Let f: R rightarrow R and g: R rightarrow R b | vedclass

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

Question

(B) Given $(f \circ g)(x) = x$,$g(x)$ is the inverse function of $f(x)$.Let $g(1 + (2n - 1) \pi) = x_n$. Then $f(x_n) = 1 + (2n - 1) \pi$.Substituting

Vedclass · Accepted Answer

$(99)^2 \frac{\pi}{2}$

Vedclass · Answer

(B) Given $(f \circ g)(x) = x$,$g(x)$ is the inverse function of $f(x)$.Let $g(1 + (2n - 1) \pi) = x_n$. Then $f(x_n) = 1 + (2n - 1) \pi$.Substituting $f(x) = 2x + \cos x + \sin^2 x$:$2x_n + \cos x_n + \sin^2 x_n = 1 + (2n - 1) \pi$$2x_n + \cos x_n + 1 - \cos^2 x_n = 1 + (2n - 1) \pi$$2x_n + \cos x_n - \cos^2 x_n = (2n - 1) \pi$.For $x_n = (2n - 1) \frac{\pi}{2}$,we have $\cos x_n = 0$.Substituting this into the equation: $2((2n - 1) \frac{\pi}{2}) + 0 - 0 = (2n - 1) \pi$,which holds true.Thus,$g(1 + (2n - 1) \pi) = (2n - 1) \frac{\pi}{2}$.Now,$\sum_{n=1}^{99} g(1 + (2n - 1) \pi) = \sum_{n=1}^{99} (2n - 1) \frac{\pi}{2} = \frac{\pi}{2} \sum_{n=1}^{99} (2n - 1)$.The sum of the first $99$ odd numbers is $99^2$.Therefore,the sum is $(99)^2 \frac{\pi}{2}$.

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be differentiable functions such that $(f \circ g)(x) = x$. If $f(x) = 2x + \cos x + \sin^2 x$,then the value of $\sum_{n=1}^{99} g(1 + (2n - 1) \pi)$ is

Similar Questions

If $f: R \rightarrow R$ is defined by $f(x)=x^{3}$,then $f^{-1}(8)$ is equal to

If $y = f(x) = \frac{ax + b}{cx - a}$,then $x$ is equal to

If $f(x) = \frac{2x - 1}{x + 5}$ $(x \ne -5)$,then $f^{-1}(x)$ is equal to

$f: (-\infty, 0] \rightarrow [0, \infty)$ is defined as $f(x) = x^2$. The domain and range of its inverse are

If $[\cdot]$ denotes the greatest integer function and if $f:(5,10) \rightarrow(7,12)$ is a function defined by $f(x)=x+2\left[\frac{x}{5}\right]$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module