Solve for x , |x+3|+x x+2 >1zZ - Vidyadip

Question

Solve for $x , \frac{|x+3|+x}{x+2}>1$zZ

✓

Answer

$\begin{array}{l}\text { We have, } \frac{|x+3|+x}{x+2}>1 \\ \Rightarrow \frac{|x+3|+x}{x+2}-1>0 \\ \Rightarrow \frac{|x+3|+x-x-2}{x+2}>0 \\ \Rightarrow \frac{|x+3|-2}{x+2}>0\end{array}$
$\begin{array}{l}\Rightarrow x =-3 \\ \therefore x =-3 \text { is a critical point. }\end{array}$
So, here we have two intervals $(-\infty,-3)$ and $[-3, \infty)$
Case I: When - $3 \leq x<\infty$, then $|x+3|=(x+3)$
$\begin{array}{l}\therefore \frac{|x+3|-2}{x+2}>0 \\ \Rightarrow \frac{x+3-2}{x+2}>0 \\ \Rightarrow \frac{x+1}{x+2}>0 \\ \Rightarrow \frac{(x+1)(x+2)^2}{(x+2)}>0 \times(x+2)^2 \\ \Rightarrow(x+1)(x+2)>0\end{array}$
Product of (x + 1) and (x + 2) will be positive, if both are of same sign.
$\begin{array}{l}\therefore(x+1)>0 \text { and }(x+2)>0 \\ \text { or }(x+1)<0 \text { and }(x+2)<0 \\ \Rightarrow x>-1 \text { and } x>-2 \\ \text { or } x<-1 \text { and } x<-2\end{array}$
On number line, these inequalities can be represented as,

Thus, $-1< x< \infty$ or $-\infty< x< -2$
But, here - $3 \leq x <\infty$
$\therefore-1< x <\infty$ or $-3 \leq x <-2$
Then, solution set in this case is
$x \in[-3,-2) \cup(-1, \infty)$
Case II: When $x<-3$, then $|x+3|=-(x+3)$
$\begin{array}{l}\therefore \frac{|x+3|-2}{x+2}>0 \\ \Rightarrow \frac{-x-3-2}{x+2}>0 \\ \Rightarrow \frac{-(x+5)}{x+2}>0 \\ \Rightarrow \frac{x+5}{x+2}<0 \\ \Rightarrow \frac{(x+5)(x+2)^2}{x+2}<0 \times( x +2)^2 \\ \Rightarrow( x +5)( x +2)<0\end{array}$
Product of (x + 5) and (x + 2) will be negative, if both are of opposite sign.
$\begin{array}{l}\therefore(x+5)>0 \text { and }(x+2)<0 \\ \text { or }(x+5)<0 \text { and }(x+2)>0 \\ \Rightarrow x>-5 \text { and } x<-2 \\ \text { or } x<-5 \text { and } x>-2\end{array}$
On number line, these inequalities can be represented as,

Thus, - 5 < x < - 2 i.e., solution set in the case is x∈ (- 5, - 2).
On combining cases I and II, we get the required solution set of given inequality, which is
$x \in(-5,-2) \cup(-1, \infty)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Model Paper 5→Model Paper 5 — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Prove the following:
$\sin(\text{n}+1)\text{x}\sin(\text{n}+2)\text{x}+2\cos(\text{n}+1)\text{x}\cos(\text{n}+2)\text{x}=\cos\text{x}$

Find the equation of the circle which circumscribes the triangle formed by the lines
y = x + 2, 3y = 4x and 2y = 3x.

Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests:

Ravi:	25	50	45	30	70	42	36	48	35	60
Hashina:	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent?

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Differentiate the following from first principle:

$\frac{\text{x}^2+1}{\text{x}}$

Solve the following equations:
$3\sin2\text{x}-5 \sin\text{x}\cos \text{x} + 8 \cos2\text{x = 2}$

$\text{If}\ \frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}+\frac{\cos(\text{C+D})}{\cos(\text{C}-\text{D})}=0,$ prove that $\tan\text{A}\tan\text{B}\tan\text{C}\tan\text{D}=-1$

In each of the followin find the equation of the hyperbola satisying the given conditions:
vertices $(0, \pm5), $ foci $(0,\pm8)$ [NCERT]

If $\alpha\text{ and }\beta$ are two solutions of the equation $\text{a}\tan\text{x+b}\sec\text{x}=\text{c},$ the find the value of $\sin(\alpha+\beta)\text{ and }\cos(\alpha+\beta).$

Prove that $\cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ}=\cos 24^{\circ}+\cos 48^{\circ}$