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⁻ f(x) exists. Statement 2: f is continu

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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⁻ f(x) exists. Statement 2: f is continuous at x=2 .

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Correct answer: D. Statement 1 is true, Statement 2 is false.

aligned & Consider x [x] f(x) 6-x & _ x 2⁻ x [x] = 2 1 =2 aligned aligned & _ x 2⁻ 6-x =2 & _ x 2⁻ f(x)=2 aligned [By Sandwich theorem] Now _ x 2⁺ x [x] =1, _ x 2⁺ 6-x =2 Hence by Sandwich theorem _ x 2⁺ f(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 largest integer less than or equal to x . Statement 1: _ x 2⁻ f(x) exists. Statement 2: f is continuous at x=2 .

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