Download App

Maths 4 Marks

Maxima and Minima · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 88 questions.

Questions · Page 2 of 2

4 Marks

Question 514 Marks
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
f(x) = (x - 1)2 + 3 in [-3, 1]
Answer
The given function is f(x) = (x - 1)2 + 3.
$\therefore$ f'(x) = 2(x - 1)
Now, f'(x) = 0 ⇒ 2(x - 1) = 0 ⇒ x = 1
Then, we evalute the value of f at critical point x = 1 and at the end point of the interval [-3, 1].
f(1) = (1 - 1)2 + 3 = 0 + 3 = 3
f(-3) = (-3 - 1)2 + 3 = 16 + 3 = 19
Hence, we can conclude that the absolute maximum value of f on [-3, 1] is occurring at x = -3 and the minimum value of f on [-3, 1] is 3 occurring at x = 1.
View full question & answer→
Question 524 Marks
Find the maximum and the minimum values, if any, without using derivaives of the following functions:

f(x) = sin2x + 5 on R.

Answer
$\text{f}(\text{x})=\sin2\text{x}+5$
We know that $-1\leq\sin2\text{x}\leq1$
$\Rightarrow-1+5\leq\sin2\text{x}+5\leq1+5$
$\Rightarrow4\leq\sin2\text{x}+5\leq6$

Here, the maximum and minimum value of f are 4 and 2 respectively.
View full question & answer→
Question 534 Marks
Find the maximum value of 2x3 - 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].
Answer
Let f(x) = 2x3 - 24x + 107.
$\therefore$ f'(x) = 6x2 - 24 = 6(x2 - 4)
Now, f'(x) = 0 ⇒ 6(x2 - 4) = 0 ⇒ x2 = 4
$\Rightarrow\text{x}=\pm2$
We first consider the interval [1, 3]
Then, We evalute the value of f the critical point $\text{x}=2\in[1,3]$ and at the end point of the interval [1, 3].
f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75
f(1) = 2(1) - 24(1) + 107 = 2 - 24 + 107 = 85
f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, We consider the interval [-3, -1].
Evaluate the value of f at the critical point $\text{x}=2\in[1,3]$ and at the end point of the interval [1, 3].
f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 129
f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129
f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x) in the interval [-3, -1] is 139 occurring at x = -2
View full question & answer→
Question 544 Marks
A particle is moving in a straigth line such that its distance at any time t is given by $\text{S}=\frac{\text{t}^{4}}{4}-2\text{t}^{3}+4\text{t}^{2}-7$. Find when its velocity is maximum and acceleration minimum.
Answer
Given,

$\Rightarrow \text{s}=\frac{\text{t}^{4}}{4}-2\text{t}^{3}+4\text{t}^{2}-7$

$\Rightarrow\text{V}=\frac{\text{ds}}{\text{dt}}=\text{t}^{3}-6\text{t}^{2}+8\text{t}$

$\Rightarrow\text{V}=\frac{\text{ds}}{\text{dt}}=3\text{t}^{2}-12\text{t}^{2}+8$

For maximum or minimum values of V, we must have $\frac{\text{dv}}{\text{dt}}=0$

$\Rightarrow3\text{t}^{2}-12\text{t}+8$

On solving the equation, we get 

$\text{t}=2\pm\frac{2}{\sqrt{2}}$

Now, $\frac{\text{d}^{2}\text{v}}{\text{dt}^{2}}=6\text{t}-12$

At $\text{t}=2\pm\frac{2}{\sqrt{2}}$,

 $\frac{\text{d}^{2}\text{v}}{\text{dt}^{2}}=6\Big(2-\frac{2}{\sqrt{3}}\Big)-12$

$\Rightarrow \frac{-12}{\sqrt{3}}<0$

So, velocity is maximum at $\text{t}=2\pm\frac{2}{\sqrt{2}}$

Again, $\frac{\text{dv}}{\text{dt}}=6\text{t}-12$ 

For maximum or minimum values of a, we must have $\frac{\text{da}}{\text{dt}}=0$

$\Rightarrow6\text{t}-12=0$

$\Rightarrow\text{t}=2$

Now, $\frac{\text{d}^{2}\text{a}}{\text{dt}^{2}}=6>0$

So, acceleration is minimum at t = 2.

View full question & answer→
Question 554 Marks
Find the absolute maximum and minimum values of a function f given by
$\text{f}(\text{x})=12\text{x}^\frac{4}{3}-6\text{x}^\frac{1}{3},\text{x}\in[-1,1]$
Answer
$\text{f}(\text{x})=12\text{x}^\frac{4}{3}-6\text{x}^\frac{1}{3}$
$\therefore\ \text{f}'(\text{x})=16\text{x}^\frac{1}{3}-\frac{2}{\text{x}^{\frac{2}{3}}}=\frac{2(8\text{x}-1)}{\text{x}^{\frac{2}{3}}}$
Thus, f'(x) = 0
$\Rightarrow\ \text{x}=\frac{1}{8}$
Further note that f'(x) is not defined at x = 0.
So, the critical points are x = 0 and $\text{x}=\frac{1}{8}.$
Evaluating the value of f at critical points x = 0, $\frac{1}{8}$ and at end points of the interval x = -1 and x = 1.
$\text{f}(-1)=12(-1)^{\frac{4}{3}}-6(-1)^{-\frac{1}{3}}=18$
$\text{f}(0)=12(0)-6(0)=0$
$\text{f}\Big(\frac{1}{8}\Big)=12(\frac{1}{8})^{\frac{4}{3}}-6(\frac{1}{8})^{\frac{1}{3}}=\frac{-9}{4}$
$\text{f}(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}}=16$
Hence, we can clude thet absolute maximum value of f is 18 at x = -1 and absolute minimum value f is $\frac{-9}{4}\text{ at x}=\frac{1}{8}.$
View full question & answer→
Question 564 Marks
The function y = a log x + bx2 + x has extreme values at x = 1 and x = 2. Find a and b.
Answer
We have, $\text{y}=\text{a}\log\text{x}+\text{b}\text{x}^{2}=\text{x}$ 
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{\text{x}}+2\text{b}\text{x}+1$
and $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{-\text{a}}{\text{x}^{2}}+2\text{b}$
For maxima and minimum value,
$\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{a}}{\text{x}}+2\text{bx}+1=0$
Given that extreme value exist at x = 1, 2
$\Rightarrow\text{a}+2\text{b}=-1\ ...(\text{i})$
$\frac{\text{a}}{2}+4\text{b}=-1$
$\Rightarrow\text{a}+8\text{b}=-2\ ...(\text{ii})$
Solving (i) and (ii), We get
$\text{a}=\frac{-2}{3}, \ \text{b}=\frac{-1}{6}$
View full question & answer→
Question 574 Marks
Find the points of 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: 
$\text{f}(\text{x})=\sin2\text{x},0\leq\text{x}\leq\pi$ 
Answer
Given, $\text{f}(\text{x})=\sin2\text{x}$
$\Rightarrow \text{f}'(\text{x})=2\cos2\text{x}$
For a local maximum or a local minimum, We must have f'(x) = 0
$\Rightarrow 2\cos2\text{x}=0$
$\Rightarrow\cos2\text{x}=0$
$\Rightarrow\text{x}=\frac{\pi}{4}\ \text{or}\ \frac{3\pi}{4}$ 
Since f'(x) change from negative to positive when x increases through $\frac{\pi}{4}, \text{x}=\frac{\pi}{4}$is the point of iocal minima.
The local minimum value os f(x) = 3 at$\text{x}=\frac{\pi}{4}$ is given by $\sin\frac{\pi}{2}=1$ 
Since f'(x) changes from positive to negative when x increases though $\frac{3\pi}{4}, \text{x}=\frac{3\pi}{4}$ is point of local maxima.
The local maximum value of f'(x) at $\text{x}=\frac{3\pi}{4}$ is given by $\sin\Big(\frac{3\pi}{2}\Big)=-1$
View full question & answer→
Question 584 Marks
The total area of a page is 150cm2. The combined width of the margin at the top and bottom is 3cm and the side 2cm. What must be the dimensions of the page in order that the area of the printed matter may be maximum?
Answer
Let x and y be the length and breadth of the rectangular page, respectively. Then, Area of the page =150
$\Rightarrow \text{y}=\frac{150}{\text{x}} ...(\text{i})$
Area of the printed matter = (x - 3)(y - 2)
$\Rightarrow\text{A}=\text{xy}-2\text{x}-3\text{y}+6$
$\Rightarrow\text{A}=150-2\text{x}-\frac{450}{\text{x}}+6$
$\Rightarrow \frac{\text{dA}}{\text{dx}}=-2+\frac{450}{\text{x}^{2}}$
For maximum or minimum values of A, we must have $\frac{\text{dA}}{\text{dx}}=0$
$\Rightarrow -2+\frac{450}{\text{x}^{2}}=0$
$\Rightarrow -2{\text{x}^{2}}=450$
$\Rightarrow\text{x}=15$
Substituting the value of in (i), we get 
y = 10
Now, $\Rightarrow \frac{\text{d}^{2}\text{A}}{\text{dx}^{2}}=\frac{-900}{\text{x}^{3}}$
$\Rightarrow \frac{\text{d}^{2}\text{A}}{\text{dx}^{2}}=\frac{-900}{(15)^{3}}$
$\Rightarrow \frac{\text{d}^{2}\text{A}}{\text{dx}^{2}}=\frac{-900}{3375}<0$
So, area of the printed matter is maximum when x = 15 and y = 10.
View full question & answer→
Question 594 Marks
A closed cylinder has volume 2156cm3. What will be the radius of its base so that its total surface area is minimum.
Answer
$\text{Volume}=\pi\text{r}^{2}\text{h}$
$\Rightarrow 2156=\pi\text{r}^{2}\text{h}$
$\Rightarrow 2156=\frac{22}{7}\text{r}^{2}\text{h}$
$\Rightarrow \text{h}=\frac{2156\times7}{22\text{r}^{2}}$
$\Rightarrow \text{h}=\frac{686}{\text{r}^{2}} \ ...(\text{i})$
Surface area $=2\pi\text{r}\text{h}+2\pi\text{r}^{2}$
$\Rightarrow \text{S}=\frac{4312}{\text{r}}+\frac{44\text{r}^{2}}{7}$ [From eq.(i)]
For maximum or minimum values of S, we must have $\frac{\text{dS}}{\text{dr}}=0$
$\Rightarrow\frac{4312}{\text{-r}^{2}}+\frac{88\text{r}^{2}}{7}=0$
$\Rightarrow\frac{4312}{\text{r}^{2}}=\frac{88\text{r}}{7}$
$\Rightarrow \text{r}^{3}=\frac{4312\times7}{88}$
$\Rightarrow \text{r}^{3}=343$
$\Rightarrow \text{r}=7\text{cm}$
Now, $\frac{\text{d}^{2}\text{s}}{\text{dr}^{2}}=\frac{8624}{\text{r}^{3}}+\frac{88}{7}$
$\Rightarrow\frac{\text{d}^{2}\text{s}}{\text{dr}^{2}}=\frac{8624}{343}+\frac{88}{7}$
$\Rightarrow\frac{\text{d}^{2}\text{s}}{\text{dr}^{2}}=\frac{176}{7}>0$
So, the surface area is minimum when r = 7cm.
View full question & answer→
Question 604 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:
$\text{f}(\text{x})=\sin2\text{x}-{\text{x}},-\frac{\pi}{2}\leq\text{x}\leq\frac{\pi}{2}$
Answer
Given, $\text{f}(\text{x})=\sin2\text{x}-\text{x}$
$\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}-1$
For a local maximum or a local minimum, We must have f'(x)=0
$\Rightarrow2\cos2\text{x}-1=0$
$\Rightarrow\cos2\text{x}=\frac{1}{2}$
$\Rightarrow\text{x}=\frac{-\pi}{6}\ \text{or}\ \frac{\pi}{2}$
Since f'(x) changes from positive to negative when x increses through $-\frac{\pi}{6}, \text{x}=-\frac{\pi}{6}$ is the point of local maxima.
The local maximum value of f(x) at $\text{x}=-\frac{\pi}{6}$ is given by $\sin\Big(\frac{-\pi}{3}\Big)+\frac{\pi}{6}=\frac{\pi}{6}-\frac{\sqrt{3}}{2}$
View full question & answer→
Question 614 Marks
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 4x2 - 4x + 4 on R.
Answer
$\text{f}(\text{x})=4\text{x}^{2}-4\text{x}+4$ on R
$=4\text{x}^{2}-4\text{x}+1+3$
$=(2\text{x})^{2}+3$
$ \therefore(2\text{x}-1)^{2}\geq0$
$\Rightarrow(2\text{x}-1)^{2}+3\geq3$
$\Rightarrow\text{f}(\text{x})\geq\text{f}\Big(\frac{1}{2}\Big)$

Thus, the minimum value of f(x) is 3 at $\text{x}=\frac{1}{2}$ 
since, f(x) can be made as large as. Therefore maximum value does not exist. 
View full question & answer→
Question 624 Marks
Write the minimum value of f(x) = xx .
Answer
We have $\text{f}(\text{x})=\text{x}^{\text{x}}$
$\therefore\ \text{f}'(\text{x})=\text{x}^{\text{x}}(\log\text{x}+1)$
For maxima and minima, f'(x) = 0
$\Rightarrow \text{x}^{\text{x}}(\log\text{x}+1)=0$
$\Rightarrow \text{x}=\text{e}^{-1}$
Now,
$\therefore\ \text{f}''(\text{x})=\text{x}^{\text{x}}(\log\text{x}+1)^{2}+\frac{\text{x}^{\text{x}}}{\text{x}}$
At $\text{x}=\frac{1}{\text{e}}$
$\therefore\ \text{f}''(\text{x})>0\ \text{as}\ \text{x}^{\text{x}}(\log\text{x}+1)^{2}+\frac{\text{x}^{\text{3}}}{\text{x}}>0$
$\therefore \text{x}=\frac{1}{\text{e}}$ is the point of local minima.
Hence, minimum value $=\text{f}\Big(\frac{1}{\text{e}}\Big)=\Big(\frac{1}{\text{e}}\Big)^\frac{1}{\text{e}}=\text{e}^\frac{-1}{\text{e}}$ .
View full question & answer→
Question 634 Marks
Find the absolute maximum and minimum values of a function f given by
f(x) = 2x3 - 15x2 + 36x + 1 on the interval [1,5]
Answer
Given, f(x) = 2x3 = 15x2 + 36x + 1
$\therefore$ f'(x) = 6x2 - 30x + 36 
=6(x2 - 5x + 6) = 6(x - 2)(x - 3)
Now, that f'(x) = 0 gives x = 2 and x = 3
We shall now evaluate the value of f at these points and at the ens points of the interval [1,5], i.e at x = 1, 2, 3 and 5
At x = 1, f(1) = 2(13) - 15(12) + 36(1) + 1 = 24
At x = 2, f(2) = 2(23) - 15(22) + 36(2) + 1 = 29
At x = 3, f(3) = 2(33) - 15(32) + 36(3) + 1 = 28
At x = 5, f(2) = 2(53) - 15(52) + 36(5) + 1 = 56
Thus, We conclude that the absolute maximum value of on [1, 5] is 56, occurring at x = 5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1.
View full question & answer→
Question 644 Marks
Find the maximum slope of the curve y = −x3 + 3x2 + 2x − 27.
Answer
Given, y = -x3 + 3x2 + 2x - 27 ...(i)
Slope $=\frac{\text{dy}}{\text{dx}}=-3\text{x}^{2}+6\text{x}+2$
Now, $\text{M}=-3\text{x}^{2}+6\text{x}+2$
$ \Rightarrow \frac{\text{dM}}{\text{dx}}=-6\text{x}+6$
For maximum or minimum value of M, we must have $\frac{\text{dM}}{\text{dx}}=0$
$\Rightarrow -6\text{x}+6=0$
$\Rightarrow 6\text{x}=6$
$\Rightarrow \text{x}=1$
Substituing the value of x in eq.(i), we get 
$\text{y}=-1^{3}+3\times1^{2}+2\times1-27=-23$
$\frac{\text{d}^{2}\text{M}}{\text{dx}^{2}}=-6<0$
So, the slope is maximum when x = 1 and y = -23.
$\therefore$ At (1, -23)
Maximum slope = -3(1)2 + 6(1) + 2 = -3 + 6 + 2 = 5
View full question & answer→
Question 654 Marks
Two sides of a triangle have lengths 'a' and 'b' and the angle between them is θ. What value of θ will maximize the area of the triangle? Find the maximum area of the triangle also.
Answer
ABC is a given triangle with AB = a, BC = b and $\angle \text{ABC =} \theta$
 AD in perpendicular to BC.
$\therefore$ $\text{BC}=\text{a}\sin\theta$
Now, Area of $\triangle\text{AbC}=\frac{1}{2}\times\text{BC}\times\text{AD}$
$\text{A}=\frac{1}{2}\ \text{b}\times\text{a}\sin\theta...(\text{i})$
$\therefore\ \frac{\text{dA}}{\text{d}\theta}=\frac{1}{2}\ \text{ab} \cos\theta$
For maxima and minima.
$\frac{\text{dA}}{\text{d}\theta}=0$
$\Rightarrow\frac{1}{2}\ \text{ab} \cos\theta=0$
$\Rightarrow \cos\theta=0 $
$\Rightarrow\theta=\frac{\pi}{2}$
Now, $\frac{\text{d}^{2}\text{A}}{\text{d}\theta}=-\frac{1}{2}\ \text{ab} \sin\theta$
At, $\theta=\frac{\pi}{2}, \frac{\text{d}^{2}\text{A}}{\text{d}\theta^{2}}=-\frac{1}{2}\ \text{ab}<0$
$\theta=\frac{\pi}{2}$ is point of local maxima
$\therefore$ maximum area of $\triangle= \frac{1}{2}\ \text{ab}\ \sin\frac{\pi}{2}=\frac{1}{2}\ \text{ab}$.
View full question & answer→
Question 664 Marks
Write the minimum value of $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}.$
Answer
We have $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}$

Taking log on both side, we get 

$\log\text{f}(\text{x})=\frac{1}{\text{x}}\ \log\text{x}$

Differentitating W.r.t. x, we get

$\frac{1}{\text{f}(\text{x})}\text{f}'(\text{x})=\frac{-1}{\text{x}^{2}}\ \log\text{x}+\frac{1}{\text{x}^{2}}$

$\Rightarrow\text{f}'(\text{x})=\text{f}'(\text{x})\frac{1}{\text{x}^{2}}\ (1-\log\text{x})$

$\Rightarrow\text{f}'(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big) ...(\text{i})$

$\Rightarrow\text{f}'(\text{x})=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x})$

For a local maximum or a local minima, we must have f'(x) = 0

$=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x}) =0$

$\Rightarrow \log\text{x}=1$

$\therefore \text{x}=\text{e}$

Now, $\text{f}''(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$

$=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$

At x = e

$\text{f}''(\text{e})=\text{e}^\frac{1}{\text{e}}\Big(\frac{1}{\text{e}^{2}}-\frac{1}{\text{e}^{2}}\log\text{e}\Big)^{2}+\text{e}^\frac{1}{\text{e}}\Big(\frac{-3}{\text{e}^{3}}+\frac{2}{\text{e}^{3}}\log\text{e}\Big)$

$=-\text{e}^\frac{1}{\text{e}}(\frac{1}{\text{e}^{3}})<0$

So, x = e is a point of local maximum.

Thus, the maximum value is given by 

$\text{f}(\text{e})=\text{e}^\frac{1}{\text{e}}$

View full question & answer→
Question 674 Marks
Find the point on the curve y2 = 4x which is nearest to the point (2, -8).
Answer
Let point (x, y) be the nearest to the point (2, -8). Then,
y2 = 4x
$\Rightarrow \text{x}=\frac{\text{y}^{2}}{4}\ ...(\text{i})$
$\text{d}^{2}=(\text{x}-2)^{2}+(\text{y}+8)^{2}$
Now, $\text{Z}=\text{d}^{2}=(\text{x}-2)^{2}+(\text{y}+8)^{2}$
$\Rightarrow \text{Z}=\Big(\frac{\text{y}^{2}}{4}-2\Big)^{2}+(\text{y+8})^{2}$
$\Rightarrow \text{Z}=\frac{\text{y}^{4}}{16}+4-\text{y}^{2}+\text{y}^{2}+64+16\text{y}$
For maximum or minimum value of Z, we must have $\frac{\text{dZ}}{\text{dy}}=0$ 
$\Rightarrow \frac{4\text{y}^{3}}{16}+16=0$
$\Rightarrow \frac{4\text{y}^{3}}{16}=16$
$\Rightarrow \text{y}^{3}=-64$
$\Rightarrow \text{y}=-4$
Substituting the valur of x in eq.(i), we get x = 4
Now, $\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=12>0$
So, the nearest point is (4, -4).
View full question & answer→
Question 684 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) = x3(x - 1)2

Answer
Given, $\text{f}(\text{x})=\text{x}^{3}(\text{x}-1)^{2}$
$\Rightarrow\text{f}'(\text{x})=3\text{x}^{2}(\text{x - 1})^{2}+2\text{x}^{3}(\text{x - 1})$
For a local maximum or a local minimum, We must have f'(x)=0
$\Rightarrow3\text{x}^{2}(\text{x}-1^{2})+2\text{x}^{3}(\text{x}-1)=0$
$\Rightarrow\text{x}(\text{x}-1)\left\{3\text{x}-3+2\text{x}\right\}=0$
$\Rightarrow\text{x}^{2}(\text{x - 1})(5\text{x}-3)=0$
$\Rightarrow\text{x}=0, 1, \frac{3}{5}$
Since f'(x) changes from negative to positive when x increases through 1, x = 1 is the point of local minima.
The local minimum value of f(x) at x = 1 is given by (1)3(1 - 1)2 = 0
Since f(x) changes from positive to negative when x increases through $\frac{3}{5},\text{x}=\frac{3}{5}$ is the ponit of local maxima.
The local minimum value of f(x) at $\text{x}=\frac{3}{5}$ is given by $\Big(\frac{3}{5}\Big)^{3}\Big(\frac{3}{5}-1\Big)^{2}=\frac{27}{125}\times\frac{4}{25}=\frac{108}{3125}$
Since f"(x) does not change from positive as x increases through 0, x = 0 is a point of inflexion.
View full question & answer→
Question 694 Marks
A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.
Answer
Let the dimensions of the rectangular part be x and y.
Perimeter of the window = x + y + x + x + y = 12
⇒ 3x + 2y = 12
$\text{y}=\frac{12-3\text{x}}{2}....(\text{i})$
Area of the window $=\text{xy}+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\text{A}=\text{x}\Big(\frac{12-3\text{x}}{2}\Big)+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\text{A}=6\text{x}-\frac{3\text{x}^{2}}{2}+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\frac{\text{dA}}{\text{dx}}=6\text{x}-\frac{6\text{x}}{2}+\frac{2\sqrt{3}}{4}\text{x}$
For maximum or minimum values of A, We must have $\frac{\text{dA}}{\text{dx}}=0$
$\Rightarrow 6=\text{x}\Big(3-\frac{\sqrt{3}}{2}\Big) $
$\Rightarrow \text{x}=\frac{12}{6\sqrt{3}} $
Substituting the values of x in eq.(i), We get
$\Rightarrow \text{y}=\frac{12-3\Big(\frac{12}{6-\sqrt{3}}\Big)}{2} $
$\Rightarrow \text{y}=\frac{18-6\sqrt{3}}{6-\sqrt{3}}$.
View full question & answer→
Question 704 Marks
The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.
Answer
The surface area of a sphere is given by $\text{SA}=4\pi\text{r}^{2}$
The surface area of a sphere is given by $\text{SA}=6\text{S}^{2}$
Where r is the radius of the sphere and S is the side lenght of the cube.
Let the sum of surface areas be K, so $\text{K}=4\pi\text{r}^{2}+6\text{S}^{2}$. Then we can solve for S getting $\text{S}=\sqrt{\frac{\text{K}-4\pi\text{r}^{2}}{6}}$
Now, the sum of volume is $\frac{4}{3}\pi\text{r}^{2}+\text{S}^{3}$.
Substituting for yields  $\frac{4}{3}\pi\text{r}^{2}+\Big(\frac{\text{K}-4\pi\text{r}^{2}}{6}\Big)^\frac{3}{2}$ .  
Setting this eual to zero we find:
$4\pi\text{r}^{2}+\frac{3}{2}\Big(\frac{\text{k}-4\pi\text{r}^{2}}{6}\Big)^\frac{1}{2}\Big(-\frac{4}{3}\pi\text{r}\Big)=0$
$4\pi\text{r}^{2}=2\pi\text{r}\Big(\frac{\text{k}-4\pi\text{r}^{2}}{6}\Big)$
$2\text{r}=\Big(\frac{\text{K}-4\pi\text{r}^{2}}{6}\Big)^\frac{1}{2}$
$4\text{r}^{2}=\frac{\text{K}-4\pi\text{r}^{2}}{6}$
$24\text{r}^{2}=(4\pi\text{r}^{2}+6\text{S}^{2})$
$24\text{r}^{2}=6\text{s}^{2}$
$4\text{r}^{2}=\text{S}^{2}$
Since, we are dealing with lenghts, we only need the positive root.
Since, 2r is the diameter ot the sphere, we have shown that the sum of the volumes is minimum when the diametet of the sphare length of the cube.
View full question & answer→
Question 714 Marks
The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius a.
Answer
Let the breadth height and strength of the beam be b, h and S, respectively.
$\text{a}^{2}=\frac{\text{h}^{2}+\text{b}^{2}}{4}$
$\Rightarrow4\text{a}^{2}-\text{b}^{2}=\text{h}^{2}\ ...(\text{i})$
Here, Strength of beam, $\text{S}=\text{Kbh}^{2}$
$\Rightarrow \text{S}=\text{kb}(4\text{R}^{2}-\text{b}^{2})$
$\Rightarrow \text{S}=\text{k}(\text{b}4\text{R}^{2}-\text{b}^{3})$
$\Rightarrow \frac{\text{dS}}{\text{db}}=\text{k}(4\text{a}^{2}-3\text{b}^{2})$
For maximum or minimum values of S, we must have $\frac{\text{dS}}{\text{db}}=0$
$\Rightarrow \text{k}(4\text{a}^{2}-3\text{b}^{2})=0$
$\Rightarrow 4\text{a}^{2}-3\text{b}^{2}=0$
$\Rightarrow 4\text{a}^{2}-=3\text{b}^{2}$
$\Rightarrow \text{b}=\frac{2\text{a}}{\sqrt{3}}$
Substituting the value of b in eq.(i), we get 
$\Rightarrow 4\text{a}^{2}-(\frac{2\text{a}}{\sqrt{3}})^{2}=\text{h}^{2}$
$\Rightarrow \frac{12\text{a}^{2}-4\text{a}^{2}}{3}=\text{h}^{2}$
$\Rightarrow \text{h}=\frac{2\sqrt{2}}{\sqrt{3}}\text{a}$
Now, $\frac{\text{d}^{2}\text{s}}{\text{db}^{2}}=-6\text{Kb}$
$\Rightarrow\frac{\text{d}^{2}\text{s}}{\text{db}^{2}}=-6\text{K}\frac{2\text{a}}{\sqrt{3}}$
$\Rightarrow\frac{\text{d}^{2}\text{s}}{\text{db}^{2}}=\frac{-12\text{Ka}}{\sqrt{3}}<0$
So, the strength of beam is maximum when $ \text{b}=\frac{2\text{a}}{\sqrt{3}}$ and $\text{h}=\frac{2\sqrt{2}}{\sqrt{3}}\text{a}$.
View full question & answer→
Question 724 Marks
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$\text{f}(\text{x})=4\text{x}-\frac{\text{x}^{2}}{2}\ \text{in}\ [2,4,5]$
Answer
We have, $\text{f}(\text{x})=4\text{x}-\frac{\text{x}^{2}}{2}$
$\Rightarrow\text{f}'(\text{x})=4-\text{x}$
For a local maximum or a local minimum value, We must have f'(x) = 0
$\Rightarrow4-\text{x}=0$
$\Rightarrow\text{x}=4$
Thus, the critical points of f are -2, 4 and 4.5.
Now, $\text{f}(-2)=4(-2)-\frac{(2)^{2}}{2}=-8-2=-10$
$\text{f}(4)=4(4)-\frac{(4)^{2}}{2}=16-8=8$
$\text{f}(4.5)=4(4.5)-\frac{(4.5)^{2}}{2}=18-10.125=7.875$
Hence, the absolute maximum value when x = 4 is 8 and the absolute minimum value when x = -2 is -10.
View full question & answer→
Question 734 Marks
Find the absolute maximum and minimum values of the function of given by 
$\text{f}(\text{x})=\cos^{2}\text{x}+\sin\text{x}, \text{x}\in[0,\pi]$
Answer
Given, $\text{f}(\text{x})=\cos^{2}\text{x}+\sin\text{x}$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(-\sin\text{x})+\cos\text{x}$
$=-2\sin\text{x}\cos\text{x}+\cos\text{x}$
For a local maximum or a local minimum, We must have f'(x) = 0
 $\Rightarrow-2\sin\text{x}\cos\text{x}+\cos\text{x}=0$
$\Rightarrow\cos\text{x}(2\sin\text{x}-1)=0$
$\Rightarrow\sin\text{x}=\frac{1}{2}\ \text{or}\ \cos\text{x}=0$
$\Rightarrow\text{x}=\frac{\pi}{6}\ \text{or}\ \frac{\pi}{2}\ [\therefore \text{x}\in(0,\pi)]$
Thus, the critical points of f are $0,\frac{\pi}{6},\frac{\pi}{2}\ \text{and}\ \pi$.
Now, $\text{f}(0)=\cos^{2}(0)+\sin(0)=1$
$\text{f}\Big(\frac{\pi}{6}\Big)=\cos^{2}\Big(\frac{\pi}{6}\Big)+\sin\Big(\frac{\pi}{6})=\frac{5}{4}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\cos^{2}\Big(\frac{\pi}{2}\Big)+\sin\Big(\frac{\pi}{2}\Big)=1$
$\text{f}(\pi)=\cos^{2}(\pi)+\sin(\pi)=1$
Hence, the absolute maximum value when $\text{x}=\frac{\pi}{6}$ is $\frac{5}{4}$ and the absolute minimum value when $\text{x}=0,\frac{\pi}{2},\pi$ is 1.
View full question & answer→
Question 744 Marks
A box of constant volume c is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square metre as that in the bottom. What are the most economic dimensions?                                                     
Answer
Let l, b and h be the length, breadth and height of the box, respectively.
Volume of the box = C
Given, $\text{l} =2\text{b}\ ..(\text{i})$
$\Rightarrow \text{c}=\text{Ib}\text{h}$
$\Rightarrow \text{c}=2\text{b}^{2}\text{h}$
$\Rightarrow \text{h}=\frac{\text{c}}{2\text{b}^{2}}\ ...(\text{ii})$
Let cost of the material required for bottom be K/ m2.
Cost of the material reqired for 4 walls and top = Rs 3K/ m2
Total cost, $\text{T} =\text{K}(l\text{b})+3\text{k}(2l\text{h}+2\text{bh}+l\text{b})$
$\Rightarrow\text{T}=2\text{K}\text{b}^{2}+3\text{K}\Big(\frac{4\text{bc}}{2\text{b}^{2}}+\frac{2\text{bc}}{2\text{b}^{2}}+2\text{b}^{2}\Big) $
$\Rightarrow\frac{\text{dT}}{\text{db}}=4\text{Kb}+3\text{K}\Big(\frac{-3\text{c}}{\text{b}^{2}}+4\text{b}\Big)$
For maximum or minimum values of T, we must have $\frac{\text{dT}}{\text{bd}}=0$
$\Rightarrow=4\text{Kb}+3\text{K}\Big(\frac{-3\text{c}}{\text{b}^{2}}+4\text{b}\Big)=0$
$\Rightarrow 4\text{b}=3\Big(\frac{3\text{c}}{\text{b}^{2}}-4\text{b}\Big)$
$\Rightarrow 4\text{b}=\Big(\frac{9\text{c}}{\text{b}^{2}}-12\text{b}\Big)$
$\Rightarrow 4\text{b}=\frac{9\text{c}-12\text{b}^{3}}{\text{b}^{2}}$
$\Rightarrow4 \text{b}^{3}=9\text{c}-12\text{b}^{3}$
$\Rightarrow16\text{b}^{3}=9\text{c}$
$\Rightarrow\text{b}=\Big(\frac{9\text{c}}{1}\Big)^\frac{1}{3}$
Now, $\frac{\text{d}^{2}\text{T}}{\text{db}^{2}}=4\text{K}+3\text{K}\Big(\frac{6\text{c}}{\text{b}^{2}}+4\Big)$
$\Rightarrow\frac{\text{d}^{2}\text{T}}{\text{db}^{2}}=4\text{K}+3\text{K}\Big(\frac{6\text{c}}{\text{b}^{2}}\times16+4\Big)$
$\Rightarrow \text{K}(4+3\times\frac{44}{3})$
$\Rightarrow 48 \text{K}>0$
$\therefore$ cost is minimum when $\text{b}=(\frac{9\text{c}}{16})^\frac{1}{3}$
Substituting $\text{b}=(\frac{9\text{c}}{16})^\frac{1}{3}$ in eq.(i) and eq.(ii)
$\Rightarrow l=2(\frac{9\text{c}}{16})^\frac{1}{3}$
$\text{h}=\frac{\text{c}}{2\text{b}^{2}}$
$\Rightarrow\text{h}=\frac{\text{c}}{2\Big(\frac{9\text{c}}{16}\Big)^\frac{2}{3}}$
$\Rightarrow \text{h}=\Big(\frac{32\text{c}}{81}\Big)^\frac{1}{3}$
Thus, the most economic dimensions of the box are l $=2(\frac{9\text{c}}{16})^\frac{1}{3}$, $\text{b}=(\frac{9\text{c}}{16})^\frac{1}{3}$ and $\text{h}=\Big(\frac{32\text{c}}{81}\Big)^\frac{1}{3}$ .
View full question & answer→
Question 754 Marks
Prove that $\text{f}(\text{x})=\sin\text{x}+\sqrt{3}\cos\text{x}$ has maximum value at $\text{x}=\frac{\pi}{6}$.
Answer
We have, $\text{f}(\text{x})=\sin\text{x}+\sqrt{3}\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+\sqrt{3}(-\sin\text{x})$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sqrt{3}-\sin\text{x}$
For f(x) to have maximum to minimum value, We must have f'(x) = 0
$\Rightarrow\cos\text{x}-\sqrt{3}\sin\text{x}=0$
$\Rightarrow\cot\text{x}=\sqrt{3}$
$\Rightarrow\text{x}=\frac{\pi}{6}$
Also, $\text{f}''(\text{x})=-\sin\text{x}-\sqrt{3}\cos\text{x}$
$\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{-\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}$
$=-\frac{1}{2}-\sqrt{3}\Big(\frac{\sqrt{3}}{2}\Big)=-\frac{1}{2}-\frac{3}{2}=-2<0$
So, $\text{x}=\frac{\pi}{6}$ is point of maxima.
View full question & answer→
Question 764 Marks
Find the maximum and the minimum values, if any, without using derivaives of the following functions:

f(x) = 2x3 + 5 on R.

Answer
We can observe yhat f(x) increases when the values of x are incrwased and f(x) decreases when the values of x aer decreased. Also, f(x) can be reduced by giving small values of x.
Similarly, f(x) can be enlaeged by giving large values of x. so, f(x) does not have a minimum or maximum value.
View full question & answer→
Question 774 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 - 1)(x + 2)2

Answer
Given,
f(x) = (x - 1)(x + 2)2
⇒ f'(x) = (x - 2)2 + 2(x + 2)(x - 1) 
For a local maximum or a local minimum, We must have f'(x)=0
⇒ (x + 2)(x + 2 + 2x - 2) = 0
⇒ (x + 2)(3x) = 0
⇒ x = 0, -2
Since f'(x) changes from negative to positive when x increases through 1, x=0 is the point of local minima.
The local minimum value of f(x) at x=1 is given by (0 - 1)(0 + 2)2 = -4
Since f(x) changes from positive to negative when x increases through -2, x = -2 is the ponit of local maxima.
The local minimum value of f(x) at x = -2 is given by (-2 - 1)(-2 + 2)2 = 0
View full question & answer→
Question 784 Marks
Find the maximum and the minimum values, if any, without using derivaives of the following functions:

f(x) = |x + 2| + 2 on R.

Answer
Given: $\text{f}(\text{x})=|\text{x}+2| +2$
Now, $|\text{x}+2|\geq0$  for all $\text{x}\in\text{R}$
Thus, $\text{f}(\text{x})\geq0$ for all $\text{x}\in\text{R}$

Threrefore, the minimum value of f at x = -2 is 0
since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, the function f does not have a maximum value.
View full question & answer→
Question 794 Marks
Find the maximum and minimum values of the function $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$
Answer
Given, $\text{f}(\text{x})=\frac{4}{\text{x}+2}+\text{x}$
$\Rightarrow\text{f}'(\text{x})=-\frac{4}{(\text{x}+2)^{2}}+1$
For a local maxima or a local minima, We must have f'(x) = 0
$\Rightarrow-\frac{4}{(\text{x}+2)^{2}}+1=0$
$\Rightarrow-\frac{4}{(\text{x}+2)^{2}}=-1$
$(\text{x}+2)^{2}=\pm2$
$\Rightarrow \text{x}=0 \ \text{and} -4$
Thus, x = 0 and x = -4 are the possible of local maxima or local minima.
Now, $\text{f}'(\text{x})=\frac{8}{(\text{x}+2)^{3}}$
At x = 0
$\text{f}''(0)=\frac{8}{(2)^{3}}=1>0$
So, x = 0 is a point of local minimum.
The local minimum value is given by 
$\text{f}(0)=\frac{4}{(0+2)}+0=2$
View full question & answer→
Question 804 Marks
Find the point on the parabolas x2 = 2y which is closest to the point (0,5).
Answer
Let the required point be (x, y). Then, 
x2 = 2y
$\Rightarrow \text{y}=\frac{\text{x}^{2}}{2} ...(\text{i})$
The distance between points (x, y) and (0, 5) is given by
d2 = (x)2 + (y - 5)2
Now, d2 = Z
$\Rightarrow \text{Z}=(\text{x}^{2})+\Big(\frac{\text{x}^{2}}{2}-5\Big)^{2}$
$\Rightarrow \text{Z}=\text{x}^{2}+\frac{\text{x}^{4}}{4}+25-5\text{x}^{2}$
$\Rightarrow \frac{\text{dZ}}{\text{dy}}=2\text{x}+\text{x}^{3}-10\text{x}$
For maximum or a minimum valurs of Z, we must have $\frac{\text{dZ}}{\text{dy}}=0$
$\Rightarrow \text{x}^{3}-8\text{x}=0$
$\Rightarrow \text{x}^{3}=8\text{x}$
$\Rightarrow \text{x}=\pm2\sqrt{2}$
Substituting the value of x in eq.(i), we get 
y = 4
$\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=3\text{x}^{2}-8$
$\Rightarrow\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=24-8=16>0$
So, the nearest pointis $(\pm2\sqrt{2}, 4)$.
View full question & answer→
Question 814 Marks
Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum.
Answer
Let the two positive numbers be x and y. Then,
x + y = 15 ....(1)
Now, z = x2 + y2
⇒ z = x2 + (15 - x)2  [from eq.(1)]
⇒ z = x2 + x2 + 225 - 30x
⇒ z = 2x2 + 225 - 30x
$\Rightarrow \frac{\text{dz}}{\text{dx}}=4\text{x}-30$
For maximum or minimum value of z, We must have
$\Rightarrow \frac{\text{dz}}{\text{dx}}=0$
$\Rightarrow 4\text{x}-30=0$
$\Rightarrow \text{x}=\frac{15}{2}$
$\frac{\text{d}^{2}\text{z}}{\text{dx}^{2}}=4>0$
Substituting $\text{x}=\frac{15}{2}$ in (1), We get 
$ \text{y}=\frac{15}{2}$
Thus, z is minimum when $\text{x}=\text{y}=\frac{15}{2}$.
View full question & answer→
Question 824 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,
f(x) = xex
Answer
We have, $\text{f}(\text{x})=\text{x}\text{e}^{\text{x}}$
 $\therefore \text{f }(\text{x})=\text{e}^{\text{x}}+\text{xe}^{\text{x}} =\text{e}^{\text{x}}(\text{x}+1)$
$ \text{f}''(\text{x})=\text{e}^{\text{x}}(\text{x}+1)+\text{e}^\text{x}$
$=\text{e}^\text{x}(\text{x}+2)$
For maxima and minima,
f'(x) = 0
$\Rightarrow\text{e}^{\text{x}}(\text{x}+1)=0$
⇒ x = -1
Now,  $\text{f}''(-1) =\text{e}-1 =\frac{1}{\text{e}}>0$ 
$\therefore \text{x} =-1$ is point of local minima.
Hence, local minimum value $= \text{f}(-1) = \frac{-1}{\text{e}}$
View full question & answer→
Question 834 Marks
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$\text{f}(\text{x})=(\text{x}-2)\sqrt{\text{x}-1}\ \text{in}\ [1,9]$
Answer
Given, $\text{f}(\text{x})=(\text{x}-2)\sqrt{\text{x}-1}\ $
$\Rightarrow\text{f}'(\text{x})=\sqrt{\text{x}-1}+\frac{(\text{x}-2)}{2\sqrt{\text{x}-1}} $
For a local maximum or a local minimum, We must have f'(x) = 0
$\Rightarrow\sqrt{\text{x}-1}+\frac{(\text{x}-2)}{2\sqrt{\text{x}-1}}=0 $
$ \Rightarrow2(\text{x}-1)+(\text{x}-2)=0$
$\Rightarrow2\text{x}-2+\text{x}-2=0$
$\Rightarrow3\text{x}-4=0$
$\Rightarrow3\text{x}=4$
$\Rightarrow\text{x}=\frac{4}{3}$
Thus, the critical points of f are $1,\frac{4}{3}$ and 9.
Now, $ \text{f}(1)=(1-2)\sqrt{1-1}=0$ 
$\text{f}\Big(\frac{4}{3}\Big)=\Big(\frac{4}{3}-2\Big)\sqrt{\frac{4}{3}-1}=\frac{-2}{3}\times\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$
$\text{f}(9)=(9-2)\sqrt{9-1}=14\sqrt{2}$
Hence, the absolute maximum value when x = 9 is $14\sqrt{2}$  and the absolute minimum value when $\text{x}=\frac{4}{3}$ is $-\frac{2}{3\sqrt{3}}$.
View full question & answer→
Question 844 Marks
Find the maximum and the minimum values, if any, without using derivaives of the following functions:

f(x) = -|x + 1| + 3 on R.

Answer
$\text{g}(\text{x})=-|\text{x}+1|+3$
We know that $-|\text{x}+1|\leq0$ for every $\text{x}\in\text{R}$
Thererfore, $\text{g}(\text{x})=-|\text{x}+1|+3\leq3$ for every $\text{x}\in\text{R}$
The maximum value of g is attained when $|\text{x}+1|=0$
$|\text{x}+1|=0$
$\Rightarrow\text{x}=-1$

$\therefore$ maximum value of $\text{g}=\text{g}(-1)=-|-1+1|+3=3$
Hence, function g does not have a minimum value.
View full question & answer→
Question 854 Marks
Find the coordinates of a point on the parabola y = x2 + 7x + 2 which is closest to the strainght line y = 3x -3.
Answer
Let coordinates of a point on the parabola be (x, y). Then,
y = x2 + 7x + 2 ...(i)
Let the distance of a point (x, (x2 + 7x + 2)) from the line y = 3x - 3 be S. Then 
$\text{S}=\frac{-3\text{x}+(\text{x}^{2}+7\text{x}+2)+3}{\sqrt{10}}$
$\Rightarrow \frac{\text{dS}}{\text{dt}}=\frac{-3+2\text{x}+7}{\sqrt{10}}$
For maximum or minimum values of S, we must have $\frac{\text{dS}}{\text{dt}}=0$
$\Rightarrow \frac{-3+2\text{x}+7}{\sqrt{10}} = 0$
$\Rightarrow 2\text{x}=-4$
$\Rightarrow \text{x}=-2$
Now, $\frac{\text{d}^{2}\text{S}}{\text{dt}^{2}}=\frac{2}{\sqrt{10}}>0$
So, the neaerst point is (x, (x2 + 7x + 2)).
$\Rightarrow (-2,4-14+2)$ 
$\Rightarrow (-2,-8)$
View full question & answer→
Question 864 Marks
Find the point on the curvey y2 = 2x which is at a minimum distance from the point (1, 4).
Answer
Let coordinates of a point on the parabola be (x, y). Then,
y = x2 + 7x + 2 ...(i)
Let the distance of a point (x, (x2 + 7x + 2)) from the line y = 3x - 3 be S. Then 
$\text{S}=\frac{-3\text{x}+(\text{x}^{2}+7\text{x}+2)+3}{\sqrt{10}}$
$\Rightarrow \frac{\text{dS}}{\text{dt}}=\frac{-3+2\text{x}+7}{\sqrt{10}}$
For maximum or minimum values of S, we must have $\frac{\text{dS}}{\text{dt}}=0$
$\Rightarrow \frac{-3+2\text{x}+7}{\sqrt{10}} = 0$
$\Rightarrow 2\text{x}=-4$
$\Rightarrow \text{x}=-2$
Now, $\frac{\text{d}^{2}\text{S}}{\text{dt}^{2}}=\frac{2}{\sqrt{10}}>0$
So, the neaerst point is (x, (x2 + 7x + 2)).
$\Rightarrow (-2,4-14+2)$ 
$\Rightarrow (-2,-8)$
View full question & answer→
Question 874 Marks
If f(x) = x3 + ax2 + bx + c has a maximum at x = -1 and minimum at x = 3. Determine a, b and c.
Answer
We have, f(x) = x3 + ax2 + bx + c
⇒ f'(x) = 3x2 + 2ax + b
As, f(x) is maximum at x = -1 and minimum at x = 3.
So, f(-1) = 0 and f(3) = 0
⇒ 3(-1)2 + 2a(-1) + b = 0 and 3(3)2 + 2a(3) + b = 0
⇒ 3 - 2a + b = 0 ...(i)
and 27 + 6a + b = 0 ...(ii)
(ii) - (i), We get
27 - 3 + 6a + 2a = 0
⇒ 8a = -24
⇒ a = -3
Substituting a = -3 in (i), we get
3 - 2(-3) + b = 0
⇒ 3 + 6 + b = 0
⇒ b = -9
And, $\text{C}\in$ R
View full question & answer→
Question 884 Marks
Divide 64 into two parts such that the sum of the cubes of two parts is minimum.
Answer
Let, x and y be the two parts of 64.
$\therefore$ x + y = 64 ....(i)
Let, s = x3 + y3 .......(ii)
From (i) and (ii), We get
$\text{s}=\text{x}^{3}+(64-\text{x}^{3})$
$\frac{\text{ds}}{\text{dt}}=3\text{x}^{2}+3(64-\text{x}^{2})\times(-1)$
$=3\text{x}^{2}+3(4096-128\text{x}+\text{x}^{2})$
$=3(4096-128\text{x})$
For maxima or minima, $\frac{\text{ds}}{\text{dx}}=0$
$\Rightarrow -3(4096-128\text{x})=0$
$\Rightarrow\text{x}=32$ 
Now, $\frac{\text{d}^{2}\text{s}}{\text{dx}^{2}}=384>0$
x = 2 is the point of local minima.
Thus, the two parts of 64 are (32, 32).
View full question & answer→
Generate Paper FreeAll Maxima and Minima topics