CAT Exam Model Paper 4 with solutions for free online practice

(x−7)(

x2−29x+154

x2−12x+32

)=1
⇒ (x−7)

(x−7)(x−22)

(x−8)(x−4)

=1
Let P(x) = (x - 7) and Q(x) =

(x−7)(x−22)

(x−8)(x−4)

Now, three cases arise:
Case-1P(x) = 1 and Q(x) may be anything.
∴(x-7) = 1
or x = 8
But, for x = 8, Q(x) is not defined.
Case-2
P(x) = - 1 and Q(x) is an even exponent.
(x - 7) = - 1 or x = 6
For x = 6, Q(x) = - 4, which satisfies the given equation.
Case -3Q(x) = 0 and P(x) ≠ 0
∴

(x−7)(x−22)

(x−8)(x−4)

=0
⇒ x = 7 or x = 22
But, for x = 7, P(x) = 0, for which the given equation is not defined.
So the given equation is satisfied for two values of ‘x' (6 and 22).