Consider the fraction $\frac{x^{3} - a x^{2} + 19 x - a - 4}{x^{3} - (a + 1) x^{2} + 23 x - a - 7}$

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The lowest admitted reduction form of the fraction is $\frac{x - 4}{x - 5}$

The value of $' a'$ at which the above fraction admits of reduction is 4

The value of $' a'$ at which the above fraction admits of reduction is 8

The lowest admitted reduction form of the fraction is $\frac{x - 3}{x - 4}$

answer is A, C.

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Detailed Solution

Subtracting numerator from denominator, we get
$x^{2} - 4 x + 3$ i.e $(x - 1) (x - 3)$
Thus it is concluded that numerator and denominator must be completely divisible by $(x - 1)$ or $(x - 3)$ in other words both must vanish when $x = 1$ or when $x = 3,$ if $x = 3$ we get, $a = 8$ And fraction becomes $\frac{x^{3} - 8 x^{2} + 19 x - 12}{x^{3} - 9 x^{2} + 23 x - 15} = \frac{x^{2} - 7 x + 12}{x^{2} - 8 x + 15} = \frac{x - 4}{x - 5}$
If we put $x = 1,$ we get also that $a = 8 .$