FunctionshardFree

Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

f:(0,\pi/n)\to\mathbb{R}

defined by

f(x)=\displaystyle\sum_{k=1}^{n}[1+\sin kx]

, where

[x]

denotes the integral part of

x

, then range of

f(x)

is:

\{n-1,n+1\}

\{n-1,n,n+1\}

\{n,n+1\}

correct

DNone of these

Solution

Step 1: Expand:

f(x)=\sum_{k=1}^{n}[1+\sin kx]=n+\sum_{k=1}^{n}[\sin kx]

(Since

[1+u]=[u]+1

for integer part, and summing gives

n+\sum[\sin kx]

.) Step 2: Case 1:

kx\neq\dfrac{\pi}{2}

for

k=1,2,\ldots,n

. For

x\in\left(0,\dfrac{\pi}{n}\right)

with

kx\neq\dfrac{\pi}{2}

kx\in(0,\pi)

and

kx\neq\dfrac{\pi}{2}

, so

0<\sin kx<1

, giving

[\sin kx]=0

. Therefore

f(x)=n+0=n

. Step 3: Case 2: Exactly one of

kx

equals

\dfrac{\pi}{2}

(possible only if

\dfrac{\pi}{2kx}=1

for some

k

, i.e.,

x=\dfrac{\pi}{2k}

). In this case,

\sin kx=1

, so

[\sin kx]=1

for that particular

k

, and

[\sin jx]=0

for

j\neq k

. Therefore

f(x)=n+1

. Step 4: Since not more than one of

x,2x,\ldots,nx

can equal

\dfrac{\pi}{2}

(0,\pi/n)

, the range is

\{n,n+1\}

. Answer: (3)

\{n,n+1\}

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