If A+B = {x²-8}{x+2} and A-B = {-x²+2x+4}{x+2}, then what is B equal to ?

Correct Answer is : 2x2−7x+32x−1{2x²-7x+3}{2x-1}2x−12x2−7x+3

Correct Answer is : 2x2−7x+32x−1{2x²-7x+3}{2x-1}2x−12x2−7x+3

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UPSC CDS 2 2021 Math Paper

Question : 67 of 100

Marks: +1, -0

A+B = \frac{x^{2}-8}{x+2}

A-B = \frac{-x^{2}+2x+4}{x+2}

B

equal to ?

Solution: 👈: Video Solution

Given,

A+B = \frac{x^{2}-8}{x+2}

\dots

(i)

and

A-B = \frac{-x^{2}+2x+4}{x+2}

. . . (ii)

By Eq. (i) - Eq. (ii), we get

(A+B)-A-B)

= \frac{x^{2}-8}{x+2} - \frac{(-x^{2}+2x+4)}{x+2}

\Rightarrow 2B = \frac{x^{2}-8-(-x^{2}+2x+4)}{x+2}

\Rightarrow 2B = \frac{x^{2}-8+x^{2}-2x-4}{x+2}

\Rightarrow 2B = \frac{2x^{2}-2x-12}{x+2}

\Rightarrow B = \frac{x^{2}-x-6}{x+2}

= \frac{x^{2}-3x+2x-6}{x+2}

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

\text{Option (c) is } \frac{2x^{2}-7x+3}{2x-1}

On solving this further, we get

= \frac{2x^{2}-6x-x+3}{2x-1}

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

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

Since, value of

B

and option (c) represents the same value.

Hence,

B

is equal to

\frac{2x^{2}-7x+3}{2x-1}

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