Let f(x) = sin^{-1} x and g(x) = frac{x^2 - x | vedclass

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

Question

(C) The domain of $(f \circ g)(x) = \sin^{-1}(g(x))$ requires $|g(x)| \leq 1$.First,we find $g(2) = \lim_{x 	o 2} \frac{(x+1)(x-2)}{(2x+3)(x-2)} = \l

Vedclass · Accepted Answer

$(-\infty, -2] \cup [-\frac{4}{3}, \infty)$

Vedclass · Answer

(C) The domain of $(f \circ g)(x) = \sin^{-1}(g(x))$ requires $|g(x)| \leq 1$.First,we find $g(2) = \lim_{x 	o 2} \frac{(x+1)(x-2)}{(2x+3)(x-2)} = \lim_{x 	o 2} \frac{x+1}{2x+3} = \frac{3}{7}$.Thus,we solve $|\frac{x+1}{2x+3}| \leq 1$ for $x 
eq 2$.This implies $-1 \leq \frac{x+1}{2x+3} \leq 1$.Case $1$: $\frac{x+1}{2x+3} \leq 1 \Rightarrow \frac{x+1 - (2x+3)}{2x+3} \leq 0 \Rightarrow \frac{-x-2}{2x+3} \leq 0 \Rightarrow \frac{x+2}{2x+3} \geq 0$.The solution is $x \in (-\infty, -2] \cup (-\frac{3}{2}, \infty)$.Case $2$: $\frac{x+1}{2x+3} \geq -1 \Rightarrow \frac{x+1 + 2x+3}{2x+3} \geq 0 \Rightarrow \frac{3x+4}{2x+3} \geq 0$.The solution is $x \in (-\infty, -\frac{4}{3}] \cup (-\frac{3}{2}, \infty)$.Taking the intersection of both cases and excluding $x=2$ (since $g(2)$ is defined as the limit),we get $x \in (-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$. If $g(2) = \lim_{x \to 2} g(x)$,then the domain of the function $f \circ g$ is .... .

Similar Questions

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = x^{2} - 3x + 4$ and $g(x) = 2x + 1$,then the value of $x$ for which $f(x) = (f \circ g)(x)$ is

If $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$,then

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ and the composed mapping $g \circ f: S \rightarrow U$ be defined. If $g \circ f$ is an injective mapping, then:

Let $f, g: R \rightarrow R$ be two functions defined as $f(x)=|x|+x$ and $g(x)=|x|-x$ for all $x \in R$. Then $(f \circ g)(x)$ for $x < 0$ is

If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3x-2$ and $g(x)=x^2+2$, then $[(g \circ f)+(f \circ g)](x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module