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 trueYou need to login to perform this action.
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