The domain of definition of the function f(x) | vedclass

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

Question

(D) For the function $f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x)$ to be defined:$1$. The denominator must not be zero: $4 - x^2 
eq 0$ $\Rightarro

Vedclass · Accepted Answer

$(-1, 0) \cup (1, 2) \cup (2, \infty)$

Vedclass · Answer

(D) For the function $f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x)$ to be defined:$1$. The denominator must not be zero: $4 - x^2 
eq 0$ $\Rightarrow x^2 
eq 4$ $\Rightarrow x 
eq \pm 2$.$2$. The argument of the logarithm must be positive: $x^3 - x > 0$.Factoring the inequality: $x(x^2 - 1) > 0 \Rightarrow x(x - 1)(x + 1) > 0$.Using the wavy curve method (sign scheme) on the number line with critical points $-1, 0, 1$,the expression is positive in the intervals $(-1, 0) \cup (1, \infty)$.Combining the conditions: $x \in (-1, 0) \cup (1, \infty)$ and $x 
eq \pm 2$.Since $2$ lies in the interval $(1, \infty)$,we must exclude it.Thus,the domain is $D = (-1, 0) \cup (1, 2) \cup (2, \infty)$.

The domain of definition of the function $f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x)$ is

Similar Questions

The domain of the definition of the function $f(x) = \frac{1}{4 - x^2} + \log(x^3 - x)$ is

The range of the function $f(x) = \frac{x}{1 + |x|}, x \in R,$ is

The range of the function $f(x) = \sqrt{3-x} + \sqrt{2+x}$ is

If $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ is an onto function defined by $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$, then $x=$

The domain of the function $\log _{10}(x^2-5x+6)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module