If R is the set of all real numbers and f: R | vedclass

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

Question

(B) Given the function $f(x) = 3x^2 + 1$.We need to find the set $f^{-1}([1, 6])$, which consists of all $x$ such that $f(x) \in [1, 6]$.So, $1 \le 3x

Vedclass · Accepted Answer

$[ -\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} ]$

Vedclass · Answer

(B) Given the function $f(x) = 3x^2 + 1$.We need to find the set $f^{-1}([1, 6])$, which consists of all $x$ such that $f(x) \in [1, 6]$.So, $1 \le 3x^2 + 1 \le 6$.Subtracting $1$ from all parts: $0 \le 3x^2 \le 5$.Dividing by $3$: $0 \le x^2 \le \frac{5}{3}$.Taking the square root: $-\sqrt{\frac{5}{3}} \le x \le \sqrt{\frac{5}{3}}$.Thus, the set is $[ -\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} ]$.

If $R$ is the set of all real numbers and $f: R \rightarrow R$ is defined by $f(x) = 3x^2 + 1$, then the set $f^{-1}([1, 6])$ is

Similar Questions

Let $f = \left\{ \left(x, \frac{x^2}{1+x^2} \right) : x \in R \right\}$ be a function from $R$ into $R$. Determine the range of $f$.

The domain of $f(x) = \sqrt{\left(\frac{1}{\sqrt{x}} - \sqrt{x+1}\right)}$ is

Range of the function $f(x) = \sin^2(x^4) + \cos^2(x^4)$ is

If $f:[2,3] \rightarrow R$ is defined by $f(x)=x^3+3x-2$,then the range of $f(x)$ is contained in the interval

Find the range of the following function:
$f(x) = 2 - 3x$,where $x \in R$ and $x > 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module