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Question: Range of f(x) = [1 + sinx] + \(\left[ 2 + \sin \frac { x } { 2 } \right]\)+ \(\left[ 3 + \sin \frac...

Range of f(x) = [1 + sinx] + [2+sin⁡x2]\left[ 2 + \sin \frac { x } { 2 } \right]+ [3+sin⁡x3]\left[ 3 + \sin \frac { x } { 3 } \right] + ..... + [x+sin⁡xn]\left[ x + \sin \frac { x } { n } \right]∀ x ∈ [0, π], where [.] denotes the greatest integer function, is

A

{n2+n−22,n(n+1)2}\left\{ \frac { n ^ { 2 } + n - 2 } { 2 } , \frac { n ( n + 1 ) } { 2 } \right\}

B
C

{n2+n−22,n(n+1)2,n2+n+22}\left\{ \frac { n ^ { 2 } + n - 2 } { 2 } , \frac { n ( n + 1 ) } { 2 } , \frac { n ^ { 2 } + n + 2 } { 2 } \right\}

D
Answer
Explanation

Solution

f(x)=n(n+1)2+[sin⁡x]+[sin⁡x2]f ( x ) = \frac { n ( n + 1 ) } { 2 } + [ \sin x ] + \left[ \sin \frac { x } { 2 } \right] +… +[sin⁡xn]+ \left[ \sin \frac { x } { n } \right]

Since x ∈ [0, π] . Thus range is

⇒ ((25)x+(35)x+(45)x−1)\left( \left( \frac { 2 } { 5 } \right) ^ { x } + \left( \frac { 3 } { 5 } \right) ^ { x } + \left( \frac { 4 } { 5 } \right) ^ { x } - 1 \right) = 0

Clearly g(x)=(25)x+(35)x+(45)x−1g ( x ) = \left( \frac { 2 } { 5 } \right) ^ { x } + \left( \frac { 3 } { 5 } \right) ^ { x } + \left( \frac { 4 } { 5 } \right) ^ { x } - 1 is a

decreasing function and Lim⁡x→∞g(x)\operatorname { Lim } _ { x \rightarrow \infty } g ( x ) = −1, Lim⁡x→−∞\operatorname { Lim } _ { x \rightarrow - \infty } g(x) = ∞ also g(0) = 1. Thus f(x) = 0 has exactly one root.