JEE mathematics

Sponsor Area

Question
CBSEENMA11015451

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

Solution

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

Sponsor Area

Question
CBSEENMA11015452

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

Solution

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.

Question
CBSEENMA11015453

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

Solution

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

Question
CBSEENMA11015454

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

Solution

B.

192

Question
CBSEENMA11015455

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

Solution

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