MCQ
If $\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} \right] = 2$, then
- A$a = 1$ and $b = 1$
- B$a = 1$ and $b = - 1$
- ✓$a = 1$ and $b = - 2$
- D$a = 1$ and $b = 2$
✓
Answer
Correct option: C.
$a = 1$ and $b = - 2$
c
(c) $\mathop {{\rm{lim}}}\limits_{x \to \infty } \,\left( {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} \right) = 2$
==> $\mathop {{\rm{lim}}}\limits_{x \to \infty } \,\left( {\frac{{{x^3}(1 - a) - b{x^2} - ax + (1 - b)}}{{{x^2} + 1}}} \right) = 2$
==> $\mathop {\lim }\limits_{x \to \infty } \,[{x^3}(1 - a) - b{x^2} - ax + (1 - b)] = 2\,({x^2} + 1)$.
Comparing the coefficients of both sides, $1 - a = 0$ and $ - b = 2$ or $a = 1,\,b = - 2$.
(c) $\mathop {{\rm{lim}}}\limits_{x \to \infty } \,\left( {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} \right) = 2$
==> $\mathop {{\rm{lim}}}\limits_{x \to \infty } \,\left( {\frac{{{x^3}(1 - a) - b{x^2} - ax + (1 - b)}}{{{x^2} + 1}}} \right) = 2$
==> $\mathop {\lim }\limits_{x \to \infty } \,[{x^3}(1 - a) - b{x^2} - ax + (1 - b)] = 2\,({x^2} + 1)$.
Comparing the coefficients of both sides, $1 - a = 0$ and $ - b = 2$ or $a = 1,\,b = - 2$.
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