Consider [x]x≤f(x)≤6−x⇒x→2−lim[x]x=12=2 ⇒x→2−lim6−x=2∴x→2−limf(x)=2 [By Sandwich theorem] Now x→2+lim[x]x=1,x→2+lim6−x=2 Hence by Sandwich theorem x→2+limf(x) does not exists. Therefore f is not continuous at x=2. Thus statement-1 is true but statement- 2 is not true
Let f:[1,3] → R be a function satisfying x [x] ≤ f(x) ≤ √6-x, for all x ≠ 2 and f(2)=1, where R is the set of all real numbers and [x] denotes the…
JEE Main 2012 — Mathematics Calculus
Let f:[1,3]→R be a function satisfying [x]x≤f(x)≤6−x, for all x=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: x→2−limf(x) exists. Statement 2: f is continuous at x=2.
Held on 19 May 2012 · Verified 6 Jul 2026.
Options
- 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.
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