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# JEE Mathematics Solved Question Paper 2016

#### Multiple Choice Questions

1.

Let  then log p is equal to

• 2

• 1

• 1/2

• 1/4

C.

1/2

Given,

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2.

If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:

• 46th

• 59th

• 52nd

• 58th

D.

58th

Clearly, number of words start with A = 4!/ 2! = 12
Number of words starts with SL = 3! = 6
Note that, next word will be 'SMALL'
Hence, the position of word 'SMALL' is 58th.

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3.

If A =  and A adj A = AAT, then 5a +b is equal to

• -1

• 5

• 4

• 5

B.

5

Given, A = and A adj A = AAT, Clearly, A (adj A) = |A|In|

Now, substituting the value of b in Eq. (iii) we get
5a = 2
Hence, 5a + b = 2 +3 = 5

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4.

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60o. Then the time taken (in minutes) by him, from B to reach the pillar, is:

• 6

• 10

• 20

• 5

D.

5

According to given information, we have the following figure:

Now, from ΔACD and ΔBCD, we have

∵ speed is uniform.
∵ distance y will be cover in 5 min
∵ distance x covered in 10 min
∵ Distance x/2 will be cover in 5 min

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5.

If the sum of the first ten terms of the series,

is 16/5 m, the m is equal to

• 102

• 101

• 100

• 99

B.

101

Let S10 be the sum of first ten terms of the series. Then, we have

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6.

If

• is an empty set

• contains exactly one element.

• contains exactly two elements.

• contains more than two elements.

C.

contains exactly two elements.

We have,

Hence, S contains exactly two elements.

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7.

The sum of all real values of x satisfying the equation
is:

• 3

• -4

• 6

• 5

A.

3

Given,

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8.

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

• 8/5

• 4/3

• 1

• 7/4

B.

4/3

Let a be the first term and d be a common difference. Then, we have a+d, a+4d, a+8d in GP,

ie. (a +4d)2 = (a+d)(a+8d)

8d =a  [∴ d≠0]
Now, common ratio,

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9.

A value of θ for which  is purely imaginary is

• π/3

• π/6

D.

Let z =  is purely imaginary. Then, we have Re (z) = 0
We have Re (z) = 0
Now, consider z =

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10.

If the number of terms in the expansion of  is 28, then the sum of the coefficients of all the terms in this expansion is

• 64

• 2187

• 243

• 729

D.

729

Clearly, number of terms in the expansion of

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