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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2015 Exam Questions

Multiple Choice Questions

1.

The mean of the data set comprising of 16 observations is 16. If one of the observation value 16 is deleted and three new observations valued3,4 and 5 are added to the data, then the mean of the resultant data is

  • 16.8

  • 16.0

  • 15.8

  • 14.0


D.

14.0

Given, 

fraction numerator straight x subscript 1 space plus space straight x subscript 2 space plus space straight x subscript 3 space plus space..... space plus space straight x subscript 16 over denominator 16 end fraction space equals space 16
rightwards double arrow space sum from straight i space equals space 1 to 18 of straight y subscript straight i space equals space left parenthesis 16 space straight x space 16 minus 16 right parenthesis space plus space left parenthesis 3 plus 4 plus 5 right parenthesis space equals space 252
Number space of space obsercvations space space equals space 18
therefore comma space New space mean space equals space fraction numerator begin display style sum from straight i space equals 1 to 18 space of space straight y subscript straight i end style over denominator 18 end fraction
space equals space 252 divided by 18 space equals space 14

449 Views

2.

If m is the AMN of two distinct real numbers l and n (l,n>1) and G1, G2, and G3 are three geometric means between l and n, then straight G subscript 1 superscript 4 space plus 2 straight G subscript 2 superscript 4 space plus space straight G subscript 3 superscript 4 equals

  • 4l2 mn

  • 4lm2n

  • 4 lmn2

  • 4l2m2n2


B.

4lm2n

Given, 
m is the AM of and n

l +n = 2m

and G1, G2, G3, n are in GP
Let r be the common ratio of this GP
G1 = lr
G2 =lr2
G3= lr3
n = lr4

rightwards double arrow space straight r space equals space open parentheses straight n over l close parentheses to the power of 1 divided by 4 end exponent
Now comma space straight G subscript 1 superscript 4 space plus space 2 straight G subscript 2 superscript 4 space plus space straight G subscript 3 superscript 4 space equals left parenthesis l straight r right parenthesis to the power of 4 space plus space left parenthesis l straight r squared right parenthesis to the power of 4 space plus space left parenthesis l straight r cubed right parenthesis to the power of 4
space equals space straight l to the power of 4 space straight x space straight r to the power of 4 left parenthesis 1 plus 2 straight r to the power of 4 plus straight r to the power of 8 right parenthesis
equals space straight l to the power of 4 space straight x space straight r to the power of 4 space left parenthesis straight r to the power of 4 space plus 1 right parenthesis squared
equals space straight l to the power of 4 space straight x space straight n over straight l open parentheses fraction numerator n italic plus italic 1 over denominator l end fraction close parentheses
equals space l n italic space x italic 4 m to the power of italic 2 italic space italic equals italic space italic 4 l m to the power of italic 2 n

238 Views

3.

The sum of coefficients of integral powers of x in the binomial expansion of left parenthesis 1 minus 2 square root of straight x right parenthesis to the power of 50 space is

  • 1 half left parenthesis 3 to the power of 50 space plus space 1 right parenthesis
  • 1 half left parenthesis 30 to the power of 50 right parenthesis
  • 1 half left parenthesis 30 to the power of 50 minus 1 right parenthesis
  • 1 half space left parenthesis 2 to the power of 50 plus 1 right parenthesis

A.

1 half left parenthesis 3 to the power of 50 space plus space 1 right parenthesis

Let Tr+1 be the general term in the expansion of left parenthesis 1 minus 2 square root of straight x right parenthesis to the power of 60

therefore space straight T subscript straight r plus 1 end subscript space equals space straight C presuperscript 50 subscript straight r space left parenthesis 1 right parenthesis to the power of 50 minus straight r end exponent left parenthesis negative 2 straight x to the power of 1 divided by 2 end exponent right parenthesis to the power of straight r space equals space straight C presuperscript 50 subscript straight r 2 to the power of straight r straight x to the power of straight r divided by 2 end exponent left parenthesis negative 1 right parenthesis to the power of straight r
For the integral power of x, r should be even integer,
therefore, sum of coefficients= 

sum from straight r space equals 0 to 25 of space to the power of 50 straight C subscript 2 straight r end subscript space left parenthesis 2 right parenthesis to the power of 2 straight r end exponent
space equals space 1 half left parenthesis 1 plus 2 right parenthesis to the power of 50 space plus space left parenthesis 1 minus 2 right parenthesis to the power of 50 right square bracket
space equals space 1 half left square bracket 3 to the power of 50 plus 1 right square bracket

360 Views

4.

Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A x B, each having at least three elements is:

  • 219

  • 256

  • 275

  • 510


A.

219

Given,
n(A) = 4 n (B) =2
⇒ n(A x B) = 8
Total number of subsets of set (A x B)= 28
Number of subsets of set A x B having no element (i.e, Φ) = 1
Number of subsets of set AxB having two elements = 8C1
Number of subsets of set A x B having two elements = 8C2
therefore, the number of subsets having atleast three elements,
 = 28 x (1+8C18C2)
 = 28 -1-8-28
= 28 -37
= 256-37 = 219

541 Views

5.

If the angles of elevation of the top of a tower from three collinear points A, B and C on line leading to the foot of the tower are 30o, 45o and 60o respectively, then the ratio AB: BC is 

  • square root of 3 space colon space 1
  • square root of 3 space colon space square root of 2
  • 1 colon square root of 3
  • 2:3


A.

square root of 3 space colon space 1
251 Views

6.

The sum of first 9 terms of the series 1 cubed over 1 space plus fraction numerator 1 cubed plus 2 cubed over denominator 1 plus 3 end fraction space plus space fraction numerator 1 cubed plus 2 cubed plus 3 cubed over denominator 1 plus 2 plus 3 end fraction space.... is

  • 71

  • 96

  • 142

  • 192


B.

96

194 Views

7.
Let α and  β be the roots of equations x2-6x-2 = 0. If ann- βn, for n≥1, the value of a10-2a8/2a9 is equal to 
  • 6

  • -6

  • 3

  • -3


C.

3

α and β are the roots of the equation
x2-6x-2 =0
or
α=6x+2 
α2 = 6α +2
α10= 6 α9+2α8 ... (i)
β10= 6 β9+2β8 ... (ii)
On subtracting eq (ii) from eq(i), we get
α10- β10= 6 ( α99) + 2 (α88)
a10 = 6a9 + 2a8 (∴ an = αn- βn)
⇒ a10 -2a8 = 6a9   
⇒ a10-2a8/2a9  = 3

856 Views

8.

The number of integers greater than 6000 that can be formed, using the digits 3,5,6,7 and 8 without repetition, is

  • 216

  • 192

  • 120

  • 72


B.

192

358 Views

9.

A complex number z is said to be unimodular, if |z|= 1. suppose z1 and z2 are complex numbers such that fraction numerator straight z subscript 1 minus 2 straight z subscript 2 over denominator 2 minus straight z subscript 1 begin display style stack straight z subscript 2 with minus on top end style end fraction is unimodular and z2 is not unimodular. Then, the point z1 lies on a

  • straight line parallel to X -axis

  • straight line parallel to Y -axis

  • circle of radius 2

  • circle of radius square root of 2


C.

circle of radius 2

If z unimodular, then |z| = 1, also, use property of modulus i.e. straight z top enclose straight z space equals space vertical line straight z vertical line squared
Given, z2 is not unimodular i.e |z2|≠1 and fraction numerator straight z subscript 1 minus 2 straight z subscript 2 over denominator 2 minus straight z subscript 1 begin display style stack straight z subscript 2 with minus on top end style end fraction is unimodularfraction numerator straight z subscript 1 minus 2 straight z subscript 2 over denominator 2 minus straight z subscript 1 begin display style stack straight z subscript 2 with minus on top end style end fraction space equals space 1
rightwards double arrow vertical line z subscript 1 minus 2 z subscript 2 vertical line squared space equals space vertical line 2 minus z subscript 1 top enclose z subscript 2 end enclose vertical line squared
rightwards double arrow left parenthesis z subscript 1 minus 2 z subscript 2 right parenthesis left parenthesis stack z subscript 1 with bar on top space minus 2 stack z subscript 2 with bar on top right parenthesis space equals space left parenthesis 2 minus z subscript 1 stack z subscript 2 with bar on top right parenthesis left parenthesis 2 minus top enclose z subscript 1 z subscript 2 right parenthesis
left parenthesis because space z top enclose z space equals space vertical line z squared vertical line right parenthesis
rightwards double arrow space vertical line z subscript 2 vertical line squared space plus space 4 vertical line z subscript 2 vertical line squared minus space 2 top enclose z subscript 1 end enclose z subscript 2 space minus space 2 z subscript 1 top enclose z subscript 2 end enclose
rightwards double arrow space left parenthesis vertical line z subscript 1 vertical line squared minus 1 right parenthesis left parenthesis vertical line z subscript 1 vertical line minus 4 right parenthesis space equals space 0
vertical line z subscript 2 vertical line space not equal to 1
vertical line z subscript 1 vertical line space equals space 2
z subscript 1 space equals space x plus i y
x squared plus y squared space equals space left parenthesis 2 right parenthesis squared
therefore space point space straight z subscript 1 space lies space on space straight a space circle space of space radius space 2.

264 Views

10.

The negation of tilde space straight s space logical or left parenthesis tilde straight r logical and straight s right parenthesis is equivalent to

  • straight s logical and tilde straight r
  • straight s logical and left parenthesis straight r logical and tilde straight s right parenthesis
  • straight s logical or left parenthesis straight r space straight v tilde straight s right parenthesis
  • straight s logical and straight r

D.

straight s logical and straight r
303 Views

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