Test: Differentiability(2 Sep) - JEE MCQ

Concept:
The domain is the subset of R for which all operations in the function's formula make sense.
The derivative of a function at a given point is the slope of the tangent line at that point.
Formulae

Calculation:
Statement I: Its domain is (-∞, ∞).
The domain is the subset of R for which all operations in the function's formula make sense.
Since 10 is a positive real constant, it can be raised to any real power, so the domain is not limited. It is R.
So, Its domain is (-∞, ∞).
Statement II: It is a continuous function
We know that f(x) = 10^x 
Derivative of f(x) is f'(x) = 10^x log 10
And now we have to find the points at which the derivative of f(x) is zero.
⇒ f'(x) = 10^x log 10 = 0 
As, we know that this can't happen.
So, 10x is a continuous function for all real numbers.
Statement II: It is differentiable at x = 0
The derivative of a function at a given point is the slope of the tangent line at that point.
Here, f(x) = 10^x 
⇒ f'(x) = 10^x log 10
At x = 0, f'(0) = 10º log 10 ⇒ log 10
The derivative of function f(x) = 10^x at x = 0 is log 10.
So, It is differentiable at x = 0.
∴ Statement I, II, and III are correct.

Modulus function:
It is defined as f : R → R is a function such that f(x) = |x| = 

• Modulus function is continuous everywhere.
• It is differentiable everywhere except at x = 0.

Calculation:
Given, f(x) = |sin x|
f(x) is not differentiable at sin x = 0
⇒ x = π, 2π,⋯ = nπ, n ∈ Z
∴ f(x) is not differentiable at x = nπ, n ∈ Z.
∴ If f(x) =|sin x|, then f(x) is everywhere continuous but not differentiable at x = nπ, n ∈ Z.
The correct answer is Option 2.

Concept:
A function f(x) is differentiable at a point if the left derivative and right derivative are equal at the point.
Calculation:
Given, f(x) = x|x| = 
f(0⁺) = 0
f(0^-) = 0
f(0) = 0
Left limit = Right limit = Function value = 0 (Real and finite)
∴ f(x) is continuous at x = 0
Now, f'(x) = 
f'(0⁺) = 0
f'(0^-) = 0
Left derivative = Right derivative = 0 (Well defined)
∴ f(x) is differentiable at x = 0.
∴ The function f(x) = x|x| is differentiable at the origin.
The correct answer is Option 3.

Let's check continuity at x = 0

and f(0) = 0
Since L.H.L = R.H.L = f(0) so f(x) is conitnous at x = 0
and f(x) is continuous in [-1, 1] 
Now, 
 does not exist finitely.
f(x) is not differentiable at x = 0
Hence f(x) is continuous in [-1, 1] and differentiable in (-1, 0) ∪ (0, 1)  (2) is correct

 f(x) = |cos x| 

From the figure, we can say that f(x) is continuous everywhere.
and not differentiable where cos x = 
Hence (2) is correct.

Calculation:
Given:
f(x³) = x⁵ for all x ∈ R, x ≠ 0. Then the value of 
⇒ f (x³) = x⁵ for all x ∈ R, x ≠ 0
Differentiating w.r.to x, we get:
(d/dx) f (x³) = (d/dx) x⁵
Using Product rule of differentiation, we can write:
f' (x³) (d/dx) x³ = 5x⁴ 
f' (x³) 3x² = 5x⁴ 
f'(x³) = (5/3) x² 
f' (8) can be written as f' (2³), i.e. by putting x = 2 in the above expression, we can write:
f'(8) =( 5 × (2)²)/3 = 20/3
⇒ 20/3

Calculation:
Given: f(x) = xⁿ, n ≠ 0.
F’(x) = nx^{n – 1}
For f(x) to be differentiable for all values of x, n – 1 ≥ 0
So, n ≥ 1.

Concept:
If f(x) is differential at a point x = a then f(x) is also continuous at x = a and left hand limit at that point is equal to right hand limit at that point. And if left hand limit is not equal to right hand limit then function is discontinuous and hence not differentiable at that point
Calculation:

We will find the left and right hand limit at x = 0 to check if function f(x) is continuous at x = 0
(L.H.L at x = 0)

As , we can see that square-root of negative number is not defined. So, limit does not exist.
(R.H.L at x = 0)

= 0
So. here left hand limit at x = 0 is not equal to right hand limit at x = 0.
So, the function is discontinuous and also not differentiable at x = 0
Hence, f(x) is not differential at x = 0.

Concept:
Differentiability of a function:
We will define the differentiability of the function with the help of its graph. If a graph of a function is smooth everywhere then the function is said to be differentiable everywhere in the xy - plane.
If the function has a sharp edge or a vertical asymptote (tangent) to the curve then the function is not differentiable at that point.
Calculation:

Observe the graph of the function f(x) = |x|.
At x = 1 the curve is smooth and thus the function is differentiable at x = 1.
On the other hand the function is not differentiable at x = 0 as it has a corner at x = 0.

Observe the graph of the function f(x) = e^x.
At x = 0 the curve is smooth so it is differentiable at x = 0.
Thus only the second statement is true.

Greatest Integer Function: (Floor function)
The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.
The domain of [x] is R and the range is I.

Note:
Any function is differentiable only if it is continuous.
The floor function f(x) = ⌊x⌋ is differentiable in every open interval between integers, (n, n + 1) for any integer n.
Calculation:
Given that,
y = [x]
Statement:1  Its derivative is 0 at x = 0.5
We know that the floor function is differentiable at all points except integer points.
Hence, y = [x] is differentiable at x = 0.5
⇒ y = [0.5] = 0
⇒ dy/dx = 0
Statement:2  It is continuous at x = 0
We know that y = [x] is only continuous in the open interval between integers and discontinuous at all integer values.
∴ Only statement 1 is correct.