What should be added to 1/(x-2)(x-4) to get {2x-5}{(x²-5x+6)(x-4)}?

Correct Answer is : 1(x2−7x+12){1}{(x²-7x+12)}(x2−7x+12)1

Correct Answer is : 1(x2−7x+12){1}{(x²-7x+12)}(x2−7x+12)1

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UPSC CDS 1 2021 Math Paper

Question : 82 of 100

Marks: +1, -0

\frac{1}{(x-2)(x-4)}

\frac{2x-5}{(x^{2}-5x+6)(x-4)}

Solution:

Concept:

Adding a term to one rational expression equals another rational expression; find the term by subtracting.

Explanation:

Let the required term be

y

Then

\frac{1}{(x-2)(x-4)} + y = \frac{2x-5}{(x^{2}-5x+6)(x-4)}

Factor

x^{2}-5x+6 = (x-2)(x-3)

So the right side becomes

\frac{2x-5}{(x-2)(x-3)(x-4)}

Subtract:

y = \frac{2x-5}{(x-2)(x-3)(x-4)} - \frac{1}{(x-2)(x-4)}

Common denominator:

(x-2)(x-3)(x-4)

y = \frac{(2x-5) - (x-3)}{(x-2)(x-3)(x-4)}

Simplify numerator:

2x-5 - x + 3 = x-2

y = \frac{x-2}{(x-2)(x-3)(x-4)} = \frac{1}{(x-3)(x-4)}

Multiply:

(x-3)(x-4) = x^{2} -7x +12

Therefore,

y = \frac{1}{x^{2}-7x+12}

Answer:

\frac{1}{x^{2}-7x+12}

(Option A).

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