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If b > a, then the equation (x – a) (x – b) –1 = 0 has (2000S)
  • a)
    both roots in (a, b)
  • b)
    both roots in (–∞, a)
  • c)
    both roots in (b, +∞)
  • d)
    one root in (–∞, a) and the other in (b, +∞)
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If b > a, then the equation (x – a) (x – b) –1 = ...
The given equation is ( x – a) (x – b) – 1 = 0, b > a.
or x2 – (a + b) x + ab –1= 0
Let  f(x) = x2 – (a + b) x + ab – 1
Since coeff. of x2 i.e. 1 > 0,
∴ it represents upward parabola, intersecting x - axis at two points. (corresponding to two real roots, D being +ve).
Also f (a) = f (b) = – 1 ⇒ curve is below x-axis at a and b
⇒ a and b both lie between the roots.
Thus the graph of given eqn is as shown.
from graph it is clear that one root of the equation lies in (– ∞, a) and other in (b, ∞).
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If b > a, then the equation (x – a) (x – b) –1 = 0 has (2000S)a)both roots in (a, b)b)both roots in (–∞, a)c)both roots in (b, +∞)d)one root in (–∞, a) and the other in (b, +∞)Correct answer is option 'D'. Can you explain this answer?
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