Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2016 Exam Questions

Multiple Choice Questions

21.

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

  • 3a2−26a+55=0

  • 3a2−32a+84=0

  • 3a2−34a+91=0

  • 3a2−23a+44=0


B.

3a2−32a+84=0

We know that, if x1, x2..... xn are n observations, then their standard deviation is given by


square root of 1 over straight n sum straight x subscript straight i superscript 2 space minus open parentheses fraction numerator space stack sum straight x subscript straight i with space below over denominator straight n end fraction close parentheses squared end root

we begin inline style space end style begin inline style have end style begin inline style comma end style begin inline style space end style begin inline style space end style begin inline style left parenthesis end style begin inline style 3 end style begin inline style. end style begin inline style 5 end style begin inline style right parenthesis squared end style begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator left parenthesis 2 squared space plus 3 squared space plus straight a squared plus 11 squared right parenthesis over denominator 4 end fraction begin inline style space end style begin inline style minus end style begin inline style space end style begin inline style open parentheses begin display style fraction numerator 2 plus 3 plus straight a plus 11 over denominator 4 end fraction end style close parentheses squared end style begin inline style space end style begin inline style space end style begin inline style space end style

space begin inline style rightwards double arrow end style begin inline style space end style 49 over 9 begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator 4 space plus space 9 space plus straight a squared plus 121 over denominator 4 end fraction begin inline style minus end style begin inline style open parentheses begin display style fraction numerator 16 plus straight a over denominator 4 end fraction end style close parentheses squared end style begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style
begin inline style rightwards double arrow end style begin inline style space end style 49 over 9 begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator 134 space plus straight a squared over denominator 4 end fraction begin inline style minus end style fraction numerator 256 space plus straight a squared plus 32 straight a over denominator 16 end fraction begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style

49 over 9 begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator 4 straight a squared space plus 536 minus 256 minus straight a squared minus 32 straight a over denominator 16 end fraction

space begin inline style space end style begin inline style 49 end style begin inline style space end style begin inline style straight x end style begin inline style space end style begin inline style 4 end style begin inline style space end style begin inline style equals end style begin inline style space end style begin inline style 3 end style begin inline style straight a squared end style begin inline style minus end style begin inline style 32 end style begin inline style straight a end style begin inline style plus end style begin inline style 280 end style begin inline style space end style begin inline style 3 end style begin inline style straight a squared end style begin inline style minus end style begin inline style 32 end style begin inline style straight a end style begin inline style space end style begin inline style plus end style begin inline style space end style begin inline style 84 end style begin inline style space end style begin inline style equals end style begin inline style 0 end style

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22.

If the line fraction numerator straight x minus 3 over denominator 2 end fraction space equals space fraction numerator straight y plus 2 over denominator negative 1 end fraction space equals space fraction numerator straight z plus 4 over denominator 3 end fraction lies in the plane lx +my -z = 9, then l2 +m2 is equal to 

  • 26

  • 18

  • 5

  • 2


D.

2

Since the linefraction numerator straight x minus 3 over denominator 2 end fraction space equals space fraction numerator straight y plus 2 over denominator negative 1 end fraction space equals space fraction numerator straight z plus 4 over denominator 3 end fraction lies in the plane lx my-z =9, therefore we have 2l-m-3 = 0
[∴ normal will be perpendicular to the line]
⇒ 2l-m = 3
and 3l -2m +4 = 9
[∴ point (-3,-2-4) lies on the plane
⇒ 3l-2m = 5
On solving eqs (i) adn (ii), we get
l =1 and m=-1
therefore, l2 +m2 = 2

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23.

Let bold a with hat on top space bold b with hat on top space and space bold c with hat on top space be three unit vectors such that straight a with hat on top space straight x space left parenthesis straight b with hat on top space straight x straight c with hat on top right parenthesis space equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top plus straight c with hat on top right parenthesis. space If space straight b with hat on top space is space not space parallel space to space stack straight c. with hat on top then the angle between bold a with bold hat on top bold space and space bold b with bold hat on top is

  • 3π/4

  • π/2

  • 2π/3

  • 5π/6


D.

5π/6

Given comma space vertical line straight a with hat on top vertical line space equals space vertical line stack straight b vertical line with hat on top space equals vertical line straight c with hat on top vertical line space equals space 1
and space space straight a with hat on top space straight x space left parenthesis straight b with hat on top space straight x space straight c with hat on top space right parenthesis equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top space plus space straight c with hat on top space right parenthesis
Now comma space consider space straight a with hat on top space straight x space left parenthesis straight b with hat on top space straight x space straight c with hat on top space right parenthesis space equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top space plus space straight c with hat on top space right parenthesis
rightwards double arrow space left parenthesis straight a with hat on top. straight c with hat on top right parenthesis straight b with hat on top space minus space left parenthesis straight a with hat on top. straight b with hat on top right parenthesis straight c with hat on top space equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top space plus fraction numerator square root of 3 over denominator 2 end fraction straight c with hat on top
On space comparing comma space we space get

straight a with hat on top. straight b with hat on top space equals space minus space fraction numerator square root of 3 over denominator 2 end fraction space rightwards double arrow space vertical line straight a with hat on top vertical line space vertical line straight b with hat on top space vertical line space cosθ space equals space minus fraction numerator square root of 3 over denominator 2 end fraction
rightwards double arrow space cos space straight theta space equals space minus fraction numerator square root of 3 over denominator 2 end fraction space vertical line because space vertical line straight a with hat on top vertical line space equals space vertical line straight b with hat on top vertical line space equals space 1
rightwards double arrow space cos space straight theta space equals space cos space open parentheses straight pi space minus straight pi over 6 close parentheses rightwards double arrow space straight theta space equals space fraction numerator 5 straight pi over denominator 6 end fraction space
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24.

The centres of those circles which touch the circle, x2+y2−8x−8y−4=0, externally and also touch the x-axis, lie on:

  • a circle

  • an ellipse which is not a circle

  • a hyperbola.

  • a parabola.


D.

a parabola.

245 Views

25.

If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is:

  • 5 square root of 2
  • 5 square root of 3
  • 5

  • 10


B.

5 square root of 3

Given equation of a circle is x2 + y2 -4x +6y -12 = 0, whose centre is (2,-3) and radius

space equals space square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 12 end root
equals square root of 4 plus 9 plus 12 space equals space 5 end root
Now, according to given information, we have the following figure.


x2+y2-4x +6y-12 =0

Clearly, AO perpendicular to BC, as O is mid-point of the chord.
Now in ΔAOB,. we have
OA space equals space square root of left parenthesis negative 3 minus 2 right parenthesis squared plus left parenthesis 2 plus 3 right parenthesis squared end root
equals space square root of 25 plus 25 end root space equals space square root of 50 space end root space equals space 5 square root of 2
and space OB space equals space 5
therefore comma
OB space equals space 5
AB space equals space square root of OA squared plus OB squared end root space equals space square root of 50 plus 25 end root space equals space square root of 75 space equals space 5 square root of 3

786 Views

26.

Let P be the point on the parabola, y2=8x which is at a minimum distance from the centre C of the circle, x2+(y+6)2=1. Then the equation of the circle, passing through C and having its centre at P is:

  • x2+y2−4x+8y+12=0

  • x2+y2−x+4y−12=0

  • x2+y2− 4 x +2y−24=0

  • x2+y2−4x+9y+18=0


A.

x2+y2−4x+8y+12=0

Centre of circle x2 + (y+6)2 = 1 is C (0,6)
Let the coordinates of point P be (2t2, 4t)
Now, let 

straight D space equals CP space equals space square root of left parenthesis 2 straight t squared right parenthesis squared space plus space left parenthesis 4 straight t plus 6 right parenthesis squared end root
rightwards double arrow space straight D space equals space square root of 4 straight t to the power of 4 plus 16 straight t squared plus 36 space plus 48 straight t end root
rightwards double arrow space straight D squared left parenthesis straight t right parenthesis space equals space 4 straight t to the power of 4 space plus space 16 straight t squared plus 48 straight t space plus 36
Let space straight F space left parenthesis straight t right parenthesis space equals space 4 straight t to the power of 4 space plus 16 straight t squared plus 48 straight t plus 36
For space minimum comma space straight F apostrophe left parenthesis straight t right parenthesis space equals space 0
rightwards double arrow space 16 straight t cubed space plus 32 straight t space plus 48 space equals space 0
rightwards double arrow straight t cubed space plus 2 straight t plus 3 space equals space 0
rightwards double arrow space left parenthesis straight t plus 1 right parenthesis left parenthesis straight t squared minus straight t plus 3 right parenthesis space equals space 0
straight t equals space minus 1
Thus comma space coordinate space of space point space straight P space are space left parenthesis 2 minus 4 right parenthesis
Now comma space CP space equals space square root of 2 squared plus left parenthesis negative 4 plus 6 right parenthesis squared end root space equals space square root of 4 plus 4 end root space equals space 2 square root of 2
Hence comma space the space required space space equation space of space circle space is space
left parenthesis straight x minus 2 right parenthesis squared space plus space left parenthesis straight y plus 4 right parenthesis squared space equals space left parenthesis 2 square root of 2 right parenthesis squared
rightwards double arrow space straight x squared space plus 4 minus 4 straight x space plus straight y squared space plus 16 plus 8 straight y space equals space 8
rightwards double arrow space straight x squared space plus straight y squared minus 4 straight x space plus 8 straight y space plus 12 space equals space 0

358 Views

27.

Two sides of a rhombus are along the lines, x−y+1=0 and 7x−y−5=0. If its diagonals intersect at (−1, −2), then which one of the following is a vertex of this rhombus?

  • (−3, −9)

  • (−3, −8)

  • (1/3, -8/3)

  • (-10/3, -7/3)


C.

(1/3, -8/3)

As the given lines x-y +1 =0 and 7x-y-5 = 0 are not parallel, therefore they represent the adjacent sides of the rhombus.

On solving x-y+1 = 0  adn 7x - y -5 = 0. we get x =1 and y =2
Thus, one of the vertex is A(1,2)


Let the coordinate of point C be (x,y)
Then, 
negative 1 space equals space fraction numerator straight x plus 1 over denominator 2 end fraction
and space minus 2 space equals space fraction numerator straight y plus 2 over denominator 2 end fraction
⇒ x+1 =- 2 and y =-4-2
⇒ x=-3 and y =-6
Hence, coordinates of C = (-3,-6)

Note that, vertices B and D will satisfy x-y +1 =0 and 7x - y-5 = 0, therefore the coordinate of vertex D is (1/3, -8/3)

536 Views

28.

The Boolean Expression (p∧~q)∨q∨(~p∧q) is equivalent to:

  • ~p ∧ q

  • p ∧ q

  • p ∨ q

  • p ∨ ~ q


C.

p ∨ q

Consider, (p ∧~q) ∨ q ∨(~p ∧ q)
 [(p ∧~q) ∨ q] ∨ (~p ∧ q)
[(p ∨~q) ∧ t] ∨ (~p ∧ q)
((p ∨ q) ∨ (~p ∧ q)
≡(p ∨ q ∨ ~p) ∧ (p ∨ q ∨ q)
≡(q ∨ t) ∧ (p ∨ q)
≡ t ∧ (p ∨ q)
≡ p ∨ q

246 Views

29.

The distance of the point (1, −5, 9) from the plane x−y+z=5 measured along the line x=y=z is:

  • 3 square root of 10
  • 10 square root of 3
  • fraction numerator 10 over denominator square root of 3 end fraction
  • 20 over 3

B.

10 square root of 3

the equation of the line passing through the point (1,5-9 and parallel to x =y=z is


fraction numerator straight x minus 1 over denominator 1 end fraction space equals space fraction numerator straight y plus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 9 over denominator 1 end fraction space equals space straight lambda
Thus, any point on this line is of the form
(λ +1, λ-5 ,λ+9) 
Now, if P (λ +1, λ-5, λ+9) is the point of intersection of line and plane, then
 (λ+1) - (λ-5) +λ+9 = 5
λ +15 = 5
λ = -10
therefore coordinates of point P are (-9, -15,-1)
Hence, required distance
=equals square root of left parenthesis 1 plus 9 right parenthesis squared plus left parenthesis negative 5 plus 15 right parenthesis squared plus left parenthesis 9 plus 1 right parenthesis squared end root
equals square root of 10 squared plus space 10 squared plus 10 squared end root space equals space 10 square root of 3

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30.

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is:

  • 4/3

  • fraction numerator 4 over denominator square root of 3 end fraction
  • fraction numerator 2 over denominator square root of 3 end fraction
  • square root of 3

C.

fraction numerator 2 over denominator square root of 3 end fraction

We have, 2b2/a = 8 and 2b = ae
⇒ b2 = 4a and 2b = ae
Consider, 2b = ae
Consider, 2b = ae

⇒ 4b2 = a2e2

⇒ 4a2 (e2-1)=a2e2

⇒ 4e2-4 = e23 straight e squared space equals space 4 space rightwards double arrow space straight e space equals space fraction numerator 2 over denominator square root of 3 end fraction [e>0]

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