The values of constants a and b so that l i m x → ∞ ( x 2 + 1 x + 1 − a x − b ) = 1 2 , a r e Options: A . &nbsp a = 1 , b = − 3 2 B . &nbsp a = − 1 , b = 3 2 C . &nbsp a=0, b=0 D . &nbsp a =2, b= -1

Answer: Option A : A l i m x → ∞ ( x 2 + 1 x + 1 − a x − b ) = 1 2 ⇒ l i m x → ∞ ( x 2 + 1 ) − ( a x + b ) ( x + 1 ) x + 1 = 1 2 ⇒ l i m x → ∞ x 2 ( 1 − a ) − ( a + b ) x − b + 1 x + 1 = 1 2 ⇒ 1 − a = 0 a n d a + b = − 1 2 ∴ a = 1 b = − 3 2

The values of constants a and b so thatlimx→∞(x2+1x+1−ax&

Discussion Forum : Limits Continuity And Differentiability

Question -

$The values of constants a and b so that l i m x \to \infty (\frac{x^{2} + 1}{x + 1} - a x - b) = \frac{1}{2}, a r e$

Options:

A .

a = 1, b = - \frac{3}{2}

B .

a = - 1, b = \frac{3}{2}

C . a=0, b=0

D . a =2, b= -1

Answer: Option A
:
A

$l i m x \to \infty (\frac{x^{2} + 1}{x + 1} - a x - b) = \frac{1}{2} \Rightarrow l i m x \to \infty \frac{(x^{2} + 1) - (a x + b) (x + 1)}{x + 1} = \frac{1}{2} \Rightarrow l i m x \to \infty \frac{x^{2} (1 - a) - (a + b) x - b + 1}{x + 1} = \frac{1}{2} \Rightarrow 1 - a = 0 a n d a + b = - \frac{1}{2} ∴ a = 1 b = - \frac{3}{2}$

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Discussion Forum : Limits Continuity And Differentiability

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