CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Absolute Value Function

|x+3|+x/x+2=1...

Question

If $\frac{| x + 3 | + x}{x + 2} = 1$ , then $x \in$

A

{- 5, - 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

{1, 5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

{- 1, 5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

{1, 5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A ${- 5, - 1}$
$\frac{| x + 3 | + x}{x + 2} = 1, x \neq - 2$
$\Rightarrow \frac{| x + 3 | - 2}{x + 2} = 0$
For $x \geq - 3$ , $\frac{x + 1}{x + 2} = 0 \Rightarrow x = - 1$
For $x < - 3$ , $\frac{x + 5}{x + 2} = 0 \Rightarrow x = - 5$

$∴ x = - 5, - 1$

Suggest Corrections

thumbs-up

0

Similar questions

\frac{x^{2} + 5 x + 1}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1) (x + 2)}

+ \frac{C}{(x + 1) (x + 2) (x + 3)},

B

Q. A cubic polynomial

f (x)

x = - 2

and has a relative minimum/maximum at

x = - 1

x = \frac{1}{3}

\int x^{5} (1 + x^{3})^{2 / 3} d x = A (1 + x^{3})^{8 / 3} + B (1 + x^{3})^{5 / 3} + c

x^{3} - \frac{1}{x^{3}} = 14

x - \frac{1}{x} =

(a) 5

(b) 4

(c) 3

(d) 2

\frac{x^{2} + 3 x + 5}{(x + 1) (x + 2) (x + 3)} =

\frac{A}{(x + 1)} + \frac{B}{(x + 1) (x + 2)} + \frac{c}{(x + 1) (x + 2) (x + 3)}

then ascending order of A, B, C is

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Absolute Value Function

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon