The real numbers x satisfying frac{sqrt{x+5}} | vedclass

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

Question

(D) Given the inequality $\frac{\sqrt{x+5}}{1-x} > 1$. For the expression to be defined,we must have $x+5 \ge 0 \Rightarrow x \ge -5$ and $1-x 
eq 0

Vedclass · Accepted Answer

$-1 < x < 1$

Vedclass · Answer

(D) Given the inequality $\frac{\sqrt{x+5}}{1-x} > 1$. For the expression to be defined,we must have $x+5 \ge 0 \Rightarrow x \ge -5$ and $1-x eq 0 \Rightarrow x eq 1$. Case $1$: If $1-x > 0$,i.e.,$x 1-x$. Since $x \ge -5$,$x+5$ is non-negative. If $1-x 0$. Squaring both sides: $x+5 > (1-x)^2 = 1 - 2x + x^2$. $x^2 - 3x - 4 $(x-4)(x+1) This implies $-1 Combining with $x Case $2$: If $1-x 1$,then $\sqrt{x+5} Since $\sqrt{x+5} \ge 0$ and $1-x Thus,the solution is $-1 < x < 1$.

The real numbers $x$ satisfying $\frac{\sqrt{x+5}}{1-x} > 1$ are precisely those which satisfy

Similar Questions

Solve the given inequality for real $x$: $4x + 3 < 5x + 7$

Solve the following inequality and represent it on a number line: $\frac{x}{3} + 5 \geq \frac{x}{2} + 7$

Solve the inequalities and represent the solution graphically on a number line:
$5x + 1 > -24, 5x - 1 < 24$

The solution set of $|x-1| + |x+1| < 2$ is...

Let $a, b, c, d, e$ be real numbers such that $a + b < c + d$,$b + c < d + e$,$c + d < e + a$,and $d + e < a + b$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module