If $lim x \to \infty \frac{a (2 x^{3} - x^{2}) + b (x^{3} + 5 x^{2} - 1) - c (3 x^{3} + x^{2})}{a (5 x^{4} - x) - b x^{4} + c (4 x^{4} + 1) + 2 x^{2} + 5 x} = 1$ then which of the following relations between $a$ , $b$ and $c$ must hold good?

a + 2 b + c = 1

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5 a - b + 4 c = 0

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2 a + b - 3 c = 0

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a - 5 b + c + 2 = 0

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Solution

The correct options are
A $2 a + b - 3 c = 0$
B $a - 5 b + c + 2 = 0$
D $5 a - b + 4 c = 0$
${lim}_{x \to \infty} \frac{a (2 x^{3} - x^{2}) + b (x^{3} + 5 x^{2} - 1) - c (3 x^{3} + x^{2})}{a (5 x^{4} - x) - b x^{4} + c (4 x^{4} + 1) + 2 x^{2} + 5 x} = 1 \Rightarrow {lim}_{x \to \infty} \frac{x^{3} (2 a + b - 3 c) + x^{2} (- a + 5 b - c) - b)}{x^{4} (5 a - b + 4 c) + 2 x^{2} + x (- a + c + 5) + c} = 1$
When coefficient of $x^{4}$ is $0$ , when coefficient of $x^{3}$ is $0$ and When coefficient of $x^{2}$ in numerator and denominator is equal i.e
$5 a - b + 4 c = 0$
$2 a + b - 3 c = 0$
$- a + 5 b - c = 2 \Rightarrow a - 5 b + c + 2 = 0$
Hence, options 'B', 'C' and 'D' are correct.

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