JEE mathematics

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Question
CBSEENMA11015483

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is

  • 256

  • 220

  • 219

  • 211

Solution

C.

219

Given, n(A) =2, n(B) = B
The number of subsets of AXB having 3 or more elements,
=straight C presuperscript 8 subscript 3 space plus to the power of 8 straight C subscript 4 space plus space..... plus to the power of 8 straight C subscript 8
space equals space 2 to the power of 8 space minus to the power of 8 straight C subscript 0 space minus to the power of 8 straight C subscript 1 space minus space to the power of 8 straight C subscript 2
space equals space 256 minus 1 minus 8 minus 28 space equals space 219

Sponsor Area

Question
CBSEENMA11015484

The real number k for which the equation, 2x3 +3x +k = 0 has two distinct real roots in [0,1]

  • lies between 1 and 2

  • lies between 2 and 3

  • lies between -1 and 0

  • does not exist

Solution

D.

does not exist

Let f(x) = 2x3+3x+k
On differentiating w.r.t x, we get
f'(x) = 6x2 + 3> 0, ∀ x ε R
⇒ f(x) is strictly increasing function
⇒ f(x) = 0 has only one real root, so two roots are not possible.

Question
CBSEENMA11015485

The sum of first 20  terms of the sequence 0.7,0.77,0.777...... is

  • 7 over 81 space left parenthesis 179 minus 10 to the power of negative 20 end exponent right parenthesis
  • 7 over 9 left parenthesis 99 minus 10 to the power of negative 20 end exponent right parenthesis
  • 7 over 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis
  • 7 over 9 left parenthesis 99 plus 10 to the power of negative 20 end exponent right parenthesis

Solution

C.

7 over 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis

Let S = 0.7 + 0.77 +0.777 + .... upto 20 terms
7 over 10 plus 77 over 10 squared plus 777 over 10 cubed plus..... space upto space 20 space terms
space equals space open square brackets 1 over 10 space plus 11 over 10 squared plus 111 over 10 cubed plus........ plus space upto space 20 space terms close square brackets
equals space 7 over 9 open square brackets 9 over 10 plus 99 over 100 plus 999 over 1000 plus..... plus space upto space 20 space terms close square brackets
space equals 7 over 9 open square brackets open parentheses 1 minus 1 over 10 close parentheses plus open parentheses 1 minus 1 over 10 squared close parentheses plus open parentheses 1 minus 1 over 10 cubed close parentheses plus space..... upto space 20 space terms close square brackets
7 over 9 open square brackets left parenthesis 1 plus 1 plus..... upto space 20 space terms right parenthesis minus open parentheses 1 over 10 plus fraction numerator begin display style 1 end style over denominator begin display style 10 squared end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style 10 cubed end style end fraction plus.... upto space 20 space terms close parentheses close square brackets
space equals space 7 over 9 space open square brackets 20 minus fraction numerator begin display style 1 over 10 open curly brackets 1 minus open parentheses 1 over 10 close parentheses to the power of 20 close curly brackets end style over denominator 1 minus begin display style 1 over 10 end style end fraction close square brackets
space equals space 7 over 9 open square brackets 20 minus 1 over 9 open curly brackets 1 minus open parentheses 1 over 10 close parentheses to the power of 20 close curly brackets close square brackets space equals space 7 over 9 open square brackets 179 over 9 plus 1 over 9 open parentheses 1 over 10 close parentheses to the power of 20 close square brackets
equals space 7 over 181 space open square brackets 179 plus left parenthesis 10 right parenthesis to the power of negative 20 end exponent close square brackets

Question
CBSEENMA11015486

A ray of light along straight x space plus square root of 3 straight y end root space equals space square root of 3 get reflected upon reaching X -axis, the equation of the reflected ray is 

  • straight y equals space straight x plus square root of 3
  • square root of 3 straight y end root space equals space straight x minus square root of 3
  • straight y space equals space square root of 3 straight x end root minus square root of 3
  • square root of 3 straight y end root space equals space straight x minus 1

Solution

B.

square root of 3 straight y end root space equals space straight x minus square root of 3

Given equation of line
straight x plus square root of 3 straight y space equals square root of 3 space end root space space space space space.... space left parenthesis straight i right parenthesis
straight y equals space 1 minus fraction numerator straight x over denominator square root of 3 end fraction
Slope of incident ray is space minus fraction numerator 1 over denominator square root of 3 end fraction 
So, slope of reflected ray must be fraction numerator 1 over denominator square root of 3 end fraction and the point of incident left parenthesis square root of 3 comma 0 right parenthesis
So equation of reflected ray
straight y minus 0 space equals space fraction numerator 1 over denominator square root of 3 end fraction space left parenthesis straight x minus square root of 3 right parenthesis
rightwards double arrow space square root of 3 straight y end root space equals space straight x minus square root of 3

Question
CBSEENMA11015487

The number of values of k, for which the system of equations
(k+1) x + 8y = 4k
kx + (k+3)y = 3k -1
has no solution, is 

  • infinite 

  • 1

  • 2

  • 3

Solution

B.

1

Condition for the system of equations has no solution,
straight a subscript 1 over straight a subscript 2 space equals straight b subscript 1 over straight b subscript 2 space not equal to straight c subscript 1 over straight c subscript 2
therefore space fraction numerator straight k plus 1 over denominator straight k end fraction space equals space fraction numerator 8 over denominator straight k plus 3 end fraction space not equal to fraction numerator 4 straight k over denominator 3 straight k minus 1 end fraction
Take space fraction numerator straight k plus 1 over denominator straight k end fraction space equals space fraction numerator 8 over denominator straight k plus 3 end fraction
rightwards double arrow space straight k squared space plus space 4 straight k space plus 3 space equals space 8 straight k
rightwards double arrow straight k squared minus 4 straight k space plus 3 space equals space 0
rightwards double arrow left parenthesis straight k minus 1 right parenthesis left parenthesis straight k minus 3 right parenthesis space space equals 0
straight k space equals space 1 comma 3
If space straight k space equals 1 comma space then space fraction numerator 8 over denominator 1 plus 3 end fraction space equals space fraction numerator 4.1 over denominator 2 end fraction comma space false
Therefore, k = 3
Hence, only one value of k exists.