The number of values of k, for which the system of equations(k+1

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

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

  • 219

286 Views

2.

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

  • lies between -1 and 0

183 Views

3.

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 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis
958 Views

4.

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
  • straight y space equals space square root of 3 straight x end root minus square root of 3
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5.

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

  • 2


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.

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6.

If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a, b, c ∈ R, have a common root, then a : b : c is

  • 1:2:3

  • 3:2:1

  • 1:3:2

  • 1:3:2

177 Views

7.

The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point

  • (-5,2)

  • (2,-5)

  • (5,-2)

  • (5,-2)

181 Views

8.

If x, y, z are in A.P. and tan−1 x, tan−1 y and tan−1 z are also in A.P., then

  • x= y= z

  • 2x =3y = 6z

  • 6x = 3y= 2z

  • 6x = 3y= 2z

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9.

Consider :
Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.

  • Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I 

  • Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I

  • Statement -I is True; Statement -II is False.

  • Statement -I is True; Statement -II is False.

289 Views

10. limit as straight x space rightwards arrow 0 of fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis left parenthesis 3 plus cos space straight x right parenthesis over denominator straight x space tan space 4 straight x end fraction space is space equal space to space
  • -1/4

  • 1/2

  • 1

  • 1

188 Views

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