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

If $\mathop {\lim }\limits_{x \to 1} {\sin ^{ - 1}}\left( {\frac{k}{{\ln x}} - \frac{k}{{x - 1}}} \right)$ exists, then the number of integers in the range of $k$ is:

If $f(x) = \begin{cases} 4x-5, & x \leq 2 \\ x-k, & x > 2 \end{cases}$ then the value of $k$ for which $\lim_{x \rightarrow 2} f(x)$ exists is equal to:

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$,then $\lim_{x \rightarrow \alpha} \frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2} = $

If $a_1 = 1$ and $a_{n+1} = \frac{4 + 3a_n}{3 + 2a_n}$ for $n \ge 1$,and if $\lim_{n \to \infty} a_n = a$,then the value of $a$ is:

The values of the constants $\alpha$ and $\beta$ such that $\lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module