The domain of the function f(x) = sin^{-1}[2x | vedclass

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

Question

(C) For the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$ to be defined:$1.$ For $\sin^{-1}[2x^2 - 3]$,the argument must sa

Vedclass · Accepted Answer

$\left(1, \frac{5-\sqrt{5}}{2}\right)$

Vedclass · Answer

(C) For the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$ to be defined:$1.$ For $\sin^{-1}[2x^2 - 3]$,the argument must satisfy $-1 \leq [2x^2 - 3] \leq 1$. Since the range of $\sin^{-1}$ is $[-1, 1]$,we require $-1 \leq [2x^2 - 3] Actually,for $\sin^{-1}(y)$,$y \in [-1, 1]$. Thus,$-1 \leq [2x^2 - 3] \leq 1$. This implies $-1 \leq 2x^2 - 3 $2.$ For $\log_2(\log_{1/2}(x^2 - 5x + 5))$,we need $\log_{1/2}(x^2 - 5x + 5) > 0$. Since the base $1/2 $3.$ Solving $x^2 - 5x + 5 > 0$: Roots are $\frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$. So $x \in (-\infty, \frac{5-\sqrt{5}}{2}) \cup (\frac{5+\sqrt{5}}{2}, \infty)$.$4.$ Solving $x^2 - 5x + 5 Intersection of all conditions:From $1$: $x \in [1, \sqrt{2.5}) \cup (-\sqrt{2.5}, -1]$.From $2, 3, 4$: $x \in (1, \frac{5-\sqrt{5}}{2})$.Combining these,the domain is $(1, \frac{5-\sqrt{5}}{2})$.

The domain of the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$,where $[t]$ is the greatest integer function,is:

Similar Questions

If $f$ is a real-valued function from $A$ onto $B$ defined by $f(x) = \frac{1}{\sqrt{|x - |x||}}$, then $A \cap B = $

Find the domain of the function $f(x) = \frac{x^{2}+2x+1}{x^{2}-8x+12}$.

If the real valued function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is not defined for all $x \in(-\infty, a] \cup(b, \infty)$, then $3^a+b^2=$

If the domain of the function $f(x) = \log_e(4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right)$ is $(\alpha, \beta]$,then $36|\alpha + \beta|$ is equal to:

Domain of $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module