The domain of f(x) = sqrt{left(frac{1}{sqrt{x | vedclass

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

Question

(C) For the function $f(x) = \sqrt{\frac{1}{\sqrt{x}} - \sqrt{x+1}}$ to be defined, the following conditions must be met:$1$. The expression inside th

Vedclass · Accepted Answer

$\left(0, \frac{\sqrt{5}-1}{2}\right]$

Vedclass · Answer

(C) For the function $f(x) = \sqrt{\frac{1}{\sqrt{x}} - \sqrt{x+1}}$ to be defined, the following conditions must be met:$1$. The expression inside the square root must be non-negative: $\frac{1}{\sqrt{x}} - \sqrt{x+1} \geq 0$.$2$. The term $\sqrt{x}$ in the denominator requires $x > 0$.$3$. The term $\sqrt{x+1}$ requires $x+1 \geq 0$, which means $x \geq -1$.Combining these, we need $x > 0$.Now, solve the inequality: $\frac{1}{\sqrt{x}} \geq \sqrt{x+1}$.Since $x > 0$, we can square both sides: $\frac{1}{x} \geq x+1$.$1 \geq x^2 + x \implies x^2 + x - 1 \leq 0$.The roots of $x^2 + x - 1 = 0$ are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.Since the parabola opens upward, $x^2 + x - 1 \leq 0$ for $x \in \left[\frac{-1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right]$.Intersecting this with the condition $x > 0$, we get $x \in \left(0, \frac{\sqrt{5}-1}{2}\right]$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

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

Similar Questions

If the domain of the function $f(x) = \frac{1}{\sqrt{10+3x-x^2}} + \frac{1}{\sqrt{x+|x|}}$ is $(a, b)$,then $(1+a)^2 + b^2$ is equal to :

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

Let $f(x) = \frac{1}{7 - \sin 5x}$ be a function defined on $R$. Then the range of the function $f(x)$ is equal to:

The domain of the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 2}}$ is, where $[x]$ denotes the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module