RD Sharma Class 12 Exercise 17.1 Maxima And Minima Solutions Maths

RD Sharma Class 12 Exercise 17.1 Maxima And Minima Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:21 PM IST

Students of class 12 will deal with tough competition from peers as everyone wants to score well and top the charts in board exams. However, maths has a tedious syllabus and chapters like ‘Maxima and Minima’ will prove to be extremely challenging for students. In order to get a clear understanding of the chapter, students should make use of the Class 12 RD Sharma chapter 17 exercise 17.1 solution to practice at home and keep track of their daily lessons. The RD Sharma class 12th exercise 17.1 of Maxima and Minima of class 12 will cover complex topics and concepts like:-

Maximum and minimum values.
Maximum and minimum values exist or not.
Differentiation to find maximum and minimum values.

RD Sharma Solutions Students should start their exam preparations beforehand with the RD Sharma class 12th exercise 17.1 to build a firm grip on the 17th chapter of maths and score their desired result in boards. The RD Sharma class 12 solution of Minima and maxima exercise 17.1 will contain 9 questions which are divided into various subparts.

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise

Maxima and Minima Excercise: 17.1

Maxima and Minima exercise 17.1 question 1

Answer:

The minimum value is 3 and the maxima value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Explanation:
$f(x)=4 x^{2}-4 x+4 \text { on } R$
We have,
$\begin{aligned} &f(x)=4 x^{2}-4 x+4 \text { on } R \\ &=4 x^{2}-4 x+1+3 \\ &=(2 x-1)^{2}+3 \\ &\because(2 x-1)^{2} \geq 0 \\ &(2 x-1)^{2}+3 \geq 3 \\ &f(x) \geq f\left(\frac{1}{2}\right) \end{aligned}$
Thus, at x = 1/2 , minimum value of f(x) is 3.
Since the value of f(x) is increasing rapidly. That is why it does not attain the max value.

Maxima and Minima excercise 17.1 question 2

Answer:

The maximum value is 2 and the minimum value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=-(x-1)^{2}+2 \text { on } R$
Explanation:
$f(x)=-(x-1)^{2}+2$
We can see that,
$\begin{aligned} &(x-1)^{2} \geq 0 \text { for every } x \in R \\ &-(x-1)^{2} \leq 0 \\ &-(x-1)^{2}+2 \leq 2 \text { for every } \mathrm{x} \in R \end{aligned}$
So, the maximum value of f is attained when (x-1) = 0
x-1 = 0
x = 1
Thus, maximum value of f(x) = f(1)
$\begin{aligned} &f(1)=-(1-1)^{2}+2 \\ &=2 \end{aligned}$
Therefore, maximum value is 2 and it does not have minimum value.

Maxima and Minima exercise 17.1 question 3

Answer:

Minimum value = 0 and maximum value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$\begin{aligned} &f(x)=|x+2| \text { on } R \\ &\because|x+2| \geq 0 \text { for } x \in R \\ &f(x) \geq 0 \text { for all } x \in R \end{aligned}$
Minimum value of f(x) is 0 at x = - 2 and it does not have maximum value.

Maxima and Minima excercise 17.1 question 4

Answer:

Maximum value is 6 and minimum value is 4.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=\sin x+5 \text { on } R$
Explanation:
We have,
$f(x)=\sin 2 x+5$
We know,
$\begin{aligned} &-1 \leq \sin x \leq 1 \\ &-1+5 \leq \sin 2 x+5 \leq 1+5 \\ &4 \leq \sin 2 x+5 \leq 6 \end{aligned}$
So, the minimum value is 4 and maximum value is 6.

Maxima and Minima excercise 17.1 question 5

Answer:

Maximum value is 4 and minimum value is 2.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=|\sin 4 x+3|$
Explanation:
We have,
$f(x)=|\sin 4 x+3|$
We know,
$\begin{aligned} &-1 \leq \sin 4 x \leq 1\\ &-1+\underline{3} \leq \sin 4 x+3 \leq 1+3\\ &\underline{2} \leq \sin 4 x+3 \leq 4\\ &\underline{2} \leq|\sin 4 x+3| \leq 4 \end{aligned}$
So maximum value is 4 and minimum value is 2.

Maxima and Minima exercise 17.1 question 6

Answer:

Minimum value and maximum value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=2 x^{3}+5 \text { on } R$
Explanation:
We have,
$f(x)=2 x^{3}+5 \text { on } R$
We know,
f(x) increase when value of x is increases
In this case, the value of f(x) increases rapidly, so it does not attain maximum value.
Also, f(x) can be made as small as possible. So it does not attain minimum value.
Hence, given function does not have maximum value and minimum value.

Maxima and Minima exercise 17 point 1 question 7

Answer:

Maximum value is 3 and minimum value does not exist.

Hint:

f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.

Given: Also, see,

Explanation:

We have,

$f(x)=-|x+1|+3 \text { on } R$

We have,

$\begin{aligned} &-|x+1| \leq 0 \text { for every } x \in R\\ &-|x+1|+3 \leq 3 \text { for every } x \in R\\ \end{aligned}$

Thus maximum value of f is attained when |x+1|=0

X = -1

So maximum value of f(x) = f(-1) = - |-1+1| + 3

So its maximum value is 3 and it does not have minimum value.

Maxima and Minima excercise 17.1 question 8

Answer:

Minimum value is 24 and maximum value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=16 x^{2}-16 x+28 \text { on } R$
Explanation:
We have,
$\begin{aligned} &f(x)=16 x^{2}-16 x+28 \text { on } R\\ &=16 x^{2}-16 x+4+24\\ &=(4 x-2)^{2}+24\\ &\text { Now, }(4 x-2)^{2} \geq 0 \text { for all } \mathrm{x} \in R\\ &(4 x-2)^{2}+24 \geq 24 \text { for all } \mathrm{x} \in R\\ &f(x) \geq f\left(\frac{1}{2}\right) \end{aligned}$
So, minimum value of f(x) is 24 at $x=\frac{1}{2}$
Since, value of f(x) increases rapidly. So it does not have maximum value.

Maxima and Minima excercise 17.1 question 9

Answer:

Minimum value and maximum value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=x^{3}-1 \text { on } R$
Explanation:
We have,
$f(x)=x^{3}-1$
We can see that, the value of f(x) increases rapidly. So, it does not attain maximum value.
Also, f(x) can be made as small as possible. So it does not attain minimum value.
Hence, the given function does not have maximum value and minimum value.

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