The range of the function f(x) = log_{0.5}(x^ | vedclass

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

Question

(B) Let $g(x) = x^4 - 2x^2 + 3 = (x^2 - 1)^2 + 2$. Since $(x^2 - 1)^2 \geq 0$, the minimum value of $g(x)$ is $2$ (at $x^2 = 1$) and the maximum value

Vedclass · Accepted Answer

$(-\infty, -1]$

Vedclass · Answer

(B) Let $g(x) = x^4 - 2x^2 + 3 = (x^2 - 1)^2 + 2$. Since $(x^2 - 1)^2 \geq 0$, the minimum value of $g(x)$ is $2$ (at $x^2 = 1$) and the maximum value is $\infty$. Thus, the range of $g(x)$ is $[2, \infty)$. Now, $f(x) = \log_{0.5}(g(x)) = \log_{1/2}(g(x)) = -\log_2(g(x))$. Since $g(x) \in [2, \infty)$, we have $\log_2(g(x)) \in [\log_2 2, \log_2 \infty) = [1, \infty)$. Multiplying by $-1$, we get $-\log_2(g(x)) \in (-\infty, -1]$. Therefore, the range of $f(x)$ is $(-\infty, -1]$.

The range of the function $f(x) = \log_{0.5}(x^4 - 2x^2 + 3)$ is

Similar Questions

Let $A = \{10, 11, 12, 14, 26\}$ and let $f: A \rightarrow N$ be defined such that $f(a) = \text{highest prime factor of } a$,where $a \in A$. Then the range of $f$ is:

The function $f(x) = \frac{\sec^{-1}x}{\sqrt{x - [x]}}$,where $[.]$ denotes the greatest integer less than or equal to $x$,is defined for all $x$ belonging to:

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:

If $[x]$ denotes the greatest integer $\leq x$, then the domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$ is

Range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module