Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2004 Exam Questions

Multiple Choice Questions

21.

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
  • a = 1 and b = 2


D.

a = 1 and b = 2

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 limit as straight x space rightwards arrow infinity of space open parentheses 1 plus fraction numerator begin display style straight a end style over denominator begin display style straight x end style end fraction plus fraction numerator begin display style straight b end style over denominator begin display style straight x squared end style end fraction close parentheses to the power of open parentheses fraction numerator 1 over denominator begin display style straight a over straight x plus straight b over straight x squared end style end fraction close parentheses space straight x space 2 straight x space straight x space open parentheses straight a over straight x plus straight b over straight x squared close parentheses end exponent
space equals space straight e to the power of 2 straight a end exponent
rightwards double arrow space straight a space equals 1
straight b element of straight R
117 Views

22.

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)

  • (1,3)


A.

(0,1)

Let space straight f apostrophe left parenthesis straight x right parenthesis space equals space ax squared space plus bx space plus straight c
rightwards double arrow space straight f left parenthesis straight x right parenthesis space equals space ax cubed over 3 plus bx squared over 2 space plus cx space plus straight d
straight f left parenthesis straight x right parenthesis space equals space fraction numerator 2 ax cubed space plus 3 bx squared space plus 6 cx space plus 6 straight d over denominator 6 end fraction
straight f left parenthesis 1 right parenthesis space equals space fraction numerator 2 straight a space plus 3 straight b space plus 6 straight c space plus 6 straight d over denominator 6 end fraction space equals fraction numerator 6 straight d over denominator 6 end fraction space equals space straight d
straight f left parenthesis 1 right parenthesis space space equals fraction numerator 6 straight d over denominator 6 end fraction space equals straight d
straight f left parenthesis 0 right parenthesis space equals space straight f left parenthesis 1 right parenthesis
straight f apostrophe left parenthesis straight x right parenthesis space equals space 0

∴ One of the roots of ax2 + bx + c = 0 lies between 0 and 1.
203 Views

23.

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

163 Views

24.

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)


B.

f(2 + x) = f(2 – x)

If y = f (x) is symmetric about the line x = 2 then f(2 + x) = f(2 – x).

219 Views

25.

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

  • 1

  • -1

  • 3

  • -3


D.

-3

The pair of lines is 6x2-xy +4cy2 =0
On comparing with ax2 +2hxy by2 = 0
we get a = 6, 2h =-1, b= c
therefore
m1 +m2 = -2h/b = 1/4c and m1m2 = a/b = 6/4c
On line of given pair of lines is
3x+4y = 0
slope of line = -3/4 = m1
-3/4 +m2 = 1/4c
m2 = 1/4c + 3/4
left parenthesis negative 3 divided by 4 right parenthesis open parentheses fraction numerator 1 over denominator 4 straight c end fraction plus 3 divided by 4 close parentheses space equals space fraction numerator 6 over denominator 4 straight c end fraction
space equals space 3 over 4 space open parentheses fraction numerator 1 plus 3 straight c over denominator 4 straight c end fraction close parentheses space equals space fraction numerator 6 over denominator 4 straight c end fraction
space 1 plus 3 straight c space equals space fraction numerator negative 6 space straight x space 4 over denominator 4 end fraction
1 + 3c = -8 3c = -9
⇒ c = -3

243 Views

26.

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

  • 1


B.

2

The given pair of line is x2 - 2cxy - 7y2 = 0
On comparing with ax2 + 2hxy + by2 = 0,
we get, a = 1, 2h = -2c, b = -7

straight m subscript 1 space plus space straight m subscript 2 equals space minus fraction numerator 2 straight h over denominator straight b end fraction
space equals space minus fraction numerator 2 straight c over denominator 7 end fraction
and space straight m subscript 1 straight m subscript 2 space equals space straight a over straight b space equals space fraction numerator negative 1 over denominator 7 end fraction


Given that, m1+m2 = 4m1m2
-2c/7 = - 4/7, c =4/2 =2

280 Views

27.

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 = 4px


A.

(x-p)2 = 4qy

In a circle, AB is a diameter where the co-ordinate of A is (p, q) and let the co-ordinate of B  is (x1 , y1 ).
Equation of circle in diameter form is (x - p)(x - x1 ) + (y - q)(y - y1 ) = 0
x2 - (p + x1 )x + px1 + y2 - (y1 + q)y + qy1 = 0
x2 - (p + x1 )x + y2 - (y1 + q)y + px1 + qy1 = 0
Since this circle touches X-axis

∴ y = 0
⇒ x2 - (p + x1 )x + px1 + qy1 = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x1 )2 = 4(px1 + qy1) p2 + x21 + 2px1
= 4px1 + 4qy1 x21 - 2px1 + p2 = 4qy1
Therefore the locus of point B is, (x - p)2 = 4qy

480 Views

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


B.

2ax +2by - (a2 +b2+4)=0

Let the equation of circle is
x2 + y2 + 2gx + 2fy + c = 0
It cut the circle x2 + y2 = 4 orthogonally

if 2g.0 + 2f.0 = c - 4
⇒ c = 4
∴ Equation of circle is
x2 + y2 + 2gx + 2fy + 4 = 0
Q It passes through the point (a, b)
∴ a2 + b2 + 2ag + 2f + 4 = 0
Locus of centre (-g, -f) will be a2+ b2- 2xa - 2yb + 4 = 0
2ax + 2by - (a2 + b2 + 4) = 0

450 Views

29.

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

D.

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

Let a and b be the intercepts on the co-ordinate axes. a + b = -1
⇒ b = - a - 1 = - (a + 1)

Equation of line is x/a + y/b = 1



rightwards double arrow space straight x over straight a space minus fraction numerator straight y over denominator straight a plus 1 end fraction space equals 1 space.... space left parenthesis straight i right parenthesis
Since space the space line space passes space through space left parenthesis 4 comma 3 right parenthesis
therefore 4 over straight a minus fraction numerator 3 over denominator straight a plus 1 end fraction space equals space 1
rightwards double arrow space fraction numerator 4 straight a space plus space 4 minus 3 straight a over denominator straight a left parenthesis straight a plus 1 right parenthesis end fraction space equals 1
straight a plus space 4 space equals space straight a squared space plus straight a
straight a squared space equals space 4
straight a equals space plus-or-minus 2
therefore space from space left parenthesis straight i right parenthesis
straight x over 2 minus straight y over 3 space equals 1 space or space fraction numerator straight x over denominator negative 2 end fraction space plus straight y over 1 space equals 1

150 Views

30.

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)


B.

[2, 3)

For the 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 space is
we will define

(I) −1≤ (x − 3) ≤1⇒ 2 ≤ x ≤ 4 ....... (i)
(II) 9 x 0 3 x 4 2 − ≥ ⇒ − < < ....... (ii)
From relation (i) and (ii), we get 2 < x < 3
∴ Domain of the given function = [2, 3)

139 Views

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