For x in left( 0, frac{3}{2} right),let f(x) | vedclass

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

Question

(A) Given functions are $f(x) = \sqrt{x}$,$g(x) = 	an x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$.First,find $(f \circ g)(x) = f(g(x)) = \sqrt{	an x}$.N

Vedclass · Accepted Answer

$	an \frac{11\pi}{12}$

Vedclass · Answer

(A) Given functions are $f(x) = \sqrt{x}$,$g(x) = 	an x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$.First,find $(f \circ g)(x) = f(g(x)) = \sqrt{	an x}$.Next,find $\phi(x) = (h \circ (f \circ g))(x) = h(\sqrt{	an x})$.Substituting $\sqrt{	an x}$ into $h(x)$,we get $\phi(x) = \frac{1 - (\sqrt{	an x})^2}{1 + (\sqrt{	an x})^2} = \frac{1 - 	an x}{1 + 	an x}$.Using the trigonometric identity $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we can write $\frac{1 - 	an x}{1 + 	an x} = 	an\left( \frac{\pi}{4} - x ight)$.Now,calculate $\phi\left( \frac{\pi}{3} ight) = 	an\left( \frac{\pi}{4} - \frac{\pi}{3} ight) = 	an\left( \frac{3\pi - 4\pi}{12} ight) = 	an\left( -\frac{\pi}{12} ight) = -	an\left( \frac{\pi}{12} ight)$.Since $	an(\pi - 	heta) = -	an 	heta$,we have $-	an\left( \frac{\pi}{12} ight) = 	an\left( \pi - \frac{\pi}{12} ight) = 	an\left( \frac{11\pi}{12} ight)$.

For $x \in \left( 0, \frac{3}{2} \right)$,let $f(x) = \sqrt{x}$,$g(x) = \tan x$,and $h(x) = \frac{1 - x^2}{1 + x^2}$. If $\phi(x) = ((h \circ f) \circ g)(x)$,then $\phi\left( \frac{\pi}{3} \right)$ is equal to

Similar Questions

Let $f(x) = \log_e(\sin x)$ for $0 < x < \pi$ and $g(x) = \sin^{-1}(e^{-x})$ for $x \ge 0$. If $\alpha$ is a positive real number such that $a = (fog)'(\alpha)$ and $b = (fog)(\alpha)$,then which of the following is true?

If $f(x) = 8x^3$ and $g(x) = x^{1/3}$,then $f \circ g(x)$ is:

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

Let $f(x) = \sin x$ and $g(x) = \ln |x|$. If the ranges of the composite functions $fog$ and $gof$ are $R_1$ and $R_2$ respectively,then:

If function $f: R \rightarrow R$ is defined by $f(x) = (3 - x^5)^{\frac{1}{5}}$,then $(f \circ f)(x) = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module