Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2017 Exam Questions

Multiple 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 =7


C.

4x – 2y = 1

Eccentricity of ellipse =1/2



Now, a/e = -4
⇒ a = 4 x (1/2) = 2
therefore, b2 = a2(1-e2)
= a2 (1-(1/4)) = 3
straight x squared over 4 space plus space straight y squared over 3 space equals space 1
rightwards double arrow space straight x over 2 space plus space fraction numerator 2 straight y over denominator 3 end fraction space straight x space straight y apostrophe space equals space 0
rightwards double arrow space straight y apostrophe space equals space minus fraction numerator 3 straight x over denominator 4 straight y end fraction
straight y apostrophe vertical line subscript left parenthesis 1 comma space 3 divided by 2 right parenthesis end subscript space equals space minus 3 over 4 space straight x 2 over 3 space equals space minus 1 half

Equation of normal at (1, 3/2)
y-3/2 = 2(x – 1)
⇒ 2y – 3 = 4x – 4
⇒ 4x – 2y = 1

763 Views

12.

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

  • 2/9


D.

2/9

269 Views

13.

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

  • -1


B.

-z

Here ω is complex cube of unity

straight R subscript 1 rightwards arrow with space on top straight R subscript 1 space plus straight R subscript 2 space plus space straight R subscript 3
space equals space open vertical bar table row 1 0 0 row 1 cell negative straight omega squared minus 1 end cell cell straight omega squared end cell row 1 cell straight omega squared end cell straight omega end table close vertical bar
space equals space 3 left parenthesis negative 1 minus straight omega minus straight omega right parenthesis space equals space minus space 3 straight z
rightwards double arrow space straight k space equals space minus straight z

307 Views

14.

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 84 row 63 72 end table close square brackets

C.

open square brackets table row 51 63 row 84 72 end table close square brackets
Given space straight A space equals space open square brackets table row 2 cell negative 3 end cell row cell negative 4 end cell 1 end table close square brackets
3 straight A squared space equals space open square brackets table row 16 cell negative 9 end cell row cell negative 12 end cell 13 end table close square brackets
12 space straight A space equals space open square brackets table row 24 cell negative 36 end cell row cell negative 48 end cell 12 end table close square brackets
therefore space 3 straight A squared space plus space 12 space straight A space equals space open square brackets table row 72 cell negative 63 end cell row cell negative 84 end cell 51 end table close square brackets
adj space left parenthesis 3 straight A squared space plus space 12 space straight A right parenthesis space equals space open square brackets table row 51 63 row 84 72 end table close square brackets
368 Views

15.

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

  • 7/64


C.

7/16

P(exactly one of A or B occurs)
= P(A) + P(B) – 2P(A ∩ B) =1/4
P(Exactly one of B or C occurs)
= P(B) + P(C) – 2P(B ∩ C) =1/4
P(Exactly one of C or A occurs)
= P(C) + P(A) – 2P(C ∩ A) =1/4
Adding all, we get
2ΣP(A) – 2ΣP(A ∩ B) =3/4
∴ΣP(A) – ΣP(A ∩ B) =3/8
Now, P(A ∩ B ∩ C) =1/16
(Given)
∴ P(A ∪ B ∪ C)
= ΣP(A) – ΣP(A∩B) + P(A ∩ B ∩ C)
space equals 3 over 8 space plus space 1 over 16 space equals space 7 over 16

410 Views

16.

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

  • 12/55


B.

6/55

Let A ≡ {0, 1,2 ,3 ,4 ......, 10 }
n(s) = 11C2 (where 'S' denotes sample space)
Let E be the given event
∴ E ≡ {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)} n(E) = 6
∴ n(E) = 6 = 6/55

259 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 over denominator 1 minus 9 straight x cubed end fraction

B.

fraction numerator 9 over denominator 1 plus 9 straight x cubed end fraction
Let space straight y space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 6 straight x space square root of 3 over denominator 1 minus 9 straight x cubed end fraction close parentheses space where space straight x space element of open parentheses 0 comma 1 fourth close parentheses
equals tan to the power of negative 1 end exponent space open parentheses fraction numerator 2. left parenthesis 3 straight x to the power of 3 divided by 2 end exponent right parenthesis over denominator 1 minus left parenthesis 3 straight x to the power of 3 divided by 2 end exponent right parenthesis end fraction close parentheses space space equals space 2 space tan to the power of negative 1 end exponent space left parenthesis 3 straight x to the power of 3 divided by 2 end exponent right parenthesis

As space 3 straight x to the power of 3 divided by 2 end exponent space element of space open parentheses 0 comma 3 over 8 close parentheses
therefore space dy over dx space equals space 2 space straight x space fraction numerator 1 over denominator 1 plus 9 straight x cubed end fraction space straight x space 3 space straight x 3 over 2 space straight x space space left parenthesis straight x to the power of 1 divided by 2 end exponent right parenthesis
space equals space fraction numerator 9 over denominator 1 plus 9 straight x cubed end fraction square root of straight x
therefore space straight g left parenthesis straight x right parenthesis space equals space fraction numerator 9 over denominator 1 plus 9 straight x cubed end fraction
211 Views

18.

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 3 over 2 close parentheses

A.

open parentheses 2 comma 1 half close parentheses

We have,

1 half open vertical bar table row straight k cell negative 3 straight k end cell 1 row 5 straight k 1 row cell negative straight k end cell 2 1 end table close vertical bar space equals space 28

⇒ 5k2 + 13k – 46 = 0
or
5k2 + 13k + 66 = 0 (no real solution exist)
∴ k =–23/5
or k = 2
k is an integer, so k =2
As 
Therefore (2, 1/2)

 

857 Views

19.

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

D.

4 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

1372 Views

20.

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

  • a finite set containing two or more elements


A.

a singleton

straight D space equals space open vertical bar table row 1 1 1 row 1 straight a 1 row straight a straight b 1 end table close vertical bar space equals space 0
rightwards double arrow straight a space equals space 1
and space at space straight a space equals space 1
straight D subscript 1 space equals straight D subscript 2 space equals space straight D subscript 3 space equals space 0

but at a = 1 and b =1
First two equations are x +y+ z =1
and third equations is x + y +z = 0
⇒ There is no solution
therefore, b = {1}
⇒ it is a singleton set

297 Views

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