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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    Let f: [1, 3]\[\to \] R be a function satisfying \[\frac{x}{[x]}\le f\left( x \right)\le \sqrt{6-x},\]for all \[x\ne 2\]and f(2) = 1, where R is the set of all real numbers and [x] denotes the largest integer less than or equal to x. Statement 1: \[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)\] exists. Statement 2: f is continuous at x = 2.     JEE Main  Online Paper (Held On 19  May  2012)

    A) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

    B) Statement 1 is false. Statement 2 is true.

    C) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

    D) Statement 1 is true, Statement 2 is false. 

    Correct Answer: D

    Solution :

    Consider\[\frac{x}{[x]}\le f(x)\le \sqrt{6-x}\] \[\Rightarrow \]\[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{x}{[x]}=\frac{2}{1}=2\]\[\Rightarrow \]\[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\sqrt{6-x}=2\] \[\therefore \]\[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f(x)=2\]  [By Sandwich theorem] Now \[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{x}{[x]}=1,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\sqrt{6-x}=2\] Hence by Sandwich theorem \[\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f(x)\]does not exists. Therefore f is not continuous at x = 2. Thus statement-1 is true but statement-2 is not true


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