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If $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$$, then the ordered pair (a, b) is :

JEE · Math · previous-year question

  1. A.$$\left( {1,{1 \over 2}} \right)$$
  2. B.$$\left( {1, - {1 \over 2}} \right)$$correct
  3. C.$$\left( { - 1,{1 \over 2}} \right)$$
  4. D.$$\left( { - 1, - {1 \over 2}} \right)$$

Answer

B. $$\left( {1, - {1 \over 2}} \right)$$

Explanation

$$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$$ ($$\infty$$ $$-$$ $$\infty$$) Now, $$\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b$$ $$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {\sqrt {{x^2} - x + 1} + ax}} = b$$ $$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$ $$ \Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1$$ Now, $$\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$ $$ \Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}$$ $$(a,b) = \left( {1, - {1 \over 2}} \right)$$

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