Let f(x) = sqrt{frac{x}{1-x}} + sqrt{frac{1-x | vedclass

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

Question

(C) Given $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$.Let $t = \sqrt{\frac{x}{1-x}}$. Then $f(x) = t + \frac{1}{t}$.We are given $\lim_{x \ri

Vedclass · Accepted Answer

$\{1/5, 4/5\}$

Vedclass · Answer

(C) Given $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$.Let $t = \sqrt{\frac{x}{1-x}}$. Then $f(x) = t + \frac{1}{t}$.We are given $\lim_{x \rightarrow m} (t + \frac{1}{t}) = 5/2$.This implies $t + \frac{1}{t} = 5/2$,which simplifies to $2t^2 - 5t + 2 = 0$.Solving for $t$: $(2t - 1)(t - 2) = 0$,so $t = 1/2$ or $t = 2$.Case $1$: $\sqrt{\frac{m}{1-m}} = 1/2$ $\Rightarrow \frac{m}{1-m} = 1/4$ $\Rightarrow 4m = 1 - m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$.Case $2$: $\sqrt{\frac{m}{1-m}} = 2$ $\Rightarrow \frac{m}{1-m} = 4$ $\Rightarrow m = 4 - 4m$ $\Rightarrow 5m = 4$ $\Rightarrow m = 4/5$.Thus,the set of possible values for $m$ is $\{1/5, 4/5\}$.

Let $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$. If $\lim_{x \rightarrow m} f(x) = 5/2$,then the set of all possible finite values of $m$ is:

Similar Questions

Let $a$ be an integer such that $\lim \limits_{x \rightarrow 7} \frac{18-[1-x]}{[x]-3a}$ exists,where $[t]$ denotes the greatest integer function $\leq t$. Then $a$ is equal to:

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0$,where $a, b \in R$,then:

The integral value of $n$ for which $\lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is a finite non-zero real number is

Define $f(x) = \begin{cases} b - ax & \text{if } x < 2 \\ 3 & \text{if } x = 2 \\ a + 2bx & \text{if } x > 2 \end{cases}$. If $\lim_{x \rightarrow 2} f(x)$ exists,then find the value of $\frac{a}{b}$.

If the function $f(x)$ satisfies $\mathop {\lim }\limits_{x \to 1} \frac{f(x)-2}{x^{2}-1}=\pi,$ evaluate $\mathop {\lim }\limits_{x \to 1} f(x).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module