CBSE
If 2a + 3b + 6c =0, then at least one root of the equation ax^{2 }+ bx+ c = 0 lies in the interval
(0,1)
(1,2)
(2,3)
(1,3)
A.
(0,1)
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
x^{2} + y^{2}- 2x +2y -23 = 0
x^{2} - y^{2}- 2x -2y -23 = 0
x^{2} - y^{2}- 2x -2y +23 = 0
x^{2} + y^{2}+ 2x +2y -23 = 0
A.
x^{2} + y^{2}- 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 - x_{1} )^{2} + (y - y_{1} )^{2} = r_{2}
(x - 1)^{2} + (y + 1)^{2} = 52
x^{2} + 1 - 2x + y^{2} + 2y + 1 = 25
x^{2} + y^{2} - 2x + 2y - 23 = 0
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).
If one of the lines given by 6x^{2} -xy +4cy^{2} = 0 is 3x + 4y = 0, then c equals
1
-1
3
-3
D.
-3
The pair of lines is 6x^{2}-xy +4cy^{2} =0
On comparing with ax^{2} +2hxy by^{2} = 0
we get a = 6, 2h =-1, b= c
therefore
m_{1} +m_{2} = -2h/b = 1/4c and m_{1}m_{2} = a/b = 6/4c
On line of given pair of lines is
3x+4y = 0
slope of line = -3/4 = m1
-3/4 +m_{2} = 1/4c
m_{2} = 1/4c + 3/4
1 + 3c = -8 3c = -9
⇒ c = -3
If the sum of the slopes of the lines given by x^{2} -2cxy -7y^{2} =0 is four times their product, then c has the value
-1
2
-2
1
B.
2
The given pair of line is x^{2} - 2cxy - 7y^{2} = 0
On comparing with ax^{2} + 2hxy + by^{2} = 0,
we get, a = 1, 2h = -2c, b = -7
Given that, m_{1}+m_{2} = 4m_{1}m_{2}
-2c/7 = - 4/7, c =4/2 =2
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 (x_{1} , y_{1} ).
Equation of circle in diameter form is (x - p)(x - x_{1} ) + (y - q)(y - y_{1} ) = 0
x^{2} - (p + x_{1} )x + px_{1} + y^{2} - (y_{1} + q)y + qy_{1} = 0
x^{2} - (p + x_{1} )x + y^{2} - (y_{1} + q)y + px_{1} + qy_{1} = 0
Since this circle touches X-axis
∴ y = 0
⇒ x^{2} - (p + x1 )x + px_{1} + qy_{1} = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x_{1} )^{2} = 4(px_{1} + qy_{1}) p^{2} + x^{2}_{1} + 2px_{1}
= 4px_{1} + 4qy_{1} x^{2}_{1} - 2px_{1} + p^{2} = 4qy1
Therefore the locus of point B is, (x - p)^{2} = 4qy
If a circle passes through the point (a, b) and cuts the circle x^{2} +y^{2}= 4 orthogonally, then the locus of its centre is
2ax +2by + (a^{2} +b^{2}+4)=0
2ax +2by - (a^{2} +b^{2}+4)=0
2ax -2by - (a^{2} +b^{2}+4)=0
2ax -2by + (a^{2} +b^{2}+4)=0
B.
2ax +2by - (a^{2} +b^{2}+4)=0
Let the equation of circle is
x^{2} + y^{2} + 2gx + 2fy + c = 0
It cut the circle x^{2} + y^{2} = 4 orthogonally
if 2g.0 + 2f.0 = c - 4
⇒ c = 4
∴ Equation of circle is
x^{2} + y^{2} + 2gx + 2fy + 4 = 0
Q It passes through the point (a, b)
∴ a^{2} + b^{2} + 2ag + 2f + 4 = 0
Locus of centre (-g, -f) will be a^{2}+ b^{2}- 2xa - 2yb + 4 = 0
2ax + 2by - (a^{2} + b^{2} + 4) = 0
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
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)