Let [x] denote the greatest integer less than or equal to x for any real number x. The range of the function f : R → R, defined by f(x) = sin(π[x])/x^2+5 , is (a) (-1,1) (b) [-1,1] © {-1,1} (D) {0}

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Deepak Kumar Shringi · Answer

sunny · Answer

Hii budy this is an easy question dont get tricked of denomination Finding range of numerator =sin(π[x]) Through substitution as sin is an increasing function from -1 to 1 Sin3π/2=-1 Puting x=3π/2=3×π/2=4.34 so x=4.34and [x]=4 so from question sin(4π)=0 Now similarily checking greatest integer for x=π/2,π...or any it , Range of sin(π[x])=0 and since numerator is always zero hence function range is 0. Quite tired of typing i m new hey if it helped do not forget to give an approval

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Let [x] denote the greatest integer less than or equal to x for any real number x. The range of the function f : R → R, defined by f(x) = sin(π[x])/x^2+5 , is (a) (-1,1) (b) [-1,1] © {-1,1} (D) {0}

Let [x] denote the greatest integer less than or equal to x for any real number x. The range of the function f : R → R, defined by f(x) = sin(π[x])/x^2+5 , is

(a) (-1,1)
(b) [-1,1]

© {-1,1}
(D) {0}

2 Answers

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Let [x] denote the greatest integer less than or equal to x for any real number x. The range of the function f : R → R, defined by f(x) = sin(π[x])/x^2+5 , is (a) (-1,1) (b) [-1,1] © {-1,1} (D) {0}

Let [x] denote the greatest integer less than or equal to x for any real number x. The range of the function f : R → R, defined by f(x) = sin(π[x])/x^2+5 , is (a) (-1,1)(b) [-1,1] © {-1,1}(D) {0}

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Let [x] denote the greatest integer less than or equal to x for any real number x. The range of the function f : R → R, defined by f(x) = sin(π[x])/x^2+5 , is

(a) (-1,1)
(b) [-1,1]

© {-1,1}
(D) {0}