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Answers
Answer:
1/(x-5)(x-7)
= A/(x-5) + B/(x-7)
= [A(x-7) + B(x-5)]/(x-5)(x-7)
= [Ax - 7A + Bx - 5B]/(x-5)(x-7)
= [x(A+B) - (7A+5B)]/(x-5)(x-7)
Equate the numerator on both sides to get
1 = x(A+B) - (7A+5B). Form we get
A + B = 0 or
7A + 7B = 0 …(1)
7A + 5B = -1 …(2)
Subtract (2) from (1),
7B - 5B = 1, or
2B = -1, or B = -1/2 and so A = 1/2.
So the partial fractions are
= 1/2(x-5) - 1/2(x-7). Answer
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Answer:
Pair of lines:
x2+2xy+3x+6y=0
⇒x(x+2y)+3(x+2y)=0
⇒(x+2y)(x+3)=0
⇒x+2y=0
and x+3=0
Are the two normals and their point of intersection must be the centre of the circle.
⇒x=−3
and y=2−x=23
⇒(−3,23) is the centre of required circle.
(∵x(x−4)+y(y−3)=0 is the diametric form of the circle. Hence, (0,0) and (4,3) are the diametric end points.)
⇒ Centre →(20+4,20+3)→(2,23)
radius =21(16+9)
=25
The required circle with centre (−3,23) is just sufficient to contain the circle
x(x−4)+y(y−3)=0
∴ radius of required circle