Let f(x) = frac{tan^n x}{sum_{r=0}^{2n} tan^r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $	an x = t$. Since $x \in [0, \frac{\pi}{2})$,$t \in [0, \infty)$.$f(x) = \frac{t^n}{1 + t + t^2 + \dots + t^{2n}}$.For $t=0$,$f(x) = 0$.For

Vedclass · Accepted Answer

$f(x)$ is bounded and it takes both of its bounds and the range of $f(x)$ contains exactly one integral point.

Vedclass · Answer

(A) Let $	an x = t$. Since $x \in [0, \frac{\pi}{2})$,$t \in [0, \infty)$.$f(x) = \frac{t^n}{1 + t + t^2 + \dots + t^{2n}}$.For $t=0$,$f(x) = 0$.For $t > 0$,divide numerator and denominator by $t^n$:$f(x) = \frac{1}{t^n + t^{n-1} + \dots + 1 + \dots + \frac{1}{t^{n-1}} + \frac{1}{t^n}}$.Using the $AM$-$GM$ inequality,$t^k + \frac{1}{t^k} \ge 2$ for $t > 0$.Thus,the denominator is $\ge 2 + 2 + \dots + 2 + 1 = 2n + 1$.Therefore,$f(x) \le \frac{1}{2n+1}$.The equality holds when $t=1$,i.e.,$x = \frac{\pi}{4}$.Since $f(0) = 0$ and the maximum value is $\frac{1}{2n+1}$,the range is $[0, \frac{1}{2n+1}]$.Since $n \in N$,$2n+1 \ge 3$,so $\frac{1}{2n+1} < 1$. The only integer in the range $[0, \frac{1}{2n+1}]$ is $0$. Thus,it contains exactly one integral point.

Let $f(x) = \frac{\tan^n x}{\sum_{r=0}^{2n} \tan^r x}$,$n \in N$,where $x \in [0, \frac{\pi}{2})$.

Similar Questions

If $D \subseteq R$ and $f: D \rightarrow R$ defined by $f(x) = \frac{x^2+x+a}{x^2-x+a}$ is a surjection, then '$a$' lies in the interval

The domain of the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ is

The domain of $f(x) = \sqrt{\left(\frac{1}{\sqrt{x}} - \sqrt{x+1}\right)}$ is

Find the range of the following function:
$f(x) = 2 - 3x$,where $x \in R$ and $x > 0$.

If $f: R \rightarrow A$,defined by $f(x) = \cos x + \sqrt{3} \sin x - 1$,is an onto function,then $A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module