Question

If $\underset{-1}{\overset{1}{\int }}\frac{3^x+3^{-x}}{1+3^x}dx=\frac{8}{\ln \left(k\right)},$ then which of the following is/are true?

• A. $k=\underset{0}{\overset{3}{\int }}\left(3x^2\right)dx$
• B. $k=\underset{-3}{\overset{3}{\int }}\left(3x^2\right)dx$
• C. highest prime factor of $k$ is $3$
• D. $k$ is not divisible by $6$

Answer 1

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

$k$ is not divisible by $6$

Let $I=\underset{-1}{\overset{1}{\int }}\frac{3^x+3^{-x}}{1+3^x}dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}\left(f\right(x)+f(-x\left)\right)dx$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}(\frac{3^x+3^{-x}}{1+3^x}+\frac{3^{-x}+3^x}{1+3^{-x}})dx$
$=\underset{0}{\overset{1}{\int }}(3^x+3^{-x})dx$
$={[\frac{3^x}{\ln 3}-\frac{3^{-x}}{\ln 3}]}_0^1=\frac{1}{\ln 3}[3-\frac{1}{3}]=\frac{8}{3\ln 3}=\frac{8}{\ln \left(27\right)}$
$\therefore k=27$
$\to \underset{0}{\overset{3}{\int }}\left(3x^2\right)dx={\left[x^3\right]}_0^3=27=k$
$\to \underset{-3}{\overset{3}{\int }}\left(3x^2\right)dx={\left[x^3\right]}_{-3}^3=54=2k$
$\to k=27=3^3$
So, highest prime factor of $k$ is $3$
$\to k=27$ which is not divisible by $6$