RD Sharma Solutions Class 12 Mathematics Chapter 17 VSA

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Yes, RD Sharma class 12 chapter 17 solutions will have concepts from the entire syllabus and will include all questions in the NCERT textbooks, so that the students are not left behind in any chapter.

Q: Should I refer to NCERT Or RD Sharma's solution for self-practice?

You should refer to RD Sharma for self-practice as it contains solved questions for better performance.

Q: How much does the RD Sharma class 12 chapter 17 solution cost?

The solutions are available for zero cost and can be downloaded on the Career360 website

Q: Can I take help of the RD Sharma solution while solving homework?

Yes, as teachers provide you homework after taking assistance from this solution, therefore it helps to save time while solving.

Q: Can I trust the content present in the RD Sharma solutions?

The content present in the RD Sharma solution is 100% trustworthy as it is prepared by maths experts across the country.

RD Sharma Solutions Class 12 Mathematics Chapter 17 VSA

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 03:04 PM IST

Class 12 RD Sharma chapter 17 exercise VSA solution is the best practice material in comparison to all the CBSE textbooks available for the subject of maths. It comes with only benefits as the students get enough study material and concepts that enrich them with the best knowledge about the difficult chapter of Class 12. RD Sharma Solutions As the board exams are an important venture for class 12 students for their further career options, it is important that they get the best to ace the subject of maths.

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RD Sharma Class 12 Solutions Chapter 17 VSA Maxima and Minima - Other Exercise
Maxima and Minima Excercise: VSA
RD Sharma Chapter-wise Solutions

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

Maxima and Minima Excercise: VSA

Maxima and Minima exercise Very short answer type question 1

Answer:
$f^{\prime}(c)=0$
Solution:
We know that at the extreme points of a function f(x), the first-order derivation of the function is equal to zero.
$\begin{aligned} &f^{\prime}(x)=0 \text { at } x=c \\\\ &f^{\prime}(c)=0 \end{aligned}$

Maxima and Minima exercise Very short answer type question 2
Answer:

$f^{\prime}(c)=0 \text { and } f^{\prime \prime}(c)<0$
Solution:
We know that at the extreme points of a function f(x), the first order derivation of the function is equal to zero.
$\begin{aligned} &f^{\prime}(x)=0 \text { at } x=c \\\\ &f^{\prime}(c)=0 \end{aligned}$
Also at the point of local maximum, the second order derivative of the function at the given point must be less than zero, i.e.,
$f^{\prime \prime}(c)<0$

Maxima and Minima exercise Very short answer type question 3

Answer:
$f^{\prime}(c)=0 \text { and } f^{\prime \prime}(c)>0$
Solution:
If $f(x)$ attains a local minimum at x=c than the first order derivative of the function at the given point must be equal to zero, i.e.,
$\begin{aligned} &f^{\prime}(x)=0 \text { at } x=c \\\\ &f^{\prime}(c)=0 \end{aligned}$
The second order derivative of the function at the given point must be greater than zero, i. e.,
$f^{w}(c)>0$

Maxima and Minima exercise Very short answer type question 4

Answer: 2
Solution:
$\begin{gathered} f(x)=x+\frac{1}{x} \\\\ f^{\prime}(x)=1-\frac{1}{x^{2}} \end{gathered}$
For a local maxima or a local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &1-\frac{1}{x^{2}}=0 \\\\ &x^{2}-1=0 \end{aligned}$
$\begin{aligned} &x^{2}=1 \\\\ &x=1,-1 \end{aligned}$
$\begin{aligned} &\text { But } x>0 \\\\ &\qquad x=1 \\\\ &\text { Now } f^{\prime \prime}(x)=\frac{2}{x^{3}} \\\\ &\text { At } x=1 \end{aligned}$
$f^{\prime \prime}(1)=\frac{2}{1^{3}}=2>0$
So, x=1 is a point of local minimum. Thus the local minimum value given by
$f(1)=1+\frac{1}{1}=1+1=2$

Maxima and Minima exercise Very short answer type question 5

Answer: $-2$
Given: $f(x)=x+\frac{1}{x}$
Solution:
$\begin{aligned} &f(x)=x+\frac{1}{x} \\\\ &f^{\prime}(x)=1-\frac{1}{x^{2}} \end{aligned}$
For a local maxima or a local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &1-\frac{1}{x^{2}}=0 \\\\ &x^{2}-1=0 \\\\ &x^{2}=1 \end{aligned}$
$\begin{aligned} &x=1,-1 \\\\ &x<0 \\\\ &x=-1 \end{aligned}$
Now
$\begin{aligned} &f^{\prime \prime}(x)=\frac{-2}{x^{3}} \\\\ &\text { At } x=-1 \\\\ &f^{\prime \prime}(-1)=\frac{2}{-1^{3}}=-2<0 \end{aligned}$
So, $x = -1$ is a point of local maximum. Thus the local maximum value is given by
$\begin{aligned} &f(-1)=-1+\frac{1}{-1}=-1-1=-2 \\\\ &f(-1)=-2 \end{aligned}$

Maxima and Minima exercise Very short answer type question 6

Answer: $\left(\frac{1}{e},-\frac{1}{e}\right)$
Given: $f(x)=x \log _{\mathrm{e}} x$
Solution:
$\begin{gathered} f(x)=x \log _{\mathrm{e}} x \\\\ f^{\prime}(x)=\log _{\mathrm{e}} x+1 \end{gathered}$
For a local maxima or a local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &\log _{\mathrm{e}} x+1=0 \\\\ &\log _{\mathrm{e}} x=-1 \end{aligned}$
$\begin{aligned} &x=\frac{1}{e} \\\\ &f\left(\frac{1}{e}\right)=\frac{1}{e} \log _{e}\left(\frac{1}{\theta}\right)=-\frac{1}{e} \end{aligned}$
Now
$\begin{aligned} &f^{\prime \prime}\left(\frac{1}{e}\right)=\frac{1}{x} \\\\ &\text { At } x=\frac{1}{e} \\\\ &f^{\prime \prime}\left(\frac{1}{e}\right)=\frac{1}{\frac{1}{e}}=e>0 \end{aligned}$
So, $\left(\frac{1}{e},-\frac{1}{e}\right)$ Is a point of local maximum

Maxima and Minima exercise Very short answer type question 7

Answer: $2 \sqrt{a b}$
Given: $f(x)=a x+\frac{b}{x}$
Solution:
$\begin{aligned} &f(x)=a x+\frac{b}{x} \\\\ &f^{\prime}(x)=a-\frac{b}{x^{2}} \end{aligned}$
For a local maxima or a local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &a-\frac{b}{x^{2}}=0 \\\\ &x^{2}=\frac{b}{a} \end{aligned}$
$\begin{aligned} &x=\sqrt{\frac{b}{a}},-\sqrt{\frac{b}{a}} \\\\ &\text { But } x>0 \\\\ &x=\sqrt{\frac{b}{a}} \end{aligned}$
Now
$\begin{aligned} &f^{\prime \prime}(x)=\frac{2 b}{x^{3}} \\\\ &\text { At } x=\sqrt{\frac{b}{a}} \end{aligned}$
$f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}}=\frac{2 a^{\frac{3}{2}}}{b^{\frac{1}{2}}}>0 \ldots \ldots . . .[a>0 \text { and } b>0]$
So, $x=\sqrt{\frac{b}{a}}$ is a point of local minimum.
Hence the least value is
$f\left(\sqrt{\frac{b}{a}}\right)=a \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}$

Maxima and Minima exercise Very short answer type question 8

Answer: $(e)^{-\frac{1}{c}}$
Given: $f(x)=x^{x}$
Solution:
$f(x)=x^{x}$
Taking logarithm on both sides ,we get
$\log f(x)=\mathrm{x} \log \mathrm{x}$
Differentiating with respect to x ,we get
$\begin{aligned} &\frac{1}{f(x)} f^{\prime}(x)=\log x+1 \\\\ &f^{\prime}(x)=f(x)(\log x+1) \\\\ &f^{\prime}(x)=x^{x}(\log x+1) \ldots \ldots . .(i) \end{aligned}$
For a local maxima or local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &x^{x}(\log x+1)=0 \\\\ &\log x=-1 \\\\ &x=\frac{1}{e} \end{aligned}$
Now
$\begin{aligned} &f^{\prime \prime}(x)=x^{x}(\log x+1)^{2}+\frac{x^{x}}{x} \\\\ &=x^{x}(\log x+1)^{2}+x^{x-1} \\\\ &\text { At } x=\frac{1}{e} \end{aligned}$
$\begin{aligned} &f^{\prime \prime}\left(\frac{1}{e}\right)=\left(\frac{1}{e}\right)^{\frac{1}{e}}\left(\log \frac{1}{e}+1\right)^{2}+\left(\frac{1}{e}\right)^{\frac{1}{e}-1} \\\\ &=\left(\frac{1}{e}\right)^{\frac{1}{e}-1}>0 \end{aligned}$
So, $x=\frac{1}{e}$ is a point local minimum.
Thus the minimum value is given by
$\begin{aligned} &f\left(\frac{1}{e}\right)=\left(\frac{1}{e}\right)^{\frac{1}{e}} \\\\ &=e^{-\frac{1}{e}} \\\\ &f\left(\frac{1}{\theta}\right)=e^{-\frac{1}{e}} \end{aligned}$

Maxima and Minima exercise Very short answer type question 9

Answer: $e^{\frac{1}{e}}$
Given: $f(x)=x^{\frac{1}{x}}$
Solution:
$f(x)=x^{\frac{1}{x}}$
Taking log on both sides we get
$\log \mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}} \log \mathrm{x}$
Differentiating with respect to x , we get
$\begin{aligned} &\frac{1}{f(x)} f^{\prime}(x)=-\frac{1}{x^{2}} \log x+\frac{1}{x^{2}} \\\\ &f^{\prime}(x)=\frac{f(x)}{x^{2}}(1-\log x) \end{aligned}$
$\begin{aligned} &f^{\prime}(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\log x) \ldots \ldots \ldots(i) \\\\ &f^{\prime}(x)=x^{\frac{1}{x}-2}(1-\log x) \end{aligned}$
For a local maxima or a local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &x^{\frac{1}{x}-2}(1-\log x)=0 \\\\ &\log x=1 \\\\ &x=e \end{aligned}$
Now
$f^{\prime \prime}(x)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)^{2}+x^{\frac{1}{x}}\left(-\frac{2}{x^{3}}+\frac{2}{x^{3}} \log x-\frac{1}{x^{3}}\right)$
$=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)^{2}+x^{\frac{1}{x}}\left(-\frac{3}{x^{3}}+\frac{2}{x^{3}} \log x\right)$
$\begin{aligned} &\text { At } x=e \\ &f^{\prime \prime}(e)=e^{\frac{1}{e}}\left(\frac{1}{e^{2}}-\frac{1}{e^{2}} \text { loge }\right)^{2}+e^{\frac{1}{e}}\left(-\frac{3}{e^{3}}+\frac{2}{e^{3}} \text { loge }\right) \end{aligned}$
$=-e^{\frac{1}{e}}\left(\frac{1}{e^{3}}\right)<0$
So $x = e$ is a point of local maximum.
Thus the maximum value is given by
$f(e)=e^{\frac{1}{e}}$

Maxima and Minima exercise Very short answer type question 10

Answer: $\frac{1}{e}$
Given: $f(x)=\frac{\log x}{x}$
Solution:
$\begin{aligned} &f(x)=\frac{\log x}{x} \\\\ &f^{\prime}(x)=\frac{1-\log x}{x^{2}} \end{aligned}$
For a local maxima or a local minima we must have
$\begin{aligned} &f^{\prime}(x)=0 \\\\ &\frac{1-\log x}{x^{2}}=0 \\\\ &1-\log x=0 \end{aligned}$
$\begin{aligned} &\log x=1 \\\\ &\log x=\log e \\\\ &x=e \end{aligned}$
Now
$\begin{aligned} &f^{\prime \prime}(x)=\frac{-x-2 x(1-\log x)}{x^{4}} \\\\ &=\frac{-3 x-2 x \log x}{x^{4}} \\\\ &\text { At } x=e \quad f^{\prime \prime}(x)=\frac{-3 e-2 e \text { loge }}{e^{4}}=-\frac{5}{e^{3}}<0 \end{aligned}$
So x= e is a point local maximum.
Thus the local maximum value is given by
$\begin{aligned} &f(e)=\frac{\operatorname{loge}}{e} \\\\ &=\frac{1}{e} \end{aligned}$
$\therefore f(e)=1 / e$

RD Sharma class 12th exercise VSA will consist of 10 short answer questions from the RD Sharma class 12 solution of Minima and maxima exercise VSA of the class 12 maths book. All concepts from the Chapter will be covered in these questions which makes it essential for revision study. Some of the essential concepts covered are:-

Maximum and minimum values
Maxima and minima
Applied problems on maxima and minima

The advantages for using the RD Sharma Class 12th exercise VSA solution are listed below:-

The solutions provided in RD Sharma Class 12 Chapter 17 exercise VSA material are accurate, simple and basic according to the CBSE syllabus of class 12. Therefore a thousand students in the country trust the solutions for their understanding of maths.
If you face any problem while solving questions from the NCERT, then you can refer to the RD Sharma Class 12th exercise VSA material and can also use the solution for revision as it contains solved questions.
You can download the RD Sharma class 12th exercise VSA from the Career360 website through any device while sitting in any corner of the country.
You can use RD Sharma class 12 solutions chapter 17 exercise VSA for solving homework as teachers use these solutions to provide tasks to solve because the questions present in the solution are hand-picked by experts for better performance
RD Sharma class 12th exercise VSA is available for all students for free on the Career360 website.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Does the material of RD Sharma contain essential concepts?

Yes, RD Sharma class 12 chapter 17 solutions will have concepts from the entire syllabus and will include all questions in the NCERT textbooks, so that the students are not left behind in any chapter.

2. Should I refer to NCERT Or RD Sharma's solution for self-practice?

You should refer to RD Sharma for self-practice as it contains solved questions for better performance.

3. How much does the RD Sharma class 12 chapter 17 solution cost?

The solutions are available for zero cost and can be downloaded on the Career360 website

4. Can I take help of the RD Sharma solution while solving homework?

Yes, as teachers provide you homework after taking assistance from this solution, therefore it helps to save time while solving.

5. Can I trust the content present in the RD Sharma solutions?

The content present in the RD Sharma solution is 100% trustworthy as it is prepared by maths experts across the country.