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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

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22.

If the line  lies in the plane lx +my -z = 9, then l2 +m2 is equal to

• 26

• 18

• 5

• 2

D.

2

Since the line 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  be three unit vectors such that  then the angle between  is

• 3π/4

• π/2

• 2π/3

• 5π/6

D.

5π/6

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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.

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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

• 10

B.

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

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

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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

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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,

⇒ 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)

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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

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29.

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

B.

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

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
=

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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

C.

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 = e2 [e>0]

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