#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# If the real number x satisfying the inequality 2x-1

Ankit Sharma
26 Points
one month ago
Given: |2x-1| $\leq$  X+1
Firstly break the modulus at its breaking point.
2x-1=0
X=1/2.
NOW,
For, X $\fn_cm \leq$ 1/2, 2X-1 is a negative so, modulus open with negative sign,equation is like that,
-2x+1$\fn_cm \leq \$  X+1
On solving we get, X$\fn_cm \geq$0
Taking intersection with Above applied conditions we get 0$\fn_cm \leq$ X$\fn_cm \leq$1/2 .... (eqn 1)

Now, similarly for X$\fn_cm \geq$ ½
Modules opens with +sign.
So we get here X$\fn_cm \leq$ 2
AGAIN Taking intersection with just above applied condition
We get 1/2 $\fn_cm \leq$ X$\fn_cm \leq$ 2          ...... (eqn 2)
Now take union of above two equation
We get 0$\fn_cm \leq$ X$\fn_cm \leq$2 and this is final answer
Alternative approach :::
Just draw graph of above functions then judge.