RD Sharma Solutions Class 12 Mathematics Chapter 17 VSA

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

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