Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Find the range of $f(x) = [1+\sin x]+\left[2+\sin\frac{x}{2}\right]+\left[3+\sin\frac{x}{2}\right]+...+\left[n+\sin\frac{x}{n}\right]\, ∀\,x∈[0,π],n∈N$ ([.] denotes greatest integer function).

Options:

$\left\{\begin{matrix}\frac{n(n-2)}{2},\frac{n^2+n+n}{2}\end{matrix}\right\}$

$\left\{\begin{matrix}\frac{n(n+1)}{2},\frac{n^2-n-n}{2}\end{matrix}\right\}$

$\left\{\begin{matrix}\frac{n(n+1)}{2},\frac{n^2+n+2}{2}\end{matrix}\right\}$

None of these

Correct Answer:

$\left\{\begin{matrix}\frac{n(n+1)}{2},\frac{n^2+n+2}{2}\end{matrix}\right\}$

Explanation:

$f(x) = [1+\sin x]+\left[2+\sin\frac{x}{2}\right]+\left[3+\sin\frac{x}{3}\right]+...+\left[n+\sin\frac{x}{n}\right]$

for, $x∈(0,π)$

$\sin x, \sin \frac{x}{2},\sin\frac{x}{3}, \sin\frac{x}{4}, \sin\frac{x}{5},....\sin\frac{x}{n}∈(0,1)$

∴ for $x ∈ (0, π)$

$f(x)=1+2+3+....+n=\frac{n(n+1)}{2}$

For $x = 0, f(x)=1+2+3+....+n=\frac{n(n+1)}{2}$

for $x = π, f(x)=1+[2+1]+3+4+....+n$

$=(1+2+3+...+n)+1$

$=\frac{n(n+1)}{2}+1=\frac{n^2+n+2}{2}$

∴ Range is $\left\{\begin{matrix}\frac{n(n+1)}{2},\frac{n^2+n+2}{2}\end{matrix}\right\}$