Verified Solution • 2025–26 Session moderate

QJEEM21Ma3712: JEE Main 2021 Mathematics - Class 12 JEEM21Ma3712 Solution: If , then the ordered pair (a, b) is :

If lim⁡x→∞(x2−x+1−ax)=b\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b , then the ordered pair (a, b) is :

Multiple Choice Options

  • A (1,12)\left( {1,{1 \over 2}} \right)
  • B (1,−12)\left( {1, - {1 \over 2}} \right)
  • C (−1,12)\left( { - 1,{1 \over 2}} \right)
  • D (−1,−12)\left( { - 1, - {1 \over 2}} \right)

Step-by-Step Verified Solution

lim⁡x→∞(x2−x+1)−ax=b\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b ( ∞\infty −- ∞\infty )

Now, lim⁡x→∞(x2−x+1−a2x2)x2−x+1+ax=b\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b

⇒lim⁡x→∞(1−a2)x2−x+1x2−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

⇒lim⁡x→∞(1−a2)x2−x+1x(1−1x+1x2+a)=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

⇒1−a2=0⇒a=1\Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1

Now, lim⁡x→∞−x+1x(1−1x+1x2+a)=b\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b

⇒−11+a=b⇒b=−12\Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}

(a,b)=(1,−12)(a,b) = \left( {1, - {1 \over 2}} \right)

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