Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

The graph of the function y = f(x) is symmetrical about the line x = 2, then

  • f(x + 2)= f(x – 2)

  • f(2 + x) = f(2 – x)

  • f(x) = f(-x)

  • f(x) = f(-x)

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22.

The domain of the function straight f left parenthesis straight x right parenthesis space equals space fraction numerator sin to the power of negative 1 end exponent space left parenthesis straight x minus 3 right parenthesis over denominator square root of 9 minus straight x squared end root end fraction

  • [2, 3]

  • [2, 3)

  • [1, 2]

  • [1, 2]

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23.

If limit as straight x space rightwards arrow infinity of space open parentheses 1 plus straight a over straight x plus straight b over straight x squared close parentheses to the power of 2 straight x end exponent space equals space straight e squared then the values of a and b, are

  • straight a element of space straight R with equals below space straight b element of space straight R with equals below
  • straight a space equals 1 comma space straight b element of space straight R with equals below
  • straight a element of space straight R with equals below space comma space straight b space equals 2
  • straight a element of space straight R with equals below space comma space straight b space equals 2
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24.

If 2a + 3b + 6c =0, then at least one root of the equation ax+ bx+ c = 0  lies in the interval

  • (0,1)

  • (1,2)

  • (2,3)

  • (2,3)

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25.

If the sum of the slopes of the lines given by x2 -2cxy -7y2 =0 is four times their product, then c has the value

  • -1

  • 2

  • -2

  • -2

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26.

The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is

  • straight x over 2 space plus straight y over 3 space equals space minus space 1 space and space fraction numerator straight x over denominator negative 2 end fraction space plus straight y over 1 space equals space minus space 1
  • straight x over 2 minus straight y over 3 space equals space minus space 1 space and space fraction numerator straight x over denominator negative 2 end fraction space plus straight y over 1 space equals space minus space 1
  • straight x over 2 space plus straight y over 3 space equals space 1 space and space straight x over 2 space plus straight y over 1 space equals space 1
  • straight x over 2 space plus straight y over 3 space equals space 1 space and space straight x over 2 space plus straight y over 1 space equals space 1
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27.

If one of the lines given by 6x2 -xy +4cy2 = 0 is 3x + 4y = 0, then c equals

  • 1

  • -1

  • 3

  • 3

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28.

If a circle passes through the point (a, b) and cuts the circle x2 +y2= 4 orthogonally, then the locus of its centre is

  • 2ax +2by + (a2 +b2+4)=0

  • 2ax +2by - (a2 +b2+4)=0

  • 2ax -2by - (a2 +b2+4)=0

  • 2ax -2by - (a2 +b2+4)=0

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29.

A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is

  • (x-p)2 = 4qy

  • (x-q)2 = 4py

  • (y-p)2 = 4qx

  • (y-p)2 = 4qx

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30.

If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is

  • x2 + y2- 2x +2y -23 = 0

  • x2 - y2- 2x -2y -23 = 0

  • x2 - y2- 2x -2y +23 = 0

  • x2 - y2- 2x -2y +23 = 0


A.

x2 + y2- 2x +2y -23 = 0

The lines 2x + 3y + 1 = 0 and 3x - y - 4 = 0 are diameters of circle.
On solving these equations we get x = 1, y = -1
Therefore the centre of circle = (1, -1) and circumference = 10 π
2πr = 10π
⇒ r = 5

∴ Equation of circle (x - x1 )2 + (y - y1 )2 = r2
(x - 1)2 + (y + 1)2 = 52
x2 + 1 - 2x + y2 + 2y + 1 = 25
x2 + y2 - 2x + 2y - 23 = 0

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