Continuity MCQ Quiz - Objective Question with Answer for Continuity - Download Free PDF

Concept:

A function f(x) is continuous at x = a, if the function is defined at x = a and,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Analysis:

This function is not defined for:

x² + 3x – 4 = 0

⇒ (x + 4)(x - 1) = 0 i.e.

At x = 1 and x = - 4

∴ This function f(x) is not continuous at x = 1, -4

Concept:

A function is said to be continuous if 

Calculation:

Given:

f(x) = (x + 1)cotx 

Taking log on both sides

log (f(x)) = cot x log (x + 1)

For checking continuity at x = 0

 = 0/0 form

Using L-Hospital rule

⇒ log (f(0)) = 1

⇒ f(0) = e¹ = e

Concept:

A function is written in the form of the ratio of two polynomial functions is called a rational function.

Rational functions are continuous at all the points except for the points where the denominator becomes zero.

where P(x) and Q(x) are polynomials and Q(x) ≠ 0.

f(x) will be discontinuous at points where Q(x) = 0.

Calculation:

Given:

This is a rational function, so it will be discontinuous at points where the denominator becomes zero.

4x - x³ = 0

x(4 - x²) = 0

x(2² - x2) = 0

x(2 + x)(2 - x) = 0

x = 0, x = - 2 and x = 2

Hence the function  will be discontinuous at exactly three points 0, - 2 and 2.

Mistake Points

There may be a doubt that some factors are eliminating each other so first, we have to simplify this. 

Note that, 

If (4 - x2) = 0 then the f(x) will come indeterminant or 0/0 form.

So, the function will not have any value for x  = ± 2. So, these will also be the points of discontinuity.

Also, if  (4 - x2) ≠ 0

⇒ 

Here, x = 0 is also the point of discontinuity.

There will be  exactly three points 0, - 2 and 2.

Given that,

A function f(x) is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

i.e.,

Analysis:

f(x) = x², x ≥ 0

= -x², x ≤ 0

f^'(x) = 2x, x ≥ 0

= -2x, x

f^'(x) = 2|x|

f^’(x) is continuous but not differentiable at x = 0.

Hence f(x) is differentiable but its first derivative is not differentiable at x = 0.

Option(4) is the correct answer.

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |x|

|x| = x for x ≥ 0

|x|= -x for x

At x = 0

Left limit = 0, Right limit = 0, f(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 0.

Now

Left derivative (at x = 0) = -1

Right derivative (at x = 0) = 1

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 0

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if,

  exists or its graph is a single unbroken curve.

In other words,

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

Analysis:

0} \end{array}} \right.\)

For x ≤ 0, f(x) = 1 – x

g (x) = -x

(fog)(x) = f(g(x)) = f(-x) = 1 + x

y = 1 + x is a continuous function.

Hence, the number of discontinuities = 0

We know that, (Intermediate value theorem)

If  then f(x) has at least one root in (a, b)

Since there is no root in [a, b] this implies f(a) and f(b) are of same sign 

i.e. either they both are positive or they both are negative

In both cases  0\)

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |2 − 3x|

|2 − 3x| = 2 - 3x for x

|2 − 3x| = -2 + 3x for x ≥ 2/3

At x = 2/3,

Left limit = 0, Right limit = 0, f(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 2/3.

Now

Left derivative (at x = 2/3) = -3

Right derivative (at x = 2/3) = 3

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 2/3

Concept:

Given:

A function will be discontinuous at the point where the denominator becomes zero.

X² – 1 will become zero at x = +1 and x = -1

In the numerator, ln(x) is there and it is not defined for negative values.

∴ Function f(x) will be continuous in the interval  (0, 1),(1, ∞)

We know that

(a + b + c)² ≥ 0

a² + b² + c² + 2(ab + bc + ac) ≥ 0

1 + 2(ab + bc + ac) ≥ 0

(ab + bc + ac) ≥ -1/2

(a - b)² + (b - c)² + (c - a)² ≥ 0

2(a2 + b2 + c2) - 2(ab + bc + ac) ≥ 0

(a2 + b2 + c2) - (ab + bc + ac) ≥ 0

(1) - (ab + bc + ac) ≥ 0

(ab + bc + ac) ≤ 1

⇒ -1/2 ≤ (ab + bc + ac) ≤ 1

(ab + bc + ac) ∈ [-1/2, 1]