If the domain of the function f(x) = frac{sqr | vedclass

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

Question

(C) The function is $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$.For the square root term,$x^2-25 \geq 0 \Rightarrow x \in (-\infty, -5

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) The function is $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$.For the square root term,$x^2-25 \geq 0 \Rightarrow x \in (-\infty, -5] \cup [5, \infty)$.For the denominator,$4-x^2 
eq 0 \Rightarrow x 
eq \pm 2$.For the logarithm,$x^2+2x-15 > 0$ $\Rightarrow (x+5)(x-3) > 0$ $\Rightarrow x \in (-\infty, -5) \cup (3, \infty)$.Taking the intersection of all conditions:$x \in ((-\infty, -5] \cup [5, \infty)) \cap (-\infty, -5) \cup (3, \infty) \cap \mathbb{R} \setminus \{-2, 2\}$.This simplifies to $x \in (-\infty, -5) \cup [5, \infty)$.Comparing with $(-\infty, \alpha) \cup [\beta, \infty)$,we get $\alpha = -5$ and $\beta = 5$.Therefore,$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150$.

If the domain of the function $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$ is $(-\infty, \alpha) \cup [\beta, \infty)$,then $\alpha^2 + \beta^3$ is equal to:

Similar Questions

The domain of the real valued function $f(x) = \sqrt{\frac{x-2}{3-x}}$ is...

The domain of the real-valued function $f(x) = \frac{\sqrt{|x|-x}}{\sqrt{x-[x]}}$ is

The set of all real values of $x$ for which $f(x)=\sqrt{\frac{|x|-2}{|x|-3}}$ is a well-defined function is

If $f: R \to R$,then the range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

If $f: R \rightarrow A$ defined by $f(x) = \frac{1}{x^2+2x+2}$,$\forall x \in R$ is surjective,then $A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module