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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

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

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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

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13.

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

then k is equal to

• 1

• -z

• z

• -1

B.

-z

Here ω is complex cube of unity

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14.

If A, then adj (3A2 + 12A) is equal to

C.

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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
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)

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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

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17.

If for  x ∈ (0, 1/4)  the derivatives  then g(x) is equals to

B.

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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

A.

We have,

⇒ 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)

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19.

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

D.

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

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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

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

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