Test: Continuity - JEE MCQ

= 1 x 2
= 2 ≠ f(0)
Hence, given function is not continuous at x = 0

The three conditions required for a function f is said to be continuous on (a,b) if f is continuous at each point of (a,b), f is right continuous at x = a, f is left continuous at x = b.

and

and
f(2) = 1.
Therefore , f(x) is discontinuous at x = 2

Here, f(x) = [X] could be expressed graphically as:

From the graph it is clear that the function is discontinuous.

Set the denominator in:
equal to 0 to find where the expression is undefined.

Subtract 2 from both sides of the equation.

The domain is all values of x that make the expression defined.
Interval Notation:

Set-Builder Notation:

Since the domain is not all real numbers, f(x) is not continuous over all real numbers. 
f(x) is discontinuous at x = -2.

For continuity: LHL = RHL

At x=2,
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2

At x = 0,
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0

f(x) = |x-1| + |x+1| , x € R

► For x ≥ 1,
f(x) = x-1 + x+1
f(x) = 2x

► For -1 ≤ x ≤ 1,
f(x) = -x + 1 + x+1
f(x) = 2

► For x ≤ -1,
f(x) = -x + 1 - x-1
f(x) = -2x


As the graph of f(x) shows, the function is continuous throughout its domain.

Lim f (x) = lim (x-1)(x-2)  at x tend to k 

► So it get    k²-3k+2

► Now f (k) = k² -3k+2

► So f (x) =f (k) so continous at everywhere

Sgn(x) = |x|/x
► for x > 0, sgn(x) = 1
► for x = 0, sgn(x) = 0
► for x < 0, sgn(x) = -1
► Now see graphically or theoretically sgn(x) is discontinuous at x = 0

• [-1,1] cannot be continuous interval because log is not defined at 0.
• The value of x cannot be greater than 1 because then the function will become complex.
• (0,1) will not be considered because its continuous at 1 as well. Hence D is the correct option.