If l i m x → − ∞ ( √ x 2 − x + 1 − a x − b ) = 0 then the values of a and b are given by - Options: A . &nbsp a = − 1 , b = 1 2 B . &nbsp a = 1 , b = 1 2 C . &nbsp a = 1 , b = − 1 2 D . &nbsp a=-1, b=-1/2

Answer: Option A : A We have, l i m x → − ∞ ( √ x 2 − x + 1 − a x − b ) = 0 [ putting x=-y; x → − ∞ , y → ∞ ] ⇒ l i m y → ∞ ( √ y 2 + y + 1 + a y − b ) = 0 ⇒ l i m y → ∞ [ y ( 1 + 1 y + 1 y 2 ) 1 / 2 + a y − b ] = 0 ⇒ l i m y → ∞ [ y { 1 + 1 2 ( 1 y + 1 y 2 + . . . . . . . . . ) } + a y − b ] = 0 ⇒ l i m y → ∞ [ y ( 1 + a ) + ( 1 2 − b ) + 1 2 y + . . . . . . . . ] = 0 ⇒ 1 + a = 0 a n d ( 1 / 2 ) − b = 0 ⇒ a = − 1 a n d b = 1 / 2

Iflimx→−∞(√x2−x+1−ax−b)=0

Discussion Forum : Limits Continuity And Differentiability

Question -

$If l i m x \to - \infty (\sqrt{x^{2} - x + 1} - a x - b) = 0$ then the values of a and b are given by -

Options:

A .

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

B .

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

C .

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

D . a=-1, b=-1/2

Answer: Option A
:
A

$We have, l i m x \to - \infty (\sqrt{x^{2} - x + 1} - a x - b) = 0 [putting x=-y; x \to - \infty, y \to \infty] \Rightarrow l i m y \to \infty (\sqrt{y^{2} + y + 1} + a y - b) = 0 \Rightarrow l i m y \to \infty [y {(1 + \frac{1}{y} + \frac{1}{y^{2}})}^{1 / 2} + a y - b] = 0 \Rightarrow l i m y \to \infty [y {1 + \frac{1}{2} (\frac{1}{y} + \frac{1}{y^{2}} + . . . . . . . . .)} + a y - b] = 0 \Rightarrow l i m y \to \infty [y (1 + a) + (\frac{1}{2} - b) + \frac{1}{2 y} + . . . . . . . .] = 0 \Rightarrow 1 + a = 0 a n d (1 / 2) - b = 0 \Rightarrow a = - 1 a n d b = 1 / 2$

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