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UPSC CDS 1 2021 Math Paper

Question : 89 of 100

Marks: +1, -0

\frac{3x-6}{(x-6)(x+k)}=\frac{2}{x-6}+\frac{1}{x+k}

then what is the value of k?

Solution:

Concept:

Compare numerators after combining the fractions on the right side.

Explanation:

Start with the given equation:

\frac{3x-6}{(x-6)(x+k)} = \frac{2}{x-6} + \frac{1}{x+k}

Combine the right side over a common denominator

(x-6)(x+k)

\frac{2(x+k) + (x-6)}{(x-6)(x+k)}

Simplify the numerator:

2x + 2k + x - 6 = 3x + 2k - 6

Now the equation is

\frac{3x-6}{(x-6)(x+k)} = \frac{3x + 2k - 6}{(x-6)(x+k)}

Since the denominators are equal, equate the numerators:

3x - 6 = 3x + 2k - 6

Cancel

3x

from both sides:

-6 = 2k - 6

Add

6

to both sides:

0 = 2k

Thus,

k = 0

Answer:

k = 0

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