Let f(x) = sin x + cos x and g(x) = x^2 - 1. | vedclass

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

Question

(C) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. First,we find the composition $g(f(x))$: $g(f(x)) = (\sin x + \cos x)^2 - 1$ $= (\sin^2 x + \

Vedclass · Accepted Answer

$[ - \frac{\pi }{4}, \frac{\pi }{4} ]$

Vedclass · Answer

(C) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. First,we find the composition $g(f(x))$: $g(f(x)) = (\sin x + \cos x)^2 - 1$ $= (\sin^2 x + \cos^2 x + 2 \sin x \cos x) - 1$ $= (1 + \sin 2x) - 1 = \sin 2x$. For a function to be invertible,it must be bijective (one-to-one and onto). The function $h(x) = \sin 2x$ is invertible in the interval where the argument $2x$ lies within the principal branch of the inverse sine function,i.e.,$[ -\frac{\pi}{2}, \frac{\pi}{2} ]$. Thus,we require: $-\frac{\pi}{2} \le 2x \le \frac{\pi}{2}$ Dividing by $2$,we get: $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$. Therefore,$g(f(x))$ is invertible for $x \in [ -\frac{\pi}{4}, \frac{\pi}{4} ]$.

Let $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. Then $g(f(x))$ is invertible for $x \in $

Similar Questions

Let $f(x)=(x+1)^2-1, x \geq-1$. Then $\{x \mid f(x)=f^{-1}(x)\} =$

If $f(x) = \exp(2x^3 + 3x^2 + 6x)$ and $g(x)$ is the inverse function of $f(x)$,then the value of $g'(e^{11})$ is -

Let $f: R \rightarrow R$ be defined by $f(x) = 2x + 6$,which is a bijective mapping. Then $f^{-1}(x)$ is given by:

Let $f(x) = (x + 2)^2 - 2, x \geq - 2$. Then $f^{-1}(x) =$

Suppose $f(x)=(x+1)^{2}$ for $x \geq -1$. If $g(x)$ is a function whose graph is the reflection of the graph of $f(x)$ in the line $y=x$,then $g(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module