Question 13 Marks
Show that $\frac{\text{logx}}{\text{x}}$ has a minimum value at x = e.
AnswerHere, $\text{f}(\text{x})=\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\frac{1-\log\text{x}}{\text{x}^{2}}$
For the local maxima or minima, We have f'(x) = 0
$\Rightarrow\frac{1-\log\text{x}}{\text{x}^{2}}=0$
$\Rightarrow1=\log\text{x}$
$\Rightarrow\log\text{e}=\log\text{x}$
$\Rightarrow\text{x}=\text{e}$
Now, $\text{f}''(\text{x})=\frac{\text{x}^{2}(\frac{-1}{\text{x}})-2\text{x}(1-\log\text{x})}{\text{x}^{4}}=\frac{-3+2\log\text{x}}{\text{x}^{3}}$
$\Rightarrow\text{f}''(\text{e})=\frac{-3+2\log\text{x}}{\text{x}^{3}}=\frac{-1}{\text{e}^{3}}<0$
So, x = e is the point of local maximum.
View full question & answer→Question 23 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}}, \text{x}\geq0$
AnswerGiven, $\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}}$
$\text{f}'(\text{x})=\sqrt{1-\text{x}}-\frac{\text{x}}{2\sqrt{1-\text{x}}}=\frac{2-3\text{x}}{2\sqrt{1-\text{x}}}$
For the local maxima or minima, We must have f"(x) = 0
$\Rightarrow\frac{2-3\text{x}}{2\sqrt{1-\text{x}}}=0$
$\Rightarrow\text{x}=\frac{2}{3}$
Since, f'(x) changes from positive to negative when x increases through $\frac{1}{2}, \text{x}=\frac{2}{3}$ is of f(x) at $\text{x}=\frac{2}{3}$ is given by $\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}$
View full question & answer→Question 33 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\text{x}^{3}-6\text{x}^{2}+9\text{x}+15$
AnswerGiven,
$f(x) = x^3 - 6x^2 + 9x + 15$
$\Rightarrow f'(x) = 3x^2- 12x + 9$
For a local maximum or a local minimum, We must have $f'(x) = 0$
$\Rightarrow 3x^2- 12x + 9 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow (x - 1)(x - 3) = 0$
$\Rightarrow x = 1$ or $3$
Since$ f'(x)$ change from negative to positive when x increases through $3,$
$x = 0$ is the point of iocal minima.
The local minimum value os $f(x) = 3$ at $x = 3$ is given by $(3)^3- 6(3)^2+ 9(3) + 15 = 27 - 54 + 27 + 15 = 15$
Since f'(x) changes from positive to negative when x increases though $1, x = 1$ is point of local maxima.
The local maximum value of $f'(x)$ at $x = 1$ is given by ($1)^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$
View full question & answer→Question 43 Marks
Manufacturer can sell x items at a price of rupees $(5-\frac{\text{x}}{100})$ each. The cost price is Rs $\Big(\frac{\text{x}}{5}+500\Big)$. Find the number of items he should sell to earn maximum profit.
AnswerProfit = S.P. - C.P.
$\Rightarrow\text{P}=\text{x}(50-\frac{\text{x}}{2})-\Big(\frac{\text{x}^{2}}{4}+35\text{x}+25\Big)$
$\Rightarrow\text{P}=50\text{x}-\frac{\text{x}^{2}}{2}-\frac{\text{x}^{2}}{4}-35\text{x}+25$
$\frac{\text{dP}}{\text{dx}}=50\text{x}-\text{x}-\frac{\text{x}}{2}-35$
For maximum or minimum value of P, we must have $\frac{\text{dP}}{\text{dx}}=0$
$\Rightarrow 15 -\frac{3\text{x}}{2}=0$
$\Rightarrow 15 =\frac{3\text{x}}{2}$
$\Rightarrow \text{x}=\frac{30}{3}$
$\Rightarrow \text{x}=10$
Now, $\frac{\text{d}^{2}\text{P}}{\text{dx}^{2}}=\frac{-3}{2}<0$
So, profit is maximum if daily output is 10 items.
View full question & answer→Question 53 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$f(x) = (x - 5)^4$
AnswerGiven, $f(x) = (x - 5)^4$
$\Rightarrow f'(x) = 4(x - 5)^3$
For a local maximum or a local minimum, We must here $f'(x) = 0$
$\Rightarrow 4(x - 5)^3= 0$
$\Rightarrow x = 5$
Since f'(x) changes from negative to positive when x increases through $5, x = 5$ is the point of local minima.
The local minimum value of (x) at $x = 5$ is given by $(5 - 5) = 0$
View full question & answer→Question 63 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\frac{1}{\text{x}^{2}+2}$
Answer$\text{g}'(\text{x})=\frac{1}{\text{x}^{2}+2}$
$\therefore\text{g}'(\text{x})=\frac{-(2\text{x})}{(\text{x}^{2}+2)^{2}}$
$\Rightarrow\text{g}'(\text{x})=0$
$\Rightarrow\frac{-2\text{x}}{(\text{x}^{2}+2)^{2}}=0$
$\Rightarrow\text{x}=0$
Now, for value close to x = 0 and to the left of 0, g'(x) > 0. Also, for value close to x = 0 and to the right of 0, g'(x) < 0.
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of g(0) is $\frac{1}{0+2}=\frac{1}{2}$.
View full question & answer→Question 73 Marks
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,
$\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}},\text{x}\leq1$
Answer$\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}}$
$\therefore\ \text{f}(\text{x})=\sqrt{1-\text{x}}+\frac{\text{x}}{2\sqrt{1-\text{x}}}(-1)$
$=\frac{2(1-\text{x})-\text{x}}{2\sqrt{1-\text{x}}}$
$=\frac{2-3\text{x}}{2\sqrt{1-\text{x}}}$
$\text{f}''(\text{x})=\frac{2\sqrt{1-\text{x}}(-3)+\frac{(2-3\text{x})}{\sqrt{1-\text{x}}}}{4(1-\text{x})}$
For maximum and minimum, f'(x) = 0
$\Rightarrow\frac{(2-3\text{x})}{2\sqrt{1-\text{x}}}=0$
$\Rightarrow\text{x}=\frac{3}{2}$
Now, $\text{f}''\Big(\frac{2}{3}\Big)<0$
$\therefore\text{x}=\frac{2}{3}$ is point of maxima.
$\therefore$ local max value $=\text{f}\Big(\frac{2}{3}\Big)=\frac{2}{3\sqrt{3}}$
View full question & answer→Question 83 Marks
Write the maximum value of $\text{f}(\text{x})=\frac{\log\text{x}}{\text{x}}$, if it exists.
AnswerGiven, $\text{f}(\text{x})=\frac{\log\text{x}}{\text{x}}$
limplies that $\text{f}'(\text{x})=\frac{1-\log\text{x}}{\text{x}^{2}}$
For a local maxima or a local minima, we must have f'(x) = 0
lmplies that $\frac{1-\log\text{x}}{\text{x}^{2}}=0$
lmplies that $1-\log\text{x} = 0$
lmplies that $\log\text{x}=1$
lmplies that $\log\text{x} =\log\text{e}$
limplies that x = e
Now, $\text{f}''(\text{x})=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^{4}}$
$=\frac{-3\text{x}-2\text{x}\ \log\text{x}}{\text{x}^{4}}$
At x = e
$\text{f}''(\text{e})=\frac{-3\text{x}-2\text{x}\ \log\text{e}}{\text{e}^{4}}\text{e}$
$=\frac{-5}{\text{e}^{3}}<0$
Therefore, x = e is a point of local maximum.
Thus, the local maximum value is given by
$\text{f}(\text{e})=\frac{\log\text{e}}{\text{e}}=\frac{1}{\text{e}}$.
View full question & answer→Question 93 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=2\sin\text{x}-\text{x}, -\frac{\pi}{2}\leq\text{x}\leq\frac{\pi}{2}$
Answer$\text{f}(\text{x})=2\sin\text{x}-\text{x}, -\frac{\pi}{2}\leq\text{x}\leq\frac{\pi}{2}$
For checking the minima and maxima, We have f'(x) = 2cosx - 1 = 0
$\Rightarrow\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=-\frac{\pi}{3}, \frac{\pi}{3}$
At $\text{x}=-\frac{\pi}{3},$ f(x) changes from -ve to +ve
$\Rightarrow\text{x}=-\frac{\pi}{3}$ is point of local minima with value $=-\sqrt{3}-\frac{\pi}{3}$
At $\text{x}=\frac{\pi}{3}$ , f(x) changes from +ve to -ve
$\Rightarrow\text{x}=\frac{\pi}{3}$ is point of local maxima with value $=-\sqrt{3}-\frac{\pi}{3}$
View full question & answer→Question 103 Marks
Write the minimum value of $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}},\text{x}>0.$
AnswerGiven, $\text{f}\text{(x)}=\text{x}+\frac{1}{\text{x}}$
Implies that $\text{f}'\text{(x)}=1-\frac{1}{\text{x}^2}$
For a local maxima or a local minima, We must have
$\text{f}'\text{(x)}=0$
Implies that $1-\frac{1}{\text{x}^2}=0$
Implies that $\text{x}^2=1$
Implies that $\text{x}=1,-1$
But $\text{x}>0$
Implies that $\text{x}=1$
Now, $\text{f}''(\text{x})=\frac{1}{\text{x}^{3}}$
At x = 1
$\text{f}''(1)=\frac{2}{(1)^{3}}1=2>0$
Therefore, x = 1 is a point minimum.
Thus, the local minimum value is given by
$\text{f}(1)=1+\frac{1}{1}=1+1=2$
View full question & answer→Question 113 Marks
Write a value of $\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^7}{7}+\text{C}$
Thus, $\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
View full question & answer→Question 123 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$f(x) = x^3 - 3x$
Answer$\text{g}(\text{x})=\text{x}^{3}-3\text{x}$
$\therefore\text{g}'(\text{x})=3\text{x}^{2}-3$
Now, $\text{g}'(\text{x})=0$
$ \Rightarrow3\text{x}^{3}=3$
$\Rightarrow\text{x}=\pm1$
$\Rightarrow\text{g}'({\text{x}})=6\text{x}$
$\Rightarrow\text{g}'(1)=6\geq0$
$\Rightarrow\text{g}'(-1)=-6\leq0$
By second derivative test, x = 1 is a point of local minima and local minimum value of at $x = 1$ is $g(1) = 13 - 3 = 1 - 3 = -2$. However,
$x = -1$ is a point maxima and local maximum value of g at $x = -1$ is $g(1) = (-1)^3 - 3(-1) = -1 + 3 = 2$
View full question & answer→Question 133 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\cos{\text{x}},0\leq\text{x}\leq\pi$
AnswerGiven, $\text{f}(\text{x})=\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=-\sin\text{x}$
For a local maximum or a local minimum, We must have f'(x)=0
$\Rightarrow-\sin\text{x}=0$
$\Rightarrow\sin\text{x}=0$
$\Rightarrow\text{x}=0, \text{or}\ \pi$
Since $0\leq\text{x}\leq\pi$, none is in the interval $(0, \pi)$
View full question & answer→Question 143 Marks
Find the least value of $\text{f}(\text{x})=\text{ax}+\frac{\text{b}}{\text{x}}$, where a > 0, b > 0 and x > 0.
AnswerWe have, $\text{f}(\text{x})=\text{ax}+\frac{\text{b}}{\text{x}}$
lmplies that $\text{f}'(\text{x})=\text{a}-\frac{\text{b}}{\text{x}^{2}}$
For a local maxima or a local minima, we must have f'(x) = 0
lmplies that $\text{x}^{2}-\frac{\text{b}}{\text{a}}$
lmplies that $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}},\sqrt{-\frac{\text{b}}{\text{a}}}$
But, x > 0
lmplies that $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$
Now, $\text{f}''(\text{x})=\frac{2\text{b}}{\text{x}^{3}}$
At $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$
$\text{f}''\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\frac{2\text{b}}{\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)^{3}}=\frac{2\text{a}^\frac{3}{2}}{\text{b}^\frac{1}{2}}>0$
Therefore, x = ba is a point of local minimum.
Hence, the least value is
$\text{f}\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\sqrt[\text{a}]{\frac{\text{b}}{\text{a}}}+\frac{\text{b}}{\sqrt{\frac{\text{b}}{\text{a}}}}=\sqrt{\text{ab}}+\sqrt{\text{ab}}=2\sqrt{\text{ab}}$.
View full question & answer→Question 153 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}'(\text{x})=\frac{\text{x}}{2}+\frac{2}{\text{x}}, \text{x}\geq0$
AnswerWe have, $\text{f}'(\text{x})=\frac{\text{x}}{2}+\frac{2}{\text{x}}, \text{x}\geq0$
$\therefore\text{f}'({\text{x}})=\frac{1}{2}-\frac{2}{\text{x}^{2}}$
For the point of local maxima and minima, f'(x) = 0
$\Rightarrow\frac{1}{2}-\frac{2}{\text{x}^{2}}=0$
$\Rightarrow\text{x}^{4}-4=0$
$\Rightarrow\text{x}=\sqrt{4}, -\sqrt{4}$
$\Rightarrow\text{x}=2, -2$
At, x = 2, f'(x) changes from -ve to +ve
$\therefore$ is the point of local minima.
$\therefore$ local minimum value = f(2) = 2.
View full question & answer→Question 163 Marks
The total cost of producing x radio sets per day is Rs $\Big(\frac{\text{x}^{2}}{4}+35\text{x}+25\Big)$ and the price per set at which they may be sold is Rs $(50-\frac{\text{x}}{2})$. Find the daliy output to maximum the tatal profit.
AnswerProfit = S.P. - C.P.
$\Rightarrow\text{P}=\text{x}(50-\frac{\text{x}}{2})-\Big(\frac{\text{x}^{2}}{4}+35\text{x}+25\Big)$
$\Rightarrow\text{P}=50\text{x}-\frac{\text{x}^{2}}{2}-\frac{\text{x}^{2}}{4}-35\text{x}+25$
$\frac{\text{dP}}{\text{dx}}=50\text{x}-\text{x}-\frac{\text{x}}{2}-35$
For maximum or minimum value of P, we must have $\frac{\text{dP}}{\text{dx}}=0$
$\Rightarrow 15 -\frac{3\text{x}}{2}=0$
$\Rightarrow 15 =\frac{3\text{x}}{2}$
$\Rightarrow \text{x}=\frac{30}{3}$
$\Rightarrow \text{x}=10$
Now, $\frac{\text{d}^{2}\text{P}}{\text{dx}^{2}}=\frac{-3}{2}<0$
So, profit is maximum if daily output is 10 items.
View full question & answer→Question 173 Marks
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}'(\text{x})=\text{x}^{3}(2\text{x}-1)^{3}$
AnswerGiven, $\text{f}(\text{x})=\text{x}^{3}(2\text{x}-1)^{3}$
$\Rightarrow\text{f}'(\text{x})=3\text{x}(2\text{x}-1)^{3}+6\text{x}^{3}(2\text{x-1})^{2}$
For the local maximum or minima, we must have f'(x) = 0
$\Rightarrow3\text{x}^{2}(2\text{x}-1)^{3}+6\text{x}^{3}(2\text{x}-1)^{2}=0$
$\Rightarrow3\text{x}^{2}(2\text{x}-1)^{2}(2\text{x}-1+2\text{x})=0$
$\Rightarrow\text{x}^{2}(2\text{x}-1)^{2}(4\text{x}-1)=0$
$\text{x}=0, \frac{1}{2}\ \text{and} \frac{1}{4}$
Since f'(x) changes from negative to positive when x increases through $\frac{1}{4}, \text{x}=\frac{1}{4}$ is a point of local minima.
The local mimimum value of f(x) at $\text{x}=\frac{1}{4}$ is given by $\Big(\frac{1}{4}\Big)^{3}\Big(\frac{1}{2}-1\Big)^{3}=\frac{-1}{512}$
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