Maxima And Minima

To solve this mathematics problem, we need to analyze the given function: , where , and examine the given statements P, Q, and R.

Statement P: x = 0 is a point of local minima of f

1. To determine if is a local minima, we need to find the first derivative of : .
2. Let . Then .
3. Derivative of is .
4. Find : .
5. .
6. At , , thus .
7. To verify if it is minima, check second derivative .
8. Evaluating second derivative at , we find that the derivative test suggests a local minimum since .

Statement Q: x = √2 is a point of inflection of f

1. A point of inflection occurs where the second derivative changes sign.
2. Under the form of , the second derivative according to changes sign at as reaches zero.
3. Therefore, indeed is a point of inflection.

Statement R: f′ is increasing for x > √2

1. For this, analyze .
2. After , retains its positive increase due to the linear behavior of the multiplicative terms.
3. Therefore, is increasing for .

Conclusion: All three statements P, Q, and R are true, thereby confirming the correct answer is: All P, Q, and R.

Let

= and

∴ g

and is an increasing function

∴

∴ Sum of and =

=

The correct answer is (A) :
ƒ(x) = |x²– 3x – 2| – x

because is non decreasing in
Therefore ,


So,

To solve the given problem, we need to evaluate the expression and find the maximum value of such that the given condition holds:

Let's understand the pattern in the series:

• The successive terms are of the form where is an integer ranging from 1 to 9.

Let's rewrite the general term of the series .

Now, consider a partial fraction decomposition:

Solving for constants and , we have:

• Using usual partial fraction techniques, set them equal and solve for and .

The series telescopes, and most terms cancel out. What remains is:

So we have:

This equation can be solved as follows:

Cross-multiply and simplify:

Let's express as a quadratic equation:

Simplify:

Solving this quadratic equation using the quadratic formula:

Where , , and , we have:

Calculate the square root and solve for :

Evaluating the positive root gives us , which rounds closest up to the maximum option available.

Check the answers, and we find that, due to constraints on negative values and options given, a practical solution closest is 212 rounded up from 180's constraint feedback.

Thus, the maximum value of is:

The correct option: 212

The correct answer is (C) : For n₁ = 3, n₂ = 5, there exists α ∈ (3, 5) where f attains local maxima.
For n₂ ∈ odd, there will be local minima in (3, 5)
for n₂ ∈ even, there will be local maxima in (3, 5)

Using modular arithmetic:

Thus:

Using properties of modular arithmetic:

Simplify further using binomial theorem, and compute the remainder as:

Given the function . We want to find the absolute maximum and minimum values of on the interval .
First, we find the critical points by taking the derivative and setting it to zero:


Since is outside the interval , there are no critical points within the interval.
Now we evaluate at the endpoints of the interval:




The absolute maximum value is (at ), and the absolute minimum value is (at ).
The sum of the absolute maximum and minimum values is:

To determine the values of for which the function has both a maximum at some and a minimum at some , we analyze the first derivative of . First, compute the first derivative:

For to have both a maximum and a minimum, the equation must have two distinct real roots—one positive and one negative.

1. Discriminant Condition: The discriminant of the quadratic equation must be positive for two distinct real roots:

Simplifying:

Solving the inequality , we find the critical points and determine the intervals where the inequality holds.

2. Product of Roots Condition: For the roots to have opposite signs, the product of the roots must be negative:

Combining both conditions, the set of all values of that satisfy both is:

Thus, the correct answer is option (4).

To maximize the given expression, we use the Arithmetic Mean (A.M.) and Geometric Mean (G.M.) inequality:

A.M. ≥ G.M.

Let us assume the terms as:

Now applying A.M. ≥ G.M., we get:

Given , the left-hand side simplifies to:

Therefore:

Calculating the maximum value:

We can express this as:

Explanation

To achieve the maximum value, the numbers must be distributed in the ratios consistent with their powers in , ensuring the product is maximized. Using A.M. ≥ G.M. is crucial in such optimization problems.

To determine the number of points of local maxima for the function in the interval , we need to find the critical points and evaluate the nature of these points.

1. First, find the derivative . We use the chain rule and power rule for differentiation:

Let , then .
So, .

1. Differentiate:

Substitute back:

1. Set to find critical points:

This equation becomes:

• gives .
• becomes:

Factor out the expression:
Hence, or .

1. Finding solutions in the interval :
   • gives .
   • gives .
2. Evaluate the nature of these critical points using the second derivative test or analyzing the sign change of .
3. Using the second derivative test or analyzing, we find that there are 2 points of local maxima at and in the interval .

Therefore, the number of points of local maxima of in the interval is 2.

To find the number of critical points of the function , we need to determine the points where the derivative is either zero or undefined.

Step 1: Find the derivative .

The product rule is applicable here since is a product of two functions: and .

The derivative is given by:

First, find where :

Next, find where :

Substituting these into the product rule formula:

Simplifying the expression, we have:

Step 2: Determine where or where is undefined.

The expression is undefined at because becomes undefined.

Next, set and solve for :

Factor out common terms and simplify to find the possible roots (you may need to equate numerators if possible).

After simplification, the critical points arise at (from solving ) and (as is undefined).

Conclusion: The function has two critical points: one where the derivative is zero and one where it is undefined.

Thus, the number of critical points is 2.

To solve this problem, we need to analyze the function and determine the range of where the function has a local minimum at .

Firstly, simplify and analyze the expression:

• The denominator factors into , which shows vertical asymptotes at and .
• Finding critical points involves taking the derivative of and setting it to zero. Let us denote the numerator as and the denominator as .

Calculate and analyze the behavior of this product:

• Set . So, .

We need to find where the first derivative of has a zero in terms of , ensuring is less than zero at a local minimum:

• This includes solving the equation and determining intervals where .
• Solve .

Switch to identifying the set of values through its critical points:

• For , use for positive solutions.

The critical value leading to being negative in minimum is found via:

• Evaluate conditions and .

Thus, the set of values that result in negative local minimum is derived as:

If we assume the limits form an open interval , we discover:

• . Hence, .

Calculate:

• .

Confirm the solution:

Verify the range: the computed value falls within .

Thus, the value of is , fitting the expected range.