Question 13 Marks
Find the equations of the tangent and normal to the given curves at the indicated points:
$y = x^3 $ at $(1, 1)$
AnswerThe equation of the curve is $y = x^3.$
On differentiating with respect to x, we get:
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$ \frac{\text{dy}}{\text{dx}}\Big]_{(1,1)}=3(1)^2=3$
Thus, the slope of the tangent at $(1, 1)$ is $3$ and the equation of the tangent is given as:
$y -1 = 3(x -1)$
$\Rightarrow y = 3x - 2$
The slope of the normal at $(1, 1)$ is $\frac{-1}{\text{slope of the tangent at}(1,1)}=\frac{-1}{3}.$
Therefore, the equation of the normal at $(1, 1)$ is given as:
$\text{y}-1 =\frac{-1}{3}(\text{x} -1)$
$\Rightarrow 3y - 3 = -x + 1$
$\Rightarrow x + 3y -4 = 0$
View full question & answer→Question 23 Marks
If $y = \log_e x$, then find $\triangle\text{y}$ when $x = 3$ and $\triangle\text{x} = 0.03.$
AnswerHere
$\text{x}=3,\triangle\text{x}=0.030\ \text{and}\ \text{y}=\log_\text{e}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-3}=\frac{1}{3}$
$\triangle\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-3}\text{x}(\triangle\text{x})$
$=\Big(\frac{1}{3}\Big)(0.03)$
$\triangle\text{y}=0.01$
View full question & answer→Question 33 Marks
Find the point on the curve $y = x^2$ where the slope of the tangent is equal to the x-coordinate of the point.
AnswerThe given equation of the curve is
$y = x^2...(1)$
$\therefore$ slope of tangent to (i) is
$\frac{\text{dy}}{\text{dx}}=2\text{x}\ ...(2)$
According to the question
$\frac{\text{dy}}{\text{dx}}=\text{x}\ ...(3)$
From $(2)$ and $(3)$
$2x = x$
$\Rightarrow x = 0$ & $y = 0$
Thus, the required point is $(0, 0)$
View full question & answer→Question 43 Marks
Find the equations of the tangent and normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2a.t)$
AnswerThe equation of the given parabola is $y^2 = 4ax.$
On differentiating $y^2 = 4ax$ with respect to $x,$ we have:
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}= \frac{2\text{a}}{\text{y}}$
$\therefore$ The slope of the tangent at $(at^2, 2at)$ is $\frac{\text{dy}}{\text{dx}}\Big]_{(\text{at}^2,2\text{at)}}=\frac{2\text{a}}{2\text{at}}=\frac{1}{1}.$
Then, the equation of the tangent at $(at^2, 2at)$ is given by,
$\text{y}-2\text{at}=\frac{1}{\text{t}}(\text{x}-\text{at}^2)$
$\Rightarrow ty - 2at^2 = x - at^2$
$\Rightarrow ty = x + at^2$^
Now, the slope of the normal at $(at^2, 2at)$ is given by,
$\frac{-1}{\text{slope of the tangent at }(\text{at}^2,\ 2\text{at)}}=-\text{t}$
Thus, the equation of the normal at $(at^2, 2at)$ is given as:
$y - 2at = -1(x - at^2)$
$\Rightarrow y - 2at = -tx + at^3$
$\Rightarrow y = -tx + 2at + at^3$
View full question & answer→Question 53 Marks
Write the coordinates of the point at which the tangent to the curve $y = 2x^2 - x + 1$ is parallel to the line $y = 3x + 9.$
AnswerLet $(x_1, y_1)$ be the required point.
slope of the given line $= 3$
Since, the point lies on the curve.
Hence, $\text{y}_1=2\text{x}_1{^2}-\text{x}_1+1\ ...(1)$
Now, $\text{y}=2\text{x}^2-\text{x}+1$
$\therefore\frac{\text{dy}}{\text{dx}}=4\text{x}-1$
Now,
slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=4\text{x}_1-1$
Given:
slope of the tangent = slope of the given line
$\therefore4\text{x}_1-1=3$
$\Rightarrow\text{x}_1=1$
From 1 we get
$\text{y}_1=2-1+1=2$
$\therefore(\text{x}_1,\text{y}_1)=(1,2)$
View full question & answer→Question 63 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 73 Marks
Using differentials, find the approximate value of each of the following:
$\bigg(\frac{17}{81}\bigg)^\frac{1}{4}$
AnswerConsider $\text{Y}=\text{x}^{\frac{1}{4}},\ \text{Let}\ \text{x}=\frac{16}{81}\ \text{and}\ \Delta \text{x}=\frac{1}{81}.$
Then, $\Delta \text{y}=(\text{x}+\Delta \text{x})^{\frac{1}{4}}-\text{x}^{\frac{1}{4}}$
$\Big(\frac{17}{81}\Big)^{\frac{1}{4}}-\Big(\frac{16}{81}\Big)^{\frac{1}{4}}$
$\Big(\frac{17}{81}\Big)^{\frac{1}{4}}-\frac{2}{3}$
$\therefore\ \Big(\frac{17}{81}\Big)^{\frac{1}{4}}=\frac{2}{3}+\Delta\text{y}$
Now dy is approximately equal to $\Delta \text{y}$ and is given by.
$\text{dy}=\Big(\frac{\text{dy}}{\text{dx}}\Big)\Delta \text{x}=\frac{1}{4(\text{x})^{\frac{3}{4}}}(\Delta\text{x})$ $\Big(\text{as y}=\text{x}^{\frac{1}{4}}\Big)$
$=\frac{1}{4\Big(\frac{16}{81}\Big)^{\frac{3}{4}}}\Big(\frac{1}{81}\Big)=\frac{27}{4\times8}\times\frac{1}{81}=\frac{1}{32\times3}=\frac{1}{96}=0.010$
Hence, the approximate value of $\Big(\frac{17}{81}\Big)^{\frac{1}{4}}\text{is} \frac{2}{3}9+0.010=0.667+0.010=0.677.$
View full question & answer→Question 83 Marks
Using differentials, find the approximate value of each of the following up to 3 places of decimal:
$(0.999)^{\frac{1}{10}}$
Answer$\text{Take y}=\text{x}^\frac{1}{10}, \text{x}=1, \text{dx}=\delta \text{x}=-0.001 \text{ so that x }+\delta \text{x}=0.999$
$ \text{Now }\text{y}+\delta \text{ y}=(\text{x}+\delta \text{ x})^\frac{1} {10}$ $\Rightarrow \delta\text{ y}=\text{( x}+\delta \text{ x)}^\frac{1}{10}-\text{ y}=\ (0.999)^\frac{1}{10} -1$
$\Rightarrow (0.999)\frac{1}{10 }=\delta\text{y}+1\ \dots\text{(1)}$
$\text{Now }\delta \text{y}$ is approximately equal to dy
$\text{and dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=\frac{1}{10}\text{x}^{-\frac {9}{10}}= \frac{1}{10\text{x}^\frac{9}{10}} $
$= \frac{1}{10(1)^{\frac{9}{10}}}(-0.001)=\frac{-0.001}{10}=-0.0001$
$\therefore\text{from (1), }(0.999)^{\frac{1}{10} }=-0.0001 + 1 = 0.9999$
$\therefore\ (0.999)^{\frac{1}{10}} = 0.9999$
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:
$\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}}, \text{x}\geq0$
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 103 Marks
What is the maximum value of the function sinx + cosx?
AnswerLet $\text{f}\text{(x)} =\sin\text{x}+\cos\text{x}\ \Rightarrow\ \ \text{f}'\text{(x)}=\cos\text{x}-\sin\text{x}$
Now $\text{f}'\text{(x)}=0\ \Rightarrow\ \cos\text{x}-\sin\text{x}=0\ \Rightarrow\ \ -\sin\text{x}=-\cos\text{x}$
$\Rightarrow\ \tan\text{x}=1=\tan\frac{\pi}{4}$ $\Rightarrow\ \text{x}=\text{n}\pi+\frac{\pi}{4}\ \ [\text{Turning point}]$
$\therefore\ \text{f} \bigg(\text{n}\pi+\frac{\pi}{4}\bigg)=\sin\bigg(\text{n}\pi+\frac{\pi}{4}\bigg)+\cos \bigg(\text{n}\pi+\frac{\pi} {4}\bigg)$
$=(-1)^\text{n}\sin\frac{\pi}{4}+(-1)^\text{n}\cos\frac{\pi}{4}$
$=(-1)^\text{n}\frac{1}{\sqrt{2}}+(-1)^\text{n}\frac{1}{\sqrt{2}}$ $=2(-1)^\text{n}\frac{1}{\sqrt{2}}=\sqrt{2}(-1)^\text{n}$
If n is even, then $\ \text{f}\bigg(\text{n}\pi+\frac{\pi}{4}\bigg)=\sqrt{2}$
If n is odd, then $\ \text{f}\bigg(\text{n}\pi+\frac{\pi}{4}\bigg)=-\sqrt{2}$
Therefore, maximum value of $\text{f}\text{(x)} \text{ is }\sqrt{2}$ and minimum value of $\text{f}\text{(x)} \text{ is }-\sqrt{2}.$
View full question & answer→Question 113 Marks
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge if $10$ cm long$?$
AnswerLet $x$ cm be the edge of variable cube at time $t.$
Rate of increase of edge $= 3$ cm/sec
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}$ is positive and $= 3$ cm/sec
Let $y$ be the volume of the cube. $\Rightarrow y = x^3$
$\therefore$ Rate of change of volume of cube $= \frac{\text{dy}}{\text{dt}}= \frac{\text{d}}{\text{dt}}\text{x}^3 =3\text{x}^2 \frac{\text{dx}}{\text{dt}}=3\text{x}^2(3)=9\text{x}^2\text{cm}^3/\sec$
Putting $x = 10$ cm (given), $\frac{\text{dy}}{\text{dt}} = 9(10)^2 = 900\text{ cm}^3/\sec$
Since $\frac{\text{dx}}{\text{dt}}$ is positive, therefore volume of cube is increasing at the rate of$ 900\ cm^3/sec.$
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:
$\text{f}(\text{x})=\text{x}^{3}-6\text{x}^{2}+9\text{x}+15$
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 133 Marks
Show that the function given by f(x) $\frac{\log\text{x}}{\text{x}}$ has maximum at $\text{x}=\theta.$
AnswerThe given function is $\text{f}\text{(x)}=\frac{\log\text{x}}{\text{x}}.$
$\text{f}\text{(x)}=\frac{\text{x}\Big(\frac{1}{\text{x}}\Big)-\log\text{x}}{\text{x}^2}=\frac{1-\log\text{x}}{\text{x}^2}$
Now, f(x) = 0
$\Rightarrow\ 1-\log \text{x}=0$
$\Rightarrow\ \log \text{x}=1$
$\Rightarrow\ \log\ \text{x}=\log\text{e}$
$\Rightarrow\ \text{x}=\text{e}$
Now, $\text{f}''\text{x}=\frac{\text{x}^2\Big(-\frac{1}{\text{x}}\Big)-(1-\log\text{x})(2\text{x})}{\text{x}^4}$
$=\frac{-\text{x}-2\text{x}(1-\log\text{x})}{\text{x}^4}$
$=\frac{-32\log\text{x}}{\text{x}^3}$
Now, $\text{f}''\text{(e)}=\frac{-3+2\log\text{e}}{\text{e}^3}=\frac{-3+2}{\text{e}^3}=\frac{-1}{\text{e}^3}<0$
Therefore, by second derivative test, f is the maximum at x = e.
View full question & answer→Question 143 Marks
Find an angle $\theta$
Which increases twice as fast as its cosine.
AnswerLet $\text{x}=\cos\theta$
Differentiating both sides with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}(\cos\theta)}{\text{dt}}$
$=-\sin\theta\frac{\text{d}\theta}{\text{dt}}$
But it is given that $\frac{\text{d}\theta}{\text{dt}}=2\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\theta\Big(2\frac{\text{dx}}{\text{dt}}\Big)$
$\Rightarrow\sin\theta=-\frac{1}{2}$
$\Rightarrow\theta=\pi+\frac{\pi}{6}=\frac{7\pi}{6}$
Hence, $\theta=\frac{7\pi}{6}.$
View full question & answer→Question 153 Marks
Show that $f(x) = x^9 + 4x^7 + 11$ is an increasing function for all $\text{x}\in\text{R}.$.
Answer$f(x) = x^9 + 4x^7 + 11$
$f'(x) = 9x^8 + 28x^6$
$= x^6(9x^2 + 28)$
Now,
$\text{x}\in\text{R}$
$\Rightarrow x^6 > 0$ and $9x^2 + 28 > 0$
$\Rightarrow x^6(9x^2 + 28) > 0$
$\Rightarrow f'(x) > 0$
So, $f(x)$ is increasing on function for $\text{x}\in\text{R}.$
View full question & answer→Question 163 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 173 Marks
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
$(81. 5) ^{\frac{1}{4}}$
Answer$(81.5)^{\frac{1}{4}}$
Let y = $\text{x}^{\frac{1}{4}}\ \dots\text{(i)}$
$\therefore \ \frac{\text{dy}}{\text{dx}}=\frac{1}{4}\text{x}^{\frac{-3}{4}}=\frac{1}{4\text{x}^{\frac{3}{4}}}$
$\Rightarrow\ \text{dy}=\frac{\text{dx}}{4\Big(\text{x}^{\frac{1}{4}}\Big)^{3}}\ \dots\text{(ii)}$
$\text{Now, from eq. (i) }, \text{y} + \Delta \text{y} = \sqrt{\text{x}+\Delta \text{x}}$
$=(81.5)^{\frac{1}{4}} -(81+0.5)^{\frac{1}{4}}\ \dots\text{(iii)}$
Here $\text{x} = 81 \text{ and }\Delta \text{x} = 0.5 $
$\text{Then } \Delta \text{y} =(\text{x} + \Delta \text{x})^{\frac{1}{4} }-\text{x}^{\frac{1}{4}}=(81.5)^{\frac{1}{4}}-(81)^{\frac{1}{4}}$
$=(81.5)^{\frac{1}{4}} -3\ \Rightarrow (81.5)^{\frac{1}{4}} = 3 +\Delta \text{y}$
Since, $\Delta \text{x}\text{ and }\Delta \text{dy}$ is approximately equal to dx and dy respectively.
$\therefore \ \text{From eq.(ii),}\ \text{dy}=\frac{0.5}{4\Big(81^{\frac {1}{4}}\Big)^{3}}=\frac{0.5}{108}=0.00462$
Therefore, approximate value of $(82)^{\frac{1}{4}}\text{ is }3 + 0.00462 = 3.00462. $
View full question & answer→Question 183 Marks
A circular disc of radius 3cm is being heated. Due to expansion, its radius increases at the rate of 0.05cm/ sec. Find the rate at which its area is increasing when radius is 3.2cm.
AnswerLet r be the radius and A be the area of the circular disc at any time t. Then,
$\text{A}=\pi\text{r}^2$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\text{r}\frac{\text{dr}}{\text{dt}}$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\times3.2\times0.05$
$\Big[\therefore\text{r}=3.2\text{cm}\text{ and }\frac{\text{dr}}{\text{dt}}=0.05\text{cm}/\sec\Big]$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=0.32\pi\text{ cm}^2/\sec$
View full question & answer→Question 193 Marks
Prove that the function f given by f(x) = logsin is strictly increasing on $\Big(0,\ \frac{\pi}{2}\Big)$ and strictly decreasing on $\Big(\frac{\pi}{2},\ \pi\Big).$
AnswerGiven: $\text{f}\text{(x)} = \log\sin \text{x} \ \Rightarrow\ \text{f}'\text{(x)}=\frac{1}{\sin \text{x}}\frac{\text{d}}{\text{dx}}\sin \text{x}$ $=\frac{1}{\sin \text{x}}\cos \text{x}=\cot \text{x}$
On the interva $\Big(0,\ \frac{\pi}{2}\Big)$ i.e., in first quadrant, f'(x) = cot x > 0
Therefore, f(x) is strictly increasing on $\Big(0,\ \frac{\pi}{2}\Big).$
On the interval $\Big(\frac{\pi}{2},\ \pi\Big)$ i.e., in second quadrant, f'(x) = cot x< 0
Therefore, f(x) is strictly decreasing on $\Big(\frac{\pi}{2},\ \pi\Big).$
View full question & answer→Question 203 Marks
Find the equation of the tangent to the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=\text{a},$ at the point $\Big(\frac{\text{a}^2}{4},\frac{\text{a}^2}{4}\Big).$
Answer$\sqrt{\text{x}}+\sqrt{\text{y}}=\text{a},$
Differentiating both sides w.r.t. x,
$\Rightarrow\frac{1}{2\sqrt{\text{x}}}+\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\sqrt{\text{y}}}{\sqrt{\text{x}}}$
Given $(\text{x}_1,\text{y}_1)=\Big(\frac{\text{a}^2}{4},\frac{\text{a}^2}{4}\Big)$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\Big(\frac{\text{a}^2}{4},\frac{\text{a}^2}{4}\Big)}=\frac{-\sqrt{\frac{\text{a}^2}{4}}}{\sqrt{\frac{\text{a}^2}{4}}}$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\frac{\text{a}^2}{4}=-1\big(\text{x}-\frac{\text{a}^2}{4}\big)$
$\Rightarrow\text{y}-\frac{\text{a}^2}{4}=-\text{x}+\frac{\text{a}^2}{4}$
$\Rightarrow\text{x}+\text{y}=\frac{\text{a}^2}{2}$
View full question & answer→Question 213 Marks
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
AnswerGiven: Each side of square piece of tin is 18 cm.
Let x cm be the side of each of the four squares cut off from each corner.
Then dimensions of the open box formed by folding the flaps after cutting off squares are
(18 - 2x), (18 - 2x) and x cm.
Let z denotes the volume of the open box.
$\therefore\ \text{z}=(18-2\text{x})(18-2\text{x)}$ $\Rightarrow\ \ \text{z}=(18-2\text{x})^2\text{x}$
$\Rightarrow\ \text{z}=(324+4\text{x}^2-72\text{x)}\text{x}=4\text{x}^3-72\text{x}^2+324\text{x}$
$\Rightarrow\ \frac{\text{dz}}{\text{dx}}=12\text{x}^2-144\text{x}+324\ \text{ and }\ \frac{\text{d}^2\text{z}}{\text{dx}^2}=24\text{x}-144$
Now $\frac{\text{dz}}{\text{dx}}=0\ \Rightarrow\ \ 12\text{x}^2-144+324=0$ $\Rightarrow\ \ \text{x}^2-12\text{x}+27=0$
$\Rightarrow\ (\text{x}-9)(\text{x}-3)=0\ \Rightarrow\ \text{x}=9\ \text{ or }\text{x}=3$
x = 9 is rejected because at x = 9 length = 18 - 2x = 18 - 2 × 9 = 0 which is impossible.
$\therefore$ x = 3 is the turning point.
$\text{At }\text{x}=3,\ \frac{\text{d}^2\text{z}}{\text{dx}^2}=24\times3-144=-72\ \ [\text{Negative}]$
$\therefore$ z is minimum at x = 3 i. e., side of each square to be cut off from each corner for maximum volume is 3 cm.
View full question & answer→Question 223 Marks
Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:
$\text{g}(\text{x})\frac{\text{x}}{2}+\frac{2}{\text{x}},\text{x}\geq0$
AnswerGiven: $\text{g}(\text{x)}=\frac{\text{x}}{2}+\frac{2}{\text{x}},\text{x}>0$
$ \therefore\ \text{g }'(\text{x)}=\frac{1}{2}-\frac{2}{\text{x}^2}=\frac{\text{x}^2-4} {2\text{x}^2}$ $=\frac{(\text{x}+2)(\text{x}-2)}{2\text{x}^2}\ \ \text{and g}''\text{(x)}=\frac{4}{\text{x}^3}$
$\text{Now g}'\text{(x)}=0\ \Rightarrow\ \frac{(\text{x}+2)(\text{x}-2)}{2\text{x}^2}=0$ $\ \Rightarrow\ \text{(x}+2)(\text{x}-2)=0$
$\Rightarrow\ \text{x}=-2 \ \text{ or }=2 $
But x > 0, therefore x = 2 is only the turning point.
$\therefore\ $ x = 2 is a point of local minima and local minimum value is $\text{g}(2)=\frac{2}{2}+\frac{2}{2}=2$
View full question & answer→Question 233 Marks
Show that $f(x) = e^{2x}$ is increasing on $R.$
Answer$f(x) = e^{2x}$
$f'(x) = 2e^{2x}$
Now,
$\text{x}\in\text{R}$
Since the value of $e^{2x}$ is always positive for any real value of $x, e^{2x} > 0.$
$\Rightarrow 2e^{2x} > 0$
$\Rightarrow f'(x) > 0$
So, $f(x)$ is increasing on $R.$
View full question & answer→Question 243 Marks
A balloon which always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15cm.
AnswerLet r be the radius and V be the volume of the spherical ballon at any time t. Then,
$\text{V}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}$
$\Rightarrow\frac{\text{dr}}{\text{dt}}=\Big(\frac{1}{4\pi\text{r}^2}\Big)\frac{\text{dV}}{\text{dt}}$
$\Rightarrow\frac{\text{dr}}{\text{dt}}=\frac{900}{4\pi(15)^2}$ $\Big[\therefore\text{r}=15\text{cm}\text{ and }\frac{\text{dV}}{\text{dt}}=900\text{cm}^3/\sec\Big]$
$\Rightarrow\frac{\text{dr}}{\text{dt}}=\frac{1}{\pi}\text{cm}/\sec$
View full question & answer→Question 253 Marks
Find the least value of a such that the function f given by $f(x) = x^2 + ax + 1$ strictly increasing on $(1, 2).$
Answer$\text{f}\text{(x)} = \text{x}^{2}+\text{ax} +1 \ \Rightarrow\ \text{f}\text{(x)} = 2\text{x}+\text{a}$
Since $f(x)$ is strictly increasing on$ (1, 2)$, therefore $f'(x) = 2x + a > 0$ for all x in $(1, 2)$
$\therefore\ \text{On }(1,2)\ 1 < \text{x} < 2\ \Rightarrow\ 2 < 2\text{x}< 4$
$\Rightarrow\ 2 + \text{a} < 2\text{x} + \text{a} < 4 + \text{a}$
$\therefore $ Minimum value of $f(x) $ is $2 + a$ and maximum value is 4 + a. Since $f'(x) > 0$ for all $x$ in $(1, 2)$
$\therefore\ 2 + \text{a} > 0 \text{ and } 4 + \text{a} > 0\ \Rightarrow\ \text{a} > -2\text{ and a} > -4$
Therefore least value of a is$ -2.$
View full question & answer→Question 263 Marks
Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing with respect to the radius when the radius is 2cm?
AnswerLet V be the volume of the spherical ball.
Then,
$\text{V}=\frac{4}{3}\pi\text{r}^3$
$\frac{\text{dV}}{\text{dr}}=4\pi\text{r}^2$
Thus,
The rate of change of the volume of the sphere is $4\pi\text{r}^2.$
When $\text{r}=2\text{cm},$
$\Big(\frac{\text{dV}}{\text{dr}}\Big)_\text{r=2}=4\pi(2)^2=16\pi\text{cm}^3/\text{cm}$
View full question & answer→Question 273 Marks
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
$(26)^{\frac{1}{3}}$
Answer$(26)^{\frac{1}{3}}$
Let y = $\sqrt[3]{\text{x}}\ \dots\text{(i)}$
$\therefore \ \frac{\text{dy}}{\text{dx}}=\frac{1}{3}{\text{x}^\frac{-2}{3}}=\frac{1}{3\text{x}^{2/3}}$
$\Rightarrow \text{dy}=\frac{\text{dx}}{3\text{x}^{2/3}}=\frac{\text{dx}}{3\big({\text{x}^{1/3}\big)}^2}\ \dots\text{(ii)}$
$ \text{Now, from eq. (i), y}+\Delta \text{y}=\sqrt{\text{x}+\Delta \text{x}}=(26)^\frac{1}{3}=(27 -1)^\frac{1}{3}$
Here, x = 27 and $\Delta \text{x}$ = -1,
then $\Delta \text{y}=\sqrt{\text{x}+\Delta \text{x}}-\sqrt{\text{x}}=(26)^\frac{1}{3}-(27)^\frac{1} {3}=(26)^\frac{1}{3}-3$
$ \Rightarrow\ (26)^\frac{1}{3}=\Delta\text{ y}+3$
Since, $\Delta\text{ x} \text{ and } \Delta \text{y} $ is approximately equal to dx and dy respectively.
$\therefore \text{From eq. (ii), }\text{dy}=\frac{-1}{3\bigg((27)^{\frac{1}{3}}\bigg)^{2}}=\frac{-1}{27}$
Therefore, approximately value of $ (26)^{\frac{1}{3}}\text{ is }3-\frac{1}{27}=\frac{80}{27}=2.9629.$
View full question & answer→Question 283 Marks
The radius of the circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
AnswerLet x cm be the radius of the circle at time t.
Rate of increase of radius of circle = 3 cm/sec
$\Rightarrow \ \frac{\text{dx}}{\text{dt}}\text{ is positive and } = 3\text{ cm} /\sec $
Let y be the area of the circle. $\Rightarrow\ \text{y} = \pi \text{r}^{2}$
$\therefore$ Rate of change of area of circle $= \frac {\text{dy}}{\text{dt}}= {\pi} \frac{\text{d}}{\text{dt}}\text{x}^2 ={\pi}.2\text{x}\frac{\text{dx}}{\text{dt}}=2{\pi}\text{x}(3) = 6\pi \text{x}$
Putting x = 10 cm (given), $\frac{\text{dy}}{\text{dt}}=6{\pi}(10)= 60{\pi }\text{ cm}^2/\sec$
Since $\frac{\text{dy}}{\text{dt}}$ is positive, therefore surface area is increasing at the rate of $60\pi \text{ cm}^{2}/\sec.$
View full question & answer→Question 293 Marks
The volume of a cube is increasing at the rate of $8\ cm^3/sec.$ How fast is the surface area increasing when the length of an edge is $12\ cm?$
AnswerLet x cm be the edge of the cube.
Given: Rate of increase of volume of cube $= 8\ cm^3/sec$
$\Rightarrow\ \frac{\text{d}}{\text{dt}}\left(\text{x.x.x}\right)=\frac {\text{d}}{\text{dt}}\left(\text{x}^3\right) \text { is positive = 8}$
$\Rightarrow\ 3\text{x}^2\frac{\text{d}}{\text{dt}}\text{x}=8\ \Rightarrow \frac{\text{dx}}{\text{dt}}=\frac{8}{3\text{x}^{2}}\ \dots(1)$
Let y be the surface area of the cube, i.e., $y = 6x^2$
$\therefore$ Rate of change of surface area of the cube $= \frac{\text{dy}}{\text{dt}} = 6 \frac {\text{d}}{\text{dt}}\text{x}^2 = 6 \Big(2\text{x}\frac{\text{dx}}{\text{dt}}\Big) = 12\text{x} \Big(\frac{8}{3\text{x}^2}\Big)$
$= 4\Big(\frac{8}{\text{x}}\Big) = \frac{32}{\text{x}} \text{cm}^{2} /\sec$
Putting $x = 12$ cm (given), $\frac{\text{dy}}{\text{dt}}= \frac{32}{12}=\frac{8}{3}\text{cm}^2/ \sec$
Since $\frac{\text{dy}}{\text{dt}}$ is positive, therefore surface area is increasing at the rate of $\frac{8}{3}\text{ cm}^{2}/\sec.$
View full question & answer→Question 303 Marks
Prove that $\text{f(x)}=\sin\text{x}+\sqrt{3}\cos\text{x}$ has maximum value at $\text{x}=\frac{\pi}{6}.$
AnswerWe have, $\text{f(x)}=\sin\text{x}+\sqrt{3}\cos\text{x}$ $\therefore\ \text{f}'(\text{x})=\cos\text{x}-\sqrt{3}\sin\text{x}$ For $\text{f}'(\text{x)}=0,\cos\text{x}=\sqrt{3}\sin\text{x}$ $\Rightarrow\ \tan\text{x}=\frac{1}{\sqrt{3}}$ $\Rightarrow\ \text{x}=\frac{\pi}{6}$ Differentiating f'(x), we get $\text{f}''(\text{x})=-\sin\text{x}-\sqrt{3}\cos\text{x}$ $\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}<0$ $=-\frac{1}{2}-\sqrt{3}.\frac{\sqrt{3}}{2}$$=-\frac{1}{2}-\frac{3}{2}=-2<0$
Hence, at $\text{x}=\frac{\pi}{6},\text{f(x)}$ has maxima value and $\frac{\pi}{6}$ is the point of local maxima.
View full question & answer→Question 313 Marks
Find the approximate change in the value $V$ of a cube of side $x$ metres caused by increasing the side by $1\%.$
AnswerThe volume of a cube $(V)$ of side x is given by $V = x^3.$
$\therefore\text{DV}=\Big(\frac{\text{DV}}{\text{dx}}\Big)\triangle\text{x}$
$=(3\text{x}^2)\triangle\text{x}$
$=(3\text{x}^2)(0.01\text{x})\ [\text{as }1\%\text{ of x is }0.01\text{x}]$
$=0.03\text{x}^3$
Hence, the approximate change in the volume of the cube is $0.03x^3m^3.$
View full question & answer→Question 323 Marks
Find the maximum profit that a company can make, if the profit function is given by $p(x) = 41 - 24x - 18x^2.$
AnswerGiven: $\ \text{Profit function }p(\text{x)} =41-24\text{x}-18\text{x}^2$
$\ \therefore \ \text{p}'(\text{x)}=-24-36\text{x}\ \text{ and }\ \text{p}''\text{(x)}=-36$
$\text{Now } \text{p}'\text{(x)}=0\ \Rightarrow\ \ -24-36\text{x}=0\ \Rightarrow\ \text{x}=-\frac{24}{36}=-\frac{2}{3}$
$\text{At } \text{x}=-\frac{2}{3},\ \text{p}"\text{(x)}=-36\ \ [\text{Negative}]$
$ \therefore\ \text{p}\text{(x)}$ has a local maximum value at $\text{x}=-\frac{2}{3}.$
$\therefore\ \text{At }\text{x}=-\frac{2}{3},\\ \text{Maximum profit}=41-24\bigg(-\frac{2}{3}\bigg)-18\bigg(\frac {4}{9}\bigg) \\\text{Maximum profit}=41+16-8=49 $
View full question & answer→Question 333 Marks
A balloon which always remains spherical, has a variable diameter $\frac{3}{2} (2\text{x + 1)}.$ Find the rate of change of its volume with respect to x.
AnswerGiven: Diameter of the balloon $=\frac{3}{2}(2\text{x}+1)$
$\therefore $ Radius of the balloon $=\frac{3}{4}(2\text{x}+1)$
$\therefore$ Volume of the balloon $=\frac{4}{3}\pi\Big(\frac{3}{4}\pi2\text{x}+1\Big)=\frac{9\pi}{16}(2\text{x}+1)^3$ cu. units
$\therefore$ Rate of change of volume $\text{w.r.t.x}\ = \frac{\text{dV}}{\text{dx}}= \frac{9\pi}{16}.3(2\text{x}+1)^2 .\frac {\text{d}}{\text{dx}}(2\text{x}+\text{x})$
$=\frac{ 27\pi}{16}(2\text{x}+1 )^{2}.2=\frac{27\pi}{8}(2\text{x} + 1)^{2}$
View full question & answer→Question 343 Marks
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
$(0.0037)^{\frac{1}{2}}$
AnswerTake y = $\sqrt{\text{x}},$ x = 0.0036 dx = $\delta\text{x}$ = 0.0001 so that x + $\delta \text{x}$ = 0.0037
$\text{Now y} + \delta \text{x} = \sqrt{\text{x} + \delta\text{x}}$
$\Rightarrow\ \delta\text{y}=\sqrt{\text{x}+\delta}\text{x}-\text{y}=\sqrt{.0037}-\sqrt{.0036}$
$\Rightarrow\ \delta \text{y}=\sqrt{.0037}-.06$
$\Rightarrow \ \sqrt{.0037}=\delta\text{y}+.06\ \dots\text{(i)}$
$\text{Now }\delta \text{y}$ is approximately equal to dy
and dy $= \frac{\text{dv}}{\text{dx}}\text{dx}=\frac{1}{2\sqrt {\text{x}}}\text{dx}=\frac{1}{2\sqrt{0.0036}}(0.0001)$
$=\frac{0.0001}{2(0.06)}=\frac{0.0001}{0.12}=\frac{0.01}{12}$
= 0.000833
$\therefore \text{from (1),} \sqrt{0.0037}=0.000833 + 0.6 = 0.060833$
View full question & answer→Question 353 Marks
Find the rate of change of the volume of a sphere with respect to its diameter.
AnswerLet D be the diameter and r be the radius of sphere,
So, volume of sphere $=\frac{4}{3}\pi\text{r}^2$
$\text{v}=\frac{4}{3}\pi\Big(\frac{\text{D}}{2}\Big)^3$
$\text{v}=\frac{4}{24}\pi\text{D}^3$
Differentiating it with respect to D.
$\frac{\text{dv}}{\text{dD}}=\frac{12}{24}\pi\text{D}^2$
$\frac{\text{dv}}{\text{dD}}=\frac{\pi\text{D}^2}{2}$
View full question & answer→Question 363 Marks
Sand is pouring from a pipe at the rate of $12\ cm^3/s.$ The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm?
AnswerLet the height and radius of the sand-cone formed at time t second be y cm and x cm respectively.
According to question $\text{y}=\frac{1}{6}\text{x} \ \Rightarrow\ \text{x}=6\text{y}$
Volume of cone $\text{(V)} = \frac{1}{3}{\pi \text{x}}^2\text{y}=\frac{1}{3}(6\text{y})^2\ \text{y}=12{\pi \text{y}}^3$
$\Rightarrow\ \frac {\text{dV}}{\text{dy}} = 36{\pi}\text{y}^2$
Now, since $\frac{\text{dV}}{\text{dy}}=12 \ \Rightarrow\ \frac{\text{dV}}{\text{dy}}\times \frac{\text{dy}}{\text{dt}}=12$ $\Rightarrow \ 36{\pi}\text{y}^2\times \frac{\text{dy}}{\text{dt}}=12$
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{1}{3\pi\text{y}^2}\ \Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{1}{3\pi4^2}=\frac{1}{48\pi}\text{cm/sec}$
View full question & answer→Question 373 Marks
Write the angle between the curves $y = e^{-x}$ and $y = e^x$ at their point of intersections.
AnswerThe given equation of curve are,
$y = e^{-x}...(1)$
$y = e^x...(2)$
Solving $(1)$ and $(2)$
$\text{y}=\frac{1}{\text{e}^\text{x}}=\frac{1}{\text{y}}$
$\Rightarrow\text{y}^2=1$
$\Rightarrow\text{y}=\pm1$
From $(2)$
$\pm1=\text{e}^\text{x}$
$\Rightarrow\text{x}=0$
So, the point of intersection is $P = (0, 1)$ and $Q = (0, -1).$
Slope of $(2)$
$\text{m}_2=\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\therefore\text{m}_2\text{ at }\text{P}=1\text{ and }\text{m}_2\text{ at }\text{Q}=1$
$\therefore\text{m}_1\times\text{m}_2=-1\times1=-1$
The angle between the curves is $90^\circ $
View full question & answer→Question 383 Marks
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
$\sqrt{49.5}$
Answer$\sqrt{49.5}$
Let y = $\sqrt{\text{x}} \ \dots\text{(i)}$
$ \therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{x}^\frac{-1}{2}=\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\ \text{dy}=\frac{\text{dx}}{2\sqrt{\text{x}}}\ \dots \text{(ii)}$
$\text{Now, from eq. (i), y}+\Delta \text{y} =\sqrt{\text{x}+ \Delta \text{x}}=\sqrt{49.5 }=\sqrt{49 + 0.5}$
Here, $\text{x} =49 \text{ and }\Delta \text{x} = 0.5, $
then $\Delta \text{y} = \sqrt{\text{x}+ \Delta \text{x}}- \sqrt{\text{x}} = \sqrt{49.5 } - \sqrt{49} = \sqrt{49.5} - 7$
$\Rightarrow\ \sqrt{49.5 } = \Delta \text{y} + 7$
Since, $\Delta \text{x}\text{ and } \Delta \text{y}$ is approximately equal to dx and dy respectively.
$\therefore \ \text{From eq. (ii), }\text{dy}=\frac{0.5}{2\sqrt{49}}= 0.0357$
Therefore, approximately value of $\sqrt{49.5} $ is 7 + 0.0357 = 7.0357.
View full question & answer→Question 393 Marks
If the percentage error in the radius of a sphere is $\alpha,$ find the percentage error in its volume.
Answer$\text{V}=\frac{4}{3}\pi\ \text{x}^3$
We have
$\frac{\triangle\text{x}}{\text{x}}\times100=\alpha$
$\Rightarrow\frac{\text{dV}}{\text{dx}}=4\pi\ \text{x}^2$
$\Rightarrow\frac{\text{dV}}{\text{V}}=\frac{4\pi\ \text{x}^2}{\text{V}}\text{dx}$
$\Rightarrow\frac{\triangle\text{V}}{\text{V}}=\frac{4\pi\ \text{x}^2}{\frac{4}{3}\pi\ \text{x}^3}\times\frac{\text{x}\alpha}{100}$
$\Rightarrow\frac{\triangle\text{V}}{\text{V}}\times100=3\alpha$
Hence, the percentage error in the volume is $3\alpha$
View full question & answer→Question 403 Marks
Find the slope of the normal at the point 't' on the curve $\text{x}=\frac{1}{\text{t}},\text{y}=\text{t}.$
AnswerHere,
$\text{x}=\frac{1}{\text{t}}\text{ and }\text{y}=\text{t}$
$\frac{\text{dx}}{\text{dt}}=\frac{-1}{\text{t}^2}\text{ and }\frac{\text{dy}}{\text{dt}}=1$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{1}{\Big(\frac{-1}{\text{t}^2}\Big)}=-\text{t}^2$
Now,
Slope of the tangent = $\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\text{t}^2$
Slope of the normal $=\frac{-1}{\text{slope of the tangent}}=\frac{-1}{-\text{t}^2}=\frac{1}{\text{t}^2}$
View full question & answer→Question 413 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 423 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 433 Marks
Find the set of values of 'b' for which $\text{f}(\text{x})=\text{b}(\text{x}+\cos\text{x})+4$ is decreasing on R.
Answer$\text{f}(\text{x})=\text{b}(\text{x}+\cos\text{x})+4$
$\text{f}'(\text{x})=\text{b}(1-\sin\text{x})$
Given: f(x) is decreasing on R.
$\Rightarrow\text{f}'(\text{x})<0$
$\Rightarrow\text{b}(1-\sin\text{x})<0\ ....(1)$
We know,
$\sin\text{x}\leq1$
$\Rightarrow1-\sin\text{x}\geq0$
$\Rightarrow\text{b}<0$ $[\text{Since }(1-\sin\text{x})\geq0,\text{b}(1-\sin\text{x})<0\Rightarrow\text{b}<0]$
$\Rightarrow\text{b}\in(-\infty,0)$
View full question & answer→Question 443 Marks
Show that $\text{f}(\text{x})=\frac{1}{\text{x}}$ is decreasing function on $(0,\infty).$
AnswerWe have, $\text{f}(\text{x})=\frac{1}{\text{x}}$ Let, $\text{x}_1,\text{x}_2\in(0,\infty)$ and $\text{x}_1>\text{x}_2$$\Rightarrow\frac{1}{\text{x}_1}<\frac{1}{\text{x}_2}$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ Thus, $\text{x}_1>\text{x}_2\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ So, f(x) is decreasing function.
View full question & answer→Question 453 Marks
If the rate of change of volume of a sphere is equal to the rate of change of its radius, find the radius of the sphere.
AnswerHere,
$\frac{\text{dV}}{\text{dt}}=\frac{\text{dr}}{\text{dt}}$
We know that,
$\text{V}=\frac{4}{3}\pi\text{r}^3$
$\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}$
$\frac{\text{dr}}{\text{dt}}=4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}$ [Using equation (i)]
$4\pi\text{r}^2=1$
$\text{r}=\frac{1}{\sqrt{4\pi}}$
Radius of sphere $=\frac{1}{2\sqrt{\pi}}$ units.
View full question & answer→Question 463 Marks
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
AnswerLet A be area of the circle. Then,
$\text{A}=\pi\text{r}^2$
$\Rightarrow\frac{\text{dA}}{\text{dr}}=2\pi\text{r}$
Hence, the rate of change of the area of the circle is $2\pi\text{r}.$
When r = 5cm,
$\Big(\frac{\text{dA}}{\text{dr}}\Big)_\text{r=5}=2\pi(5)$
$=10\pi\text{cm}^2/\text{cm}$
View full question & answer→Question 473 Marks
Find the absolute maximum value and the absolute minimum value of the following function in the given intervals:
$\text{f}\text{(x)} = \sin \text{x} +\cos \text{x}, \text{x}\in[0,\ \pi]$
AnswerGiven: $\text{f}\text{(x)} =\sin \text{x}+ \cos \text{x}, \text{x}\in[0,\pi]\ \Rightarrow\ \text{f}'\text{(x)}=\cos \text{x}-\sin\text{x}$
$\text{Now }\text{ f}'\text{(x)}=0\ \Rightarrow\ \cos \text{x}-\sin \text{x}=0\ \Rightarrow\ -\sin\text{x}=-\cos \text{x}$
$\Rightarrow\ \tan \text{x}=1\ \ [\text{Posituive}]$
$\therefore\ \text{x }\text{ is in I quadrant.}\ \ [\text{ x}\in(0,\ \pi)]$
$\therefore\ \tan \text{x} =1 = \tan \frac{\pi}{4}\ \Rightarrow\ \text{x}=\frac{\pi}{4}$
$\therefore \ \text{f} \bigg(\frac{\pi}{4}\bigg)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}= \sqrt{2}$
$\text{f} (0)=\sin0+ \cos0=0+1=1$
$\text{f}(\pi)=\sin\pi+\cos\pi = 0-1=-1$
Therefore, absolute minimum value is -1 and absolute maximum value is 1.
View full question & answer→Question 483 Marks
State when a function $f(x)$ is said to be increasing on an interval $[a, b].$ Test whether the function $f(x) = x^2 - 6x + 3$ is increasing on the interval $[4, 6].$
AnswerA function f(x) is said to be increasing on an interval $[a, b]$ if it is increasing at $x = a$ and $x = b.$
Here,
$f(x) = x^2 - 6x + 3$
$f'(x) = 2x - 6$
$f'(x) = 2(x - 3)$
Now, $f'(4) = 2(4 - 3)$
$= 2$
$\therefore f'(4) > 0$
So, f(x) is increasing on $x = 4$
$f'(6) = 2(6 - 3)$
$= 6$
$\therefore f'(6) > 0$
So, $f(x)$ is increasing on $x = 6$
Hence, $f(x)$ is increasing on $[4, 6].$
View full question & answer→Question 493 Marks
Find the maximum value of $2x^3 - 24x + 107$ in the interval $[1, 3].$ Find the maximum value of the same function in $[-3, -1].$
AnswerLet $\text{f}\text{(x)}=2\text{x}^3-24\text{x}+107$ $\Rightarrow\ \text{f}'\text{(x)}=6\text{x}^2-24$
Now $\text{f}'\text{(x)}=0\ \Rightarrow\ 6\text{x}^2-24=0$
$\Rightarrow\ \text{x}^2=4\ \Rightarrow\ \text{x}=\pm2$
$\Rightarrow\ \text{x}=2\text { or }\text{ x}=-2 \ \ [\text{Turning points}]$
For Interval $[1, 3], x = 2$ is turning point.
| At $x = 1,$ |
$f(1) = 2(1) -24(1) + 107 = 129$ |
| At $x = 2,$ |
$f(2) = 2(8) -24(2) + 107 = 75$ |
| At $x = 3,$ |
$f(3) = 2(27) -24(3) + 107 = 89$ |
Therefore, maximum value of $f(x)$ is $89.$
For Interval $[-3, -1], x = -2$ is turning point.
| At $x = -1,$ |
$f(1) = 2 (-1) -24(-1) + 107 = 129$ |
| At $x = -2,$ |
$f(2) = 2 (-8) -24(-2) + 107 = 139$ |
| At $x = -3,$ |
$f(3) = 2 (-27) -24(-3) + 107 = 125$ |
Therefore, maximum value of $f(x)$ is $139.$ View full question & answer→Question 503 Marks
Find the intervals in which f(x) is increasing or decreasing:
$\text{f}(\text{x})=\sin\text{x}+|\sin\text{x}|,0<\text{x}\leq2\pi$
Answer$\text{f}(\text{x})=\sin\text{x}+|\sin\text{x}|,0<\text{x}\leq2\pi$Case I:
When $\text{x}\in(0,\pi)$ $\text{f}(\text{x})=\sin\text{x}+\sin\text{x}=2\sin\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos\text{x}$ As, $\cos\text{x}>0$ for $\text{x}\in\Big(0,\frac{\pi}{2}\Big)$ and $\cos\text{x}<0$ for $\text{x}\in\Big(\frac{\pi}{2},\pi\Big)$ So, $\text{f}'(\text{x})>0$ for $\text{x}\in\Big(0,\frac{\pi}{2}\Big)$ and $\text{f}'(\text{x})<0$ for $\text{x}\in\Big(\frac{\pi}{2},\pi\Big)$ $\therefore$ f(x) is increasing on $\Big(0,\frac{\pi}{2}\Big)$ and f(x) is decreasing on $\Big(\frac{\pi}{2},\pi\Big).$Case II:
When $\text{x}\in(\pi,2\pi)$ $\text{f}(\text{x})=\sin\text{x}-\sin\text{x}=0$ $\Rightarrow\text{f}'(\text{x})=0$ $\therefore$ f(x) is neither increasing nor decreasing on $(\pi,2\pi).$
View full question & answer→