Types Of Functions

Step 1: Differentiating the given function We are given the following condition: Differentiating both sides with respect to , we use the chain rule on the left-hand side: By the Leibniz rule for differentiation under the integral sign, we get: Step 2: Solving for Now solving for , we get: Step 3: Substituting We need to find . Using the equation , we substitute : Step 4: Final Answer The correct answer is , as calculated from the equation for .

The given inequality is:
From this, we obtain the range:

Additionally, we consider:
Breaking this into cases, we solve:
This gives: And:

Combining all conditions from (1), (2), and (3):

From this range: Finally, we conclude: is a small positive quantity and .

Let:

Take the natural logarithm on both sides:

Differentiate both sides with respect to :

Simplify the derivative:

For local maxima or minima, set :

Factorize:

For non-zero solutions:

Hence:

The corresponding local maximum value of is:

Equating powers:

Using the given conditions:

Given Function:

for , where is an integer and is the greatest integer less than or equal to .

Since , we have , and thus .

The greatest integer function is discontinuous at integer values of . So, will be discontinuous when takes integer values from 1 to 13.

For each integer from 1 to 12, the equation has two solutions in the interval .

The equation has only one solution in the interval , which is .

Therefore, the number of points of discontinuity is:

The function is not differentiable at 25 points in .

We need to check for discontinuity at doubtful points .

At :

At :

At :

From the above calculations, discontinuity occurs at two points.