Question

If $A=\sum _{r=0}^n{}^n{C}_r{\cos }^{n}x\cdot {\tan }^{r}x$, then which of the following option(s) is/are correct?

• A. Maximum value of $A$ is $(\sqrt{2}{)}^n$
• B. Minimum value of $A$ is $(-\sqrt{2}{)}^n$ when $n\in N$.
• C. Minimum value of $A$ is $(-\sqrt{2}{)}^n$ when $n$ is odd.
• D. Minimum value of $A$ is $0$ when $n$ is even.

Answer 1

Answer: (A), (C), (D)

Minimum value of $A$ is $0$ when $n$ is even.

Given : $A=\sum _{r=0}^n{}^n{C}_r{\cos }^{n}x\cdot {\tan }^{r}x$
Now,
$A=\sum _{r=0}^n{}^n{C}_r{\cos }^{n}x\cdot {\left(\frac{\sin x}{\cos x}\right)}^r\Rightarrow A=\sum _{r=0}^n{}^n{C}_r{\cos }^{n-r}x\cdot {\sin }^{r}x\therefore A=(\cos x+\sin x{)}^n$
We know that
$-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$
Then
$A_{\text{Max}}=(\sqrt{2}{)}^n$
When $n$ is even
$A_{\text{min}}=0$
When $n$ is odd
$A_{\text{min}}=(-\sqrt{2}{)}^n$