Here we provide RD Sharma Class 12 Ex 18.1 Solutions Chapter 18 Maxima and Minima for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 18.1 Solutions Chapter 18 Maxima and Minima book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 18 |
| Exercise | 18.1 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 18.1 Solutions Chapter 18 Maxima and Minima
Question 1. Find the maximum and the minimum values, if any, without using derivatives of the function f (x) = 4x2 – 4x + 4 on R
Solution:
Given in the question f(x) = 4x2 – 4x + 4 on R
= 4x2 – 4x + 1 + 3
If we group the above equation we will get,
= (2x – 1)2 + 3
⇒ (2x – 1)2 ≥ 0
= (2x – 1)2 + 3 ≥ 3
= f(x) ≥ f (½)
So we can say that, the minimum value of f(x) is 3 at x = ½
Since, f(x) can be infinite. So maximum value does not exist.
Question 2. Find the maximum and the minimum values, if any, without using derivatives of the function given by f (x) = –(x – 1)2 + 2 on R
Solution:
Given function in the question is f(x) = – (x – 1)2 + 2
We can observe that (x – 1)2 ≥ 0 ∀ x ∈ R
Therefore, f(x) = – (x – 1)2 + 2 ≤ 2 ∀ x ∈ R
The maximum value of function can be attained when
⇒ (x – 1) = 0
⇒ (x – 1) = 0, x = 1
Maximum value of function = f(1) = – (1 – 1)2 + 2 = 2
So we can say that function f does not have minimum value.
Question 3. Find the maximum and the minimum values, if any, without using derivatives of the function given by f (x) = |x + 2| on R
Solution:
Given function in the question is f(x) = |x + 2| on R
⇒ f(x) ≥ 0 ∀ x ∈ R
The minimum value of f(x) = 0, which can be attained at x = -2
So We can say that, f(x) = |x + 2| does not have the maximum value.
Question 4. Find the maximum and the minimum values, if any, without using derivatives of the function given by f (x) = sin2x + 5 on R
Solution:
Given function in the question is f(x) = sin2x + 5 on R
As we know that range of sin2x is [-1,1].
Adding 5 both side.
⇒ – 1 + 5 ≤ sin2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
So we can say that, the maximum value is 6 and minimum value is 4.
Question 5. Find the maximum and the minimum values, if any, without using derivatives of the function given by f (x) = |sin 4x + 3| on R
Solution:
Given function in the question is f(x) = |sin 4x + 3| on R
As we know that range of sin 4x is [-1,1]
Adding 3 both side:
⇒ 2 ≤ sin 4x + 3 ≤ 4
⇒ 2 ≤ |sin 4x + 3| ≤ 4
So we can say that, the maximum value is 4 and minimum value is 2.
Question 6. Find the maximum and minimum values, if any without using derivatives of the function given by f(x) = 2x3 + 5 on R.
Solution:
f(x) = 2x3 + 5 on R.
We can observe that the values of f(x) increased when the values of x are increased.
f(x) can be as large as possible.
So we can conclude that , f(x) does not have the maximum value.
Also, f(x) can be made small by giving smaller values to x.
So we can say that f(x) does not have the minimum value.
Question 7. Find the maximum and minimum values, if any without using derivatives of the function given by g(x) = -| x + 1 | + 3
Solution:
We have given this equation g(x) = -| x + 1 | + 3
As we know that -| x + 1 | ≤ 0 ∀ x ∈ R
⇒ g(x) = -| x + 1 | + 3 ≤ 3 ∀ x ∈ R
The maximum value of function g is attained at | x + 1 | = 0
| x + 1 | = 0
⇒ x = -1
⇒ Maximum value of g = g(-1) = – | -1 + 1 | + 3 = 3
So we can conclude that function g does not have minimum value.
Question 8. Find the maximum and minimum values, if any without using derivatives of the function given by f(x) = 16x2 – 16x + 28 on R.
Solution:
We have given this equation f(x) = 16x2 – 16x + 28 on R.
= 16x2 – 16x + 4 + 24
= (4x – 2)2 + 24
⇒ (4x – 2)2 ≥ 0 ∀ x ∈ R
⇒ (4x – 2)2 + 24 ≥ 24 ∀ x ∈ R
⇒ f(x) ≥ f(½)
So the minimum value of f(x) is 24 at x = ½
So we can say that, f(x) can be made as large as possible by giving different values to x.
So we can conclude that, maximum values does not exist.
Question 9. Find the maximum and minimum values, if any without using derivatives of the function given by f(x) = x3 – 1 on R.
Solution:
We have given this equation f(x) = f(x) = x3 – 1 on R.
We can observe that the values of f(x) increases when the values of x are increased.
f(x) can be made as large as possible by putting values of x.
So, we can say that f(x) does not have the maximum value.
Also, f(x) can be made as small as giving by smaller values to x.
So we can conclude that f(x) does not have the minimum value.
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