Let (2^{1-a} + 2^{1+a}), f(a), (3^a + 3^{-a}) | vedclass

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

Question

(B) Given that $(2^{1-a} + 2^{1+a})$, $f(a)$, and $(3^a + 3^{-a})$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.Since $2^{1-

Vedclass · Accepted Answer

$\frac{1}{4}\log_e(\frac{4}{3})$

Vedclass · Answer

(B) Given that $(2^{1-a} + 2^{1+a})$, $f(a)$, and $(3^a + 3^{-a})$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.Since $2^{1-a} + 2^{1+a} = 2(2^{-a} + 2^a)$, and by the $AM$-$GM$ inequality, $2^a + 2^{-a} \geq 2$, the minimum value of $2^a + 2^{-a}$ is $2$.Similarly, the minimum value of $3^a + 3^{-a}$ is $2$.Thus, $2f(a) \geq 2(2) + 2 = 6$, which implies $f(a) \geq 3$. Therefore, $\alpha = 3$.The integral becomes $I = \int_{\log_e 2}^{\log_e 3} \frac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e 2}^{\log_e 3} \frac{e^{2x} dx}{e^{4x} - 1}$.Let $u = e^{2x}$, then $du = 2e^{2x} dx$, so $e^{2x} dx = \frac{du}{2}$.When $x = \log_e 2$, $u = e^{2\log_e 2} = 4$. When $x = \log_e 3$, $u = e^{2\log_e 3} = 9$.$I = \frac{1}{2} \int_{4}^{9} \frac{du}{u^2 - 1} = \frac{1}{2} \cdot \frac{1}{2} \log_e |\frac{u-1}{u+1}| \Big|_4^9 = \frac{1}{4} (\log_e \frac{8}{10} - \log_e \frac{3}{5}) = \frac{1}{4} \log_e (\frac{4}{5} \cdot \frac{5}{3}) = \frac{1}{4} \log_e (\frac{4}{3})$.

Let $(2^{1-a} + 2^{1+a})$, $f(a)$, $(3^a + 3^{-a})$ be in $A$.$P$. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is:

Similar Questions

If $\int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$,then $(m \cdot n)$ equals

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

Let $f:(0, \infty) \rightarrow \mathbb{R}$ and $F(x)=\int_0^x t f(t) d t$. If $F(x^2)=x^4+x^5$,then $\sum_{r=1}^{12} f(r^2)$ is equal to :

$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :

$\int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module