﻿ The graph of the function y = f(x) is symmetrical about the line x = 2, then from Mathematics Class 12 JEE Year 2004 Free Solved Previous Year Papers

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# JEE Mathematics 2004 Exam Questions

#### Multiple Choice Questions

21.

If  then the values of a and b, are

• a = 1 and b = 2

D.

a = 1 and b = 2

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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)

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

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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).

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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

1 + 3c = -8 3c = -9
⇒ c = -3

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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

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

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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

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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

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

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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

D.

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

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30.

The domain of the function

• [2, 3]

• [2, 3)

• [1, 2]

• [1, 2)

B.

[2, 3)

For the the function
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)

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