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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2016 Exam Questions

Multiple Choice Questions

1.

If the number of terms in the expansion of open parentheses 1 minus 2 over straight x space plus 4 over straight x squared close parentheses to the power of straight n comma straight x not equal to 0 comma 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 


open parentheses 1 minus 2 over straight x plus 4 over straight x squared close parentheses space is space fraction numerator left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction space or space straight C presuperscript straight n plus 2 end presuperscript subscript 2
left square bracket assuming space 1 over straight x space and space 1 over straight x squared space distinct right square bracket
therefore comma space fraction numerator left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction space equals space 28
rightwards double arrow space left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis space equals space 56 space
space equals space left parenthesis 6 plus 1 right parenthesis left parenthesis 6 plus 2 right parenthesis space rightwards double arrow straight n space equals space 6
Hence comma space sum space of space coefficients equals space left parenthesis 1 minus 2 plus 4 right parenthesis to the power of 6 space equals space 3 to the power of 6 space equals space 729

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2.

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)

⇒ a2 +16d2 +8ad = a2 +8ad +ad +8d2

⇒ 8d2 = ad
8d =a  [∴ d≠0]
Now, common ratio,

fraction numerator straight a plus 4 straight d over denominator straight a plus straight d end fraction space equals space fraction numerator 8 straight d space plus 4 straight d over denominator 8 straight d space plus straight d end fraction space equals space fraction numerator 12 straight d over denominator 9 straight d end fraction space equals space 4 over 3

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3.

If 
straight f left parenthesis straight x right parenthesis space plus space 2 straight f open parentheses 1 over straight x close parentheses space equals space 3 straight x comma
straight x space not equal to space 0 space and space straight S space equals space open curly brackets straight x space straight epsilon space straight R colon space straight f space left parenthesis straight x right parenthesis space equals space straight f left parenthesis negative straight x right parenthesis close curly brackets semicolon space then space straight S

  • is an empty set

  • contains exactly one element.

  • contains exactly two elements.

  • contains more than two elements.


C.

contains exactly two elements.

We have,

straight f left parenthesis straight x right parenthesis space plus space 2 straight f space open parentheses 1 over straight x close parentheses space equals space 3 straight x comma space straight x not equal to 0 space space space left parenthesis straight i right parenthesis
On space replacing space straight x space by space 1 over straight x space in space the space above space equation comma space we space get
straight f open parentheses 1 over straight x close parentheses space space 2 straight f left parenthesis straight x right parenthesis space equals space 3 over straight x
rightwards double arrow space 2 straight f left parenthesis straight x right parenthesis space plus space straight f open parentheses 1 over straight x close parentheses space equals space 3 over straight x
2 straight f left parenthesis straight x right parenthesis space plus space straight f open parentheses 1 over straight x close parentheses space equals space 3 over straight x space space space.. left parenthesis ii right parenthesis
On space multiplying space Eq. space left parenthesis ii right parenthesis space by space 2 space and space subtracting space Eq space left parenthesis straight i right parenthesis space from space Eq space left parenthesis ii right parenthesis comma space we space get
4 space straight f left parenthesis straight x right parenthesis space plus space 2 straight f open parentheses 1 over straight x close parentheses space equals space 6 over straight x
straight f left parenthesis straight x right parenthesis space space space plus space 2 straight f open parentheses 1 over straight x close parentheses space equals space 3 straight x
minus space space space space space space minus space space space space space space space space space space space space minus
to the power of _____________________
3 space straight f space left parenthesis straight x right parenthesis space equals space 6 over straight x minus 3 straight x
to the power of ______________________
rightwards double arrow space straight f space left parenthesis straight x right parenthesis space space equals space 2 over straight x minus straight x
Now space consider space straight f left parenthesis straight x right parenthesis space equals space straight f left parenthesis negative straight x right parenthesis
rightwards double arrow 2 over straight x minus straight x equals negative 2 over straight x space plus straight x
rightwards double arrow 4 over straight x space equals space 2 straight x
2 straight x squared space equals space 4
straight x squared space equals space 2
straight x space equals space plus-or-minus square root of 2 space space
Hence, S contains exactly two elements.

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4.

The sum of all real values of x satisfying the equation
left parenthesis straight x squared minus 5 straight x space plus 5 right parenthesis to the power of straight x squared plus 4 straight x minus 60 space equals space 1 end exponent is:

  • 3

  • -4

  • 6

  • 5


A.

3

Given, 

left parenthesis straight x squared minus 5 straight x space plus 5 right parenthesis to the power of straight x squared plus 4 straight x minus 60 end exponent space equals space 1
clearly comma space this space is space possible space when
straight I. space straight x squared space plus space 4 straight x minus 60 space equals space 0 space and space straight x squared minus 5 straight x space plus 5 space not equal to space 0
II. space straight x squared minus 5 straight x plus 5 space equals space 1
III. space straight x squared minus 5 straight x space plus 5 space equals space minus 1 space and space straight x squared plus 4 straight x minus 60 space equals space Even

Case space straight I space when space straight x squared space plus space 4 space straight x minus 60 space equals space 0 comma space then
straight x squared space plus space 10 space straight x space minus 6 straight x space minus 60 equals 0
rightwards double arrow space straight x left parenthesis straight x plus 10 right parenthesis minus 6 left parenthesis straight x plus 10 right parenthesis space equals space 0
rightwards double arrow left parenthesis straight x plus 10 right parenthesis left parenthesis straight x plus 6 right parenthesis space equals space 0
rightwards double arrow straight x space equals space minus space 10 space or space straight x equals space 6
Note space that comma space for space these space two space vaues space of space straight x comma space straight x squared minus 5 straight x space plus space 5 space not equal to 0
Case space II space when space straight x squared space minus 5 straight x space plus 5 space equals space 1
straight x squared space minus 5 straight x space plus 4 space equals space 0
straight x squared minus 4 straight x minus straight x space plus 4 equals 0
straight x left parenthesis straight x minus 4 right parenthesis minus 1 left parenthesis straight x minus 4 right parenthesis space equals 0
left parenthesis straight x minus 4 right parenthesis left parenthesis straight x minus 1 right parenthesis space equals 0
straight x equals space 4 space or space straight x space equals 1

Case space III space when space straight x squared minus 5 straight x space plus 5 space equals space minus 1
straight x squared minus 5 straight x space plus 5 space equals space minus 1
straight x squared minus 5 straight x space plus space 6 space equals space 0
rightwards double arrow space straight x squared minus 2 straight x minus 3 straight x space plus 6 space equals space 0
straight x left parenthesis straight x minus 2 right parenthesis minus 3 left parenthesis straight x minus 2 right parenthesis space equals space 0
left parenthesis straight x minus 2 right parenthesis left parenthesis straight x minus 3 right parenthesis space equals 0
straight x equals 2 space or space straight x space equals space 3
Now comma space when space straight x space equals space 2 comma space straight x to the power of 2 space end exponent plus 4 straight x minus 60 space equals space 4 plus 8 minus 60
equals negative 48 comma space which space is space an space interger
when space straight x space equals space 3 comma space straight x squared plus space 4 straight x space minus 60 space equals space 9 space plus 12 minus 60 space equals space minus 39
which space is space not space an space even space interger.
Thus comma space in space this space case comma space we space get space straight x space equals space 2
Hence comma space the space sum space of space all space real space values space of space
straight x space equals space minus 10 plus 6 plus 4 plus 1 plus 2 space equals 3

654 Views

5.

Let straight p space equals space limit as straight x rightwards arrow 0 to the power of plus of left parenthesis 1 plus tan squared square root of straight x right parenthesis to the power of 1 divided by 2 straight x end exponent comma space then log p is equal to 

  • 2

  • 1

  • 1/2

  • 1/4


C.

1/2

Given,
straight p space equals limit as straight x rightwards arrow 0 to the power of plus of space left parenthesis 1 plus tan squared square root of straight x right parenthesis to the power of fraction numerator 1 over denominator 2 straight x end fraction end exponent space left square bracket 1 to the power of infinity space form right square bracket
equals space straight e to the power of straight x rightwards arrow stack 0 to the power of plus with lim on top space fraction numerator tan squared square root of straight x over denominator 2 straight x end fraction end exponent
space equals space straight e to the power of 1 half space straight x rightwards arrow stack 0 to the power of plus with lim on top space fraction numerator tan squared square root of straight x over denominator 2 straight x end fraction space space end exponent space equals space straight e to the power of 1 divided by 2 end exponent
log space straight p equals space log space straight e to the power of 1 divided by 2 end exponent space equals space 1 divided by 2

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6.

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

tan space 30 to the power of straight o space equals space fraction numerator straight h over denominator straight x plus straight y end fraction
and
tan space 60 to the power of straight o space equals space straight h over straight y
rightwards double arrow space straight h space equals space fraction numerator straight x plus straight y over denominator square root of 3 end fraction space space space space... space left parenthesis straight i right parenthesis
and space space straight h equals square root of 3 straight y end root space space space space space.. space left parenthesis ii right parenthesis
From space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
fraction numerator straight x plus straight y over denominator square root of 3 end fraction space equals space square root of 3 straight y space
rightwards double arrow space straight x plus straight y space equals space 3 straight y
rightwards double arrow space straight x minus 2 straight y space equals space 0
rightwards double arrow space straight y space equals space straight x over 2
∵ 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



182 Views

7.

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

open parentheses 1 3 over 5 close parentheses squared space plus open parentheses 2 2 over 5 close parentheses squared space plus open parentheses 3 1 fifth close parentheses squared space plus space 4 squared space plus space open parentheses 4 4 over 5 close parentheses squared space plus.....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


S subscript 10 space equals open parentheses 1 3 over 5 close parentheses squared space plus open parentheses 2 2 over 5 close parentheses squared space plus open parentheses 3 1 fifth close parentheses squared space plus space 4 squared space plus space open parentheses 4 4 over 5 close parentheses squared space plus..... space to space 10 space terms
equals open parentheses 8 over 5 close parentheses squared space plus open parentheses 12 over 5 close parentheses squared space plus open parentheses 16 over 5 close parentheses squared space plus 4 squared space plus open parentheses 24 over 5 close parentheses squared space plus.... space to space 10 space terms
equals 4 squared over 5 squared space left parenthesis 2 squared space plus 3 squared plus 4 squared space plus 5 squared plus........ space to space 10 space terms right parenthesis
equals 4 squared over 5 squared space left parenthesis 2 squared space plus 3 squared plus 4 squared space plus 5 squared plus...11 squared right parenthesis
equals space 16 over 25 space left parenthesis 1 squared space plus 2 squared space plus..... plus 11 squared right parenthesis minus 1 squared right square bracket

equals 16 over 25 space open parentheses fraction numerator 11. left parenthesis 11 plus 1 right parenthesis left parenthesis 2.11 plus 1 right parenthesis over denominator 6 end fraction minus 1 close parentheses
space equals space 16 over 25 left parenthesis 506 minus 1 right parenthesis space equals space 16 over 25 space straight x space 505
rightwards double arrow 16 over 5 space straight m space equals space 16 over 25 space straight x space 505
straight m space equals space 101

347 Views

8.

A value of θ for which fraction numerator 2 space plus space 3 straight i space sin space straight theta over denominator 1 minus 2 straight i space sin space straight theta end fraction is purely imaginary is

  • π/3

  • π/6

  • sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 4 end fraction close parentheses
  • sin to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

D.

sin to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

Let z = straight z space equals space fraction numerator 2 space plus space 3 straight i space sin space straight theta over denominator 1 minus 2 straight i space sin space straight theta end fraction is purely imaginary. Then, we have Re (z) = 0
We have Re (z) = 0
Now, consider z = 

straight z equals space fraction numerator 2 space plus space 3 straight i space sin space straight theta over denominator 1 minus 2 straight i space sin space straight theta end fraction
space equals space fraction numerator left parenthesis 2 plus space 3 straight i space sin space straight theta right parenthesis left parenthesis 1 space plus space 2 straight i space sin space straight theta right parenthesis over denominator left parenthesis 1 minus 2 straight i space sin space straight theta right parenthesis left parenthesis 1 plus space 2 straight i space sin space straight theta right parenthesis end fraction
equals space fraction numerator 2 space plus space 4 straight i space sin space straight theta space plus space 3 straight i space sin space straight theta space plus space 6 space straight i squared space sin space squared space straight theta over denominator 1 squared minus space left parenthesis 2 straight i space sin space straight theta right parenthesis squared end fraction
equals space fraction numerator 2 space plus space 7 space straight i space sin space straight theta space minus space 6 space sin squared space straight theta over denominator 1 space plus space 4 space sin squared space straight theta end fraction
equals space fraction numerator 2 minus 6 space sin squared space straight theta over denominator 1 plus 4 space sin squared space straight theta end fraction space plus space straight i fraction numerator 7 space sin space straight theta over denominator 1 space plus space 4 space sin squared space straight theta end fraction
therefore space Re space left parenthesis straight z right parenthesis space equals space 0
therefore space fraction numerator 2 minus 6 space sin squared space straight theta over denominator 1 plus space 4 space sin squared space straight theta end fraction space equals space 0
rightwards double arrow space 2 space equals space 6 space sin squared straight theta
rightwards double arrow space sin squared straight theta space equals space 1 third
rightwards double arrow space sin space straight theta space equals space plus-or-minus space fraction numerator 1 over denominator square root of 3 end fraction
rightwards double arrow space straight theta space equals space sin to the power of negative 1 end exponent space open parentheses plus-or-minus fraction numerator 1 over denominator square root of 3 end fraction close parentheses space equals space plus-or-minus space sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

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9.

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 start with L = 4! = 24
A number of words start with M = 4!/2! = 12
Number of words start with SA = 3!/2! = 3
Number of words starts with SL = 3! = 6
Note that, next word will be 'SMALL'
Hence, the position of word 'SMALL' is 58th.

394 Views

10.

If A = open square brackets table row cell 5 straight a end cell cell negative straight b end cell row 3 2 end table close square brackets and A adj A = AAT, then 5a +b is equal to

  • -1

  • 5

  • 4

  • 5


B.

5

Given, A =open square brackets table row cell 5 straight a end cell cell negative straight b end cell row 3 2 end table close square brackets and A adj A = AAT, Clearly, A (adj A) = |A|In|
space equals space open vertical bar table row cell 5 straight a end cell cell negative straight b end cell row 3 2 end table close vertical bar straight I subscript 2
space equals space left parenthesis 10 straight a space plus space 3 straight b right parenthesis straight I subscript 2 space equals space left parenthesis 10 straight a space plus space 3 straight b right parenthesis open square brackets table row 1 0 row 0 1 end table close square brackets
equals open square brackets table row cell 10 straight a plus 3 straight b end cell 0 row 0 cell 10 straight a plus 3 straight b end cell end table close square brackets space.... space left parenthesis straight i right parenthesis
and space AA to the power of straight T space equals space open square brackets table row cell 5 straight a end cell cell negative straight b end cell row 3 2 end table close square brackets open square brackets table row cell 5 straight a end cell 3 row cell negative straight b end cell 2 end table close square brackets
equals open square brackets table row cell 25 straight a squared plus straight b squared end cell cell 15 straight a minus 2 straight b end cell row cell 15 straight a minus 2 straight b end cell 13 end table close square brackets space... space left parenthesis ii right parenthesis
because space straight A space left parenthesis adj space straight A right parenthesis space equals space AA to the power of straight T
therefore space open square brackets table row cell 10 straight a plus 3 straight b end cell 0 row 0 cell 10 straight a plus 3 straight b end cell end table close square brackets space equals open square brackets table row cell 25 straight a squared plus straight b squared end cell cell 15 straight a minus 2 straight b end cell row cell 15 straight a minus 2 straight b end cell 13 end table close square brackets space
using space eqs space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
rightwards double arrow space 15 straight a minus space 2 straight b equals space 0 space
rightwards double arrow space fraction numerator 2 straight b over denominator 15 end fraction space space space.. left parenthesis iii right parenthesis
and space 10 space straight a space plus space 3 straight b space equals space 13
On space substituting space the space value space of space apostrophe straight a apostrophe space from space Eq. space left parenthesis iii right parenthesis space in space eq space left parenthesis iv right parenthesis comma space we space get
10. open parentheses fraction numerator 2 straight b over denominator 15 end fraction close parentheses plus 3 straight b space equals space 13
rightwards double arrow space fraction numerator 20 straight b space plus space 45 straight b over denominator 15 end fraction space equals space 13 space rightwards double arrow space fraction numerator 65 straight b over denominator 15 end fraction space equals space 13

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

396 Views

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