square root of 2 end fraction close vertical bar space equals space straight r space
rightwards double arrow straight r space minus space 4 space equals space plus-or-minus space straight r square root of 2
therefore space straight r space equals space fraction numerator 0 minus left parenthesis 4 minus 2 right parenthesis over denominator square root of 2 end fraction space equals space straight r space minus 4 space equals space plus-or-minus straight r square root of 2
therefore space space straight r space equals space fraction numerator 4 over denominator square root of 2 space plus 1 end fraction space equals space 4 left parenthesis square root of 2 minus 1 right parenthesis

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19.

If A, open square brackets table row 2 cell negative 3 end cell row cell negative 4 end cell 1 end table close square bracketsthen adj (3A2 + 12A) is equal to

  • open square brackets table row 72 cell negative 63 end cell row cell negative 84 end cell 51 end table close square brackets
  • open square brackets table row 72 cell negative 84 end cell row cell negative 63 end cell 51 end table close square brackets
  • open square brackets table row 51 63 row 84 72 end table close square brackets
  • open square brackets table row 51 63 row 84 72 end table close square brackets
368 Views

20.

Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point 

  • open parentheses 2 comma 1 half close parentheses
  • open parentheses 2 comma negative 1 half close parentheses
  • open parentheses 1 comma negative 3 over 2 close parentheses
  • open parentheses 1 comma negative 3 over 2 close parentheses
857 Views

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