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 $f(x_n) = 1 + (2n - 1) \pi$.Substitutin

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

Let $A = \{1, 2, 3\}$ and $B = \{1, 3, 5\}$. If a relation $R$ is defined from $A$ to $B$ as $R = \{(1, 3), (2, 5), (3, 3)\}$,then find $R^{-1}$.

If $f: R \rightarrow R$ is defined by $f(x)=|x|$,then

If $f(x) = \frac{3x+2}{5x-3}$,where $x \in R - \{\frac{3}{5}\}$,then:

If $f(x) = \int\limits_0^x {\frac{1}{{\sqrt {1 + {t^3}} }}\,} dt$ and $h(x)$ is the inverse of $f(x)$,then the value of $\frac{{h''(x)}}{{{h^2}(x)}}$ is

Which of the following functions is an invertible function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module