Differentiability

Break the function based on and Key points where non-differentiability occurs: - At : due to - At : due to piece - Additional critical points due to behavior of : test derivative left and right Total: 5 points of non-differentiability when

The function is not differentiable at the points where the functions inside the are equal or where the transitions from 0 to a non-zero value or vice versa.
First, consider where .
This occurs when , which in the interval are and .
Next, consider where .
This occurs at in .
Finally, consider where .
This occurs at in .
Now, let's analyze the sign of and in : - In , and .
, non-differentiable at .
- In , and .
, non-differentiable at (transition from 0).
- In , and .
, differentiable.
- In , and .
, non-differentiable at (transition to 0).
In the interval , the points of non-differentiability are .
There are 3 such points.
The interval is .
This interval contains 1012 cycles of .
In each cycle of , there are 3 points of non-differentiability.
So, the total number of points of non-differentiability in is .
Given that the number of points is , we have , which implies .

Assertion (A) states that if is not continuous at , then it is not differentiable at . This is true because for a function to be differentiable at a point, it must first be continuous at that point.
Reason (R) states that if is differentiable at a point, then it is continuous at that point. This is also true because differentiability implies continuity.
Thus, both assertion and reason are true, and reason correctly explains the assertion.

We are given the condition . This can be rewritten as: Substitute into this equation: This equation is trivially true, so it doesn't provide new information. But, we are given that , and we need to use the given information about . Now, consider . Since the relationship holds for , we can deduce that: Thus, the correct value is .

We are given: To apply Rolle's theorem on , the function must be continuous, differentiable on and . Check continuity and differentiability: Since is a polynomial, it is continuous and differentiable everywhere including . Check and : Given: Dividing both sides by 6: So: Therefore, Rolle's theorem is applicable on . Now, Statement-II is true. Check Statement-I: By Rolle’s theorem: If a continuous and differentiable function has equal values at the endpoints of a closed interval, then there exists at least one point in where the derivative is zero. Compute: Then by Rolle’s theorem, there exists at least one point such that: Which means the quadratic equation has at least one real root in . Thus, Statement-I is also true. And since Statement-II (Rolle’s theorem applicable) is the reasoning for Statement-I (existence of root in (0,1)), Statement-II is the correct explanation of Statement-I. Therefore, the correct answer is option (4).

We use the formula .
Term 1: .
Term 2: .
So, .
Now differentiate with respect to .
Recall .
This can be rewritten as .
This matches option (2).

Differentiating and with respect to : Computing : Differentiating again: Substituting : Multiplying by the exponential factor: Thus, the correct answer is:

Let’s find: Tedious derivation — instead verify LHS and RHS via identity: Use identity:

We use the formula .
Term 1: .
Term 2: .
So, .
Now differentiate with respect to .
Recall .
This can be rewritten as .
This matches option (2).