RD Sharma XII Vol 1 2017 Solutions for Class 12 Science Math Chapter 18 Maxima And MInima are provided here with simple step-by-step explanations. These solutions for Maxima And MInima are extremely popular among class 12 Science students for Math Maxima And MInima Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2017 Book of class 12 Science Math Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2017 Solutions. All RD Sharma XII Vol 1 2017 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 18.16:

Question 1:


f(x) = (x-5)4

Answer:

Given:fx=x-54f'x=4x-53For a local maximum or a local minimum, we must havef'x=04x-53=0x=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  f (x) at x = 5 is given by
5-54=0

Page No 18.16:

Question 2:

f(x) = x3 - 3x

Answer:

Given:fx=x3-3xf'x=3x2-3For a local maximum or a local minimum, we must havef'x=03x2-3=0x2-1=0x=±1



Since f '(x) changes from negative to positive as 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
13-31=-2

Since f '(x) changes from positive to negative when x increases through -1, x = -1 is the point of local maxima.
The local maximum value of  f (x) at x = -1 is given by
-13-3-1=2



.

Page No 18.16:

Question 3:

f(x) = x3  (x-1)2

Answer:

Given: fx=x3x-12f'x=3x2x-12+2x3x-1For a local maximum or a local minimum, we must have f'x=03x2x-12+2x3x-1=0x2x-13x-3+2x=0x2x-15x-3=0x=0, 1, 35



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
131-12=0

Since f '(x) changes from positive to negative when x increases through 35, x = 35 is the point of local maxima.
The local minimum value of  f (x) at x = 35 is given by
35335-12=27125×425=1083125

Since f '(x) does not change from positive as x increases through 0, x = 0 is a point of inflexion.

Disclaimer: The solution in the book is incorrect. The solution here is created according to the question given in the book.

Page No 18.16:

Question 4:

f(x) =  (x-1) (x+2)2

Answer:

Given: fx=x-1x+22f'x=x+22+2x+2x-1For a local maximum or a local minimum, we must havef'x=0x+22+2x+2x-1=0x+2x+2+2x-2=0x+23x=0x=0, -2

Since  f '(x) changes from negative to positive when x increases through 0, x = 0 is the point of local minima.
The local minimum value of  f (x) at x = 0 is given by
0-10+22=-4

Since  f '(x) changes sign from positive to negative when x increases through -2, x = -2 is the point of local maxima.
The local maximum value of  f (x)  at x = -2 is given by
-2-1-2+22=0

Page No 18.16:

Question 5:

f(x) = 1x2+2

Answer:

    Given:fx=1x2+2 f'x=-2xx2+22For the local maxima or minima, we must have f'x=0-2xx2+22=0x=0

Now, for values close to x = 0 and to the left of 0, f'x>0.
Also, for values close to x = 0 and to the right of 0, f'x<0.

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of fx is 12.

Page No 18.16:

Question 6:

f(x) =  x3 - 6x2 + 9x + 15

Answer:

Given: fx=x3-6x2+9x+15f'x=3x2-12x+9For a local maximum or a local minimum, we must havef'x=03x2-12x+9=0x2-4x+3=0x-1x-3=0x=1 or 3



Since f '(x) changes from negative to positive when x increases through 3, x = 3 is the point of local minima.
The local minimum value of  f (x) at x = 3 is given by
33-632+93+15=27-54+27+15=15

Since f '(x) changes from positive to negative when x increases through 1, x = 1 is the point of local maxima.
The local maximum value of  f (x) at x = 1 is given by
13-612+91+15=1-6+9+15=19

Page No 18.16:

Question 7:

f(x) = sin 2x, 0<x<π

Answer:

Given:fx=sin 2xf'x=2 cos 2xFor a local maximum or a local minimum, we must havef'x=02 cos 2x=0cos 2x=0x=π4 or 3π4



Since f '(x) changes from positive to negative when x increases through π4, x = π4 is the point of maxima.
The local maximum value of  f (x) at x = π4 is given by
sinπ2=1

Since f '(x) changes from negative to positive when x increases through 3π4, x = 3π4 is the point of minima.
The local minimum value of  f (x) at x = 3π4 is given by
sin3π2=-1

Page No 18.16:

Question 8:

f(x) =  sin x- cos x, 0 < x<2π

Answer:

Given:fx=sin x-cos xf'x=cos x+sin xFor a local maximum or a local minimum, we must havef'x=0cos x+sin x=0cos x=-sin xtan x=-1x=3π4 or 7π4


Since f '(x) changes from positive to negative when x increases through 3π4, x = 3π4 is the point of local maxima.
The local maximum value of  f (x)  at x = 3π4 is given by
sin3π4-cos3π4=2

Since f '(x) changes from negative to positive when x increases through 7π4, x = 7π4 is the point of local minima.
The local minimum value of  f (x)  at x = 7π4 is given by
sin7π4-cos7π4=-2

Page No 18.16:

Question 9:

f(x) =  cos x, 0<x<π

Answer:

Given:fx=cos xf'x=-sin xFor a local maximum or a local minimum, we must havef'x=0-sin x=0sin x=0x=0 or π.

Since 0<x<π, none is in the interval 0, π.

Page No 18.16:

Question 10:

f(x) = sin 2x-x, -π2<<x<π2

Answer:

Given:fx=sin 2x-xf'x=2 cos 2x-1For a local maximum or a local minimum, we must havef'x=02 cos 2x-1=0cos 2x=12x=-π6 or π6.


Since  f '(x) changes from positive to negative when x increases through π6, x = π6 is the point of local maxima.
The local maximum value of  f (x) at x = π6 is given by
sin π3-π6=32-π6

Since f '(x) changes from negative to positive when x increases through -π6, x = -π6 is the point of local minima.
The local minimum value of  f (x)  at x = -π6 is given by
sin -π3+π6=π6-32

Page No 18.16:

Question 11:

f(x) = 2sin x-x, -π2<<x<π2

Answer:

    Given:fx=2 sin x-xf'x=2 cos x-1For a local maximum or a local minimum, we must have f'x=02 cos x-1=0cos x=12x=π3 or -π3


 

Since f '(x) changes from positive to negative when x increases through π3, x = π3 is the point of local maxima.
The local maximum value of  f (x) at x = π3 is given by
2 sin π3-π3=3-π3

Since f '(x) changes from negative to positive when x increases through -π3, x = -π3 is the point of local minima.

The local minimum value of  f (x)  at x = -π3 is given by
2 sin -π3+π3=π3-3

Page No 18.16:

Question 12:

f(x) = x1-x , x>0

Answer:

Given: fx=x1-xf'x=1-x-x21-x=2-3x21-xFor the local maxima or minima, we must havef'x=02-3x21-x=0x=23



Since,  f '(x) changes from positive to negative when x increases through 23, x = 23 is a point of maxima.

The local maximum value of  f (x) at x = 23 is given by
231-23=233=239

Page No 18.16:

Question 13:

f(x) = x3 (2x-1)3

Answer:

Given: fx=x32x-13f'x=3x22x-13+6x32x-12For the local maxima or minima, we must have f'x=03x22x-13+6x32x-12=03x22x-122x-1+2x=0x22x-124x-1=0x=0, 12 and 14


Since f '(x) changes from negative to positive when x increases through 14, x = 14 is a point of local minima.
The local minimum value of  f (x)  at x = 14 is given by
14312-13=-1512

Page No 18.16:

Question 14:

f(x) = x2+2x , x>0

Answer:

    Given:fx=x2+2xf'x=12-2x2For the local maxima or minima, we must havef'x=012-2x2=012=2x2x2=±2


Since x > 0,  f '(x) changes from negative to positive when x increases through 2. So, x = 2 is a point of local minima.

The local minimum value of  f (x) at x = 2 is given by
22+22=2



Page No 18.31:

Question 1:

(i) f(x) = x4 - 62x2 + 120x + 9

(ii) f(x) = x3 - 6x2 + 9x + 15

(iii) f(x) = (x-1) (x+2)2

(iv) f(x) = 2/x-2/x2 , x>0

(v) f(x) = xex

(vi) f(x) = x/2+2/x, x>0

(vii) f(x) = (x+1) (x+2)1/3, x>-2

(viii) f(x) = x32-x2, -5<x<5

(ix) f(x) = x3-2ax2+a2x, a>0, x R

(x) f(x) = x+a2x, a>0, x ≠ 0

(xi) f(x) = x2-x2 - 2 x2

(xii) f(x) = x + 1-x, x1

Answer:

iGiven: fx = x4-62x2+120x+9f'x = 4x3-124x+120For the local maxima or minima, we must have f'x=04x3-124x+120=0x3-31x+30=0x-1x2+x-30=0x-1x+6x-5=0x=1, 5 and -6Thus, x=1, x=5 and x=-6 are the possible points of local maxima or local minima.Now,f''x=12x2-124At x=1:  f''1 = 1212-124=-112<0So, x=1 is the point of local maximum.The local maximum value is given byf1 = 14-6212+120×1+9=68At x=5:  f''5 = 1252-124=176>0So, x=5 is the point of local minimum.The local minimum value is given byf5 = 54-6252+120×5+9=-316At x=-6:  f''-6 = 12-62-124=308>0So, x=-6 is the point of local maximum.The local minimum value is given by f-6 =-64-62-62+120×-6+9=-1647


iiGiven: fx = x3-6x2+9x+15f'x = 3x2-12x+9For the local maxima or minima, we must have f'x=0 3x2-12x+9=0 x2-4x+3=0x-1x-3=0x=1 and 3Thus, x=1 and x=3 are the possible points of local maxima or local minima.Now, f''x = 6x-12At x=1:  f''1 = 61-12=-6<0So, x=1 is the point of local maximum.The local maximum value is given byf1 = 13-612+9×1+15=19At x=3:  f''3 = 63-12=6>0So, x=3 is the point of local minimum.The local minimum value is given byf3 =33-632+9×3+15=15


iii Given:fx =x-1x+22=x-1x2+4x+4=x3+4x2+4x-x2-4x-4=x3+3x2-4f'x = 3x2+6xFor the local maxima or minima, we must have f'x=0  3x2+6x=03xx+2=0x=0 and-2Thus, x=0 and x=-2 are the possible points of local maxima or local minima.Now, f''x = 6x+6At x=0:  f''0 = 60+6=6>0So, x=0 is the point of local minimum.The local minimum value is given byf0 = 0-10+22=-4At x=-2:  f''-2 = 6-2+6=-6<0So, x=-2 is the point of local maximum.The local maximum value is given by f-2 =-2-1-2+22=0


ivGiven: fx = 2x-2x2=2x-1-2x-2f'x =-2x-2+4x-3=4x3-2x2For the local maxima or minima, we must have f'x=04x3-2x2=04-2x=0x=2Thus, x=2 is the possible point of local maxima or local minima.Now, f''x = -12x4+4x3At x=2:  f''2 =-1216+48 =-12+816=-14<0So, x=2 is the point of local maximum.The local maximum value is given byf2 = 22-222=1-12=12


v Given:fx = xexf'x = ex+xexFor the local maxima or minima, we must have f'x=0ex+xex=0ex1+x=0ex0 , x=-1x=-1Thus, x=-1 is the possible point of local maxima or local minima.Now,f''x = ex+ex+xexAt x=-1:  f''-1 =e-1+e-1-e-1 =e-1>0So, x=-1 is the point of local minimum.The local minimum value is given byf-1 = -e-1=-1e

vi Given:fx = x2+2xf'x = 12-2x2For the local maxima or minima, we must have f'x=012-2x2=0x2=4x=2 and-2Thus, x=2 and x=-2 are the possible points of local maxima or a local minima.Since x>0, x=2Now,  f''x = 4x3At x=2:  f''2 =423 =12>0So, x=2 is the point of local minimum.The local minimum value is given by f2 = x2+2x =1+1=2

vii Given:fx =x+1x+213f'x =x+213+13x+1x+2-23For the local maxima or minima, we must have f'x=0 x+213+13x+1x+2-23=013x+1=-x+213×x+22313x+1=-x+2x+1=-3x-6x=-74Thus, x=-74 is the possible point of local maxima or local minima.Now, f''-74 =23x+2-23-29x+1x+2-53At x=-74:  f''-74 =23-74+2-23-29-74+1-74+2-53=2314-23+11814-52>0So, x=-74 is the point of local minimum.The local minimum value is given byf-74 = -74+1-74+213=-341413=-3443


viii Given: fx = x32-x2f'x = 32-x2-x232-x2For the local maxima or minima, we must have f'x=032-x2-x232-x2=032-x2=x32-x232-x2 =x2 x2=16 x=±4  Thus, x=4 and x=-4 are the possible points of local maxima or local minima.Now, f''x =-x32-x2 -2x32-x2+x332-x232-x2=-x32-x2 -2x32-x2+x332-x232-x2At x=4:  f''4 = -432-42 -832-42+4332-4232-42 =-1-19264=-3<0So, x=4 is the point of local maximum.The local maximum value is given by f4 =    432-42 =16At x=-4:  f''-4 = 432-42 +832-42-4332-4232-42 =1+2=3>0So, x=-4 is the point of local minimum.The local minimum value is given by f-4 =    -432-42 =-16


ix fx = x3-2ax2+a2xf'x = 3x2-4ax+a2For the local maxima or minima, we must have f'x=03x2-4ax+a2=0 3x2-3ax-ax+a2=03xx-a-ax-a=03x-ax-a=0x=a and a3Thus, x=a and x=a3 are the possible points of local maxima or local minima.Now, f''x = 6x-4aAt x=a: f''a = 6a-4a=2a>0So, x=a is the point of local minimum.The local minimum value is given byfa = a3-2aa2+a2a=0At x=a3:  f''a3 = 6a3-4a=-2a<0So, x=a3 is the point of local maximum.The local maximum value is given byfa3 =a33 -2aa32 +a2a3 =a327-2a39+a33=4a327


xGiven: fx = x+a2xf'x = 1-a2x2For the local maxima or minima, we must havef'x=0 1-a2x2=0x2=a2x=±a     Thus, x=a and x=-a are  the possible points of local maxima or local minima.Now, f''x = a2x3At x=a:  f''a =a2a3 =1a>0So, x=a is the point of local minimum.The local minimum value is given byfa=x+a2x=a+a=2aAt x=-a:  f''a =a2-a3 =-1a<0So, x=-a is the point of local maximum.The local maximum value is given byf-a = x+a2x =-a-a=-2a


xiGiven: fx = x2-x2f'x = 2-x2-x22-x2For the local maxima or minima, we must have f'x=02-x2-x22-x2=02-x2=x2-x22-x2 =x2 x2=1 x=±1 Thus, x=1 and x=-1 are the possible points of local maxima or local minima.Now, f''x =-x2-x2 -2x2-x2+x32-x22-x2=-x2-x2 -2x2-x2+x32-x22-x2At x=1:  f''1 = -12-12 -22-12+132-122-12 =-12-32=-2<0So, x=1 is the point of local maximum.The local maximum value is given byf4 =    12-12 =1At x=-1:  f''-1 = 12-12 +22-12-132-122-12 =1+1=2>0So, x=-1 is the point of local minimum.The local minimum value is given byf-1 = -12-12 =-1


xii Given: fx = x+1-xf'x = 1-121-xFor the local maxima or minima, we must have f'x=01-121-x=01-x=121-x=14  x=34   Thus, x=34 is the possible point of local maxima or local minima.Now,  f''x = -141-x41-xAt x=34: f''34 = -141-3441-34 =-12<0So, x=34 is  the point of local maximum.The local maximum value is given byf34 =   34+1-34 =54

Page No 18.31:

Question 2:

(i) f(x) = (x-1) (x-2)2

(ii) f(x) = x1-x , x<1

(iii) f(x) = -(x-1)3 (x+1)2

Answer:

iGiven: fx=x-1x-22=x-1x2-4x+4=x3-4x2+4x-x2+4x-4=x3-5x2+8x-4f'x= 3x2-10x+8For the local maxima or minima, we must have f'x=0  3x2-10x+8=03x2-6x-4x+8=0x-23x-4=0x=2 and43Thus, x=2 and x=43 are the possible points of local maxima or local minima.Now, f''x = 6x-10At x=2:  f''2 = 62-10=2>0So, x=2 is the point of local minimum.The local minimum value is given byf2 = 2-12-22=0At x=43:  f''43 = 643-10=-2<0So, x=4 3is the point of local maximum.The local maximum value is given byf4 3 =4 3-14 3-22=13×49=427


iiGiven: fx = x1-xf'x = 1-x-x21-xFor the local maxima or minima, we must have f'x=01-x-x21-x=01-x=x21-x2-2x=x3x=2 x=23 Thus, x=23 is the possible point of local maxima or local minima.Now, f''x =-11-x -121-x+x21-x1-x=-11-x -122-x1-x1-xAt x=23:  f''23 = -11-23 -122-231-231-23 =-3-4313×3=-3-43<0So, x=23 is the point of local maximum.The local maximum value is given byf23 =    231-23 =233


iiiGiven: fx =-x-13x+12f'x =- 3x-12x+12+2x+1x-13For the local maxima or minima, we must have f'x=0 -3x-12x+12-2x+1x-13=0x-12x+1-3x+1-2x-1=0x-12x+1-3x-3-2x+2=0x-12x+1-5x-1=0x=1, -1 and -15Thus, x=1, x=-1 and x=-15 are the possible points of local maxima or local minima.Now, f''x = - 32x-1x+12+2x+1x-12+2x-13+3x-12x+1                          = -6x-1x+12+6x+1x-12-2x-13-6x-12x+1At x=1:  f''1 =-61-11+12+61+11-12-21-13-61-121+1=0So, it is a point of inflexion.At x=-1:  f''-1 =-6-1-1-1+12+6-1+1-1-12-2-1-13-6-1-12-1+1=16>0So, x=-1 is the point of local minimum.The local minimum value is given by f-1 =-1-13-1+12=0At x=-15:  f''-15 =-6-15-1-15+12+6-15+1-15-12+2-15-13-6-15-12-15+1=576125+384125-432125-864125=-336125<0So, x=-15 is the point of local maximum.The local maximum value is given by f-15 =--15-13-15+12=--2161251625=34653125

Page No 18.31:

Question 3:

The function y = a log x+bx2 + x has extreme values at x=1 and x=2. Find a and b

Answer:

Given: fx=y=a log x+bx2+xf'x=ax+2bx+1Since, f'x has extreme values at x=1 and x=2, f'1=0.a1+2b1+1=0a=-1-2b             ...1f'2=0a2+2b2+1=0a+8b=-2   a=-2-8b              ...2From eqs. 1 and 2, we get-2-8b=-1-2b6b=-1b=-16Substituting b=-16 in eq. 1, we geta=-1+13=-23

Page No 18.31:

Question 4:

Show that log xx has a maximum value at x = e.

Answer:

Here,fx = log xxf'x = 1-log xx2For the local maxima or minima, we must have f'x=01-log xx2=01=log xlog e=log xx=eNow, f''x = x2-1x-2x1-log xx4=-3+2 log xx3f''e = -3+2 log ee3=-1e3<0So, x=e is the point of local maximum.

Page No 18.31:

Question 5:

Find the maximum and minimum values of the function f(x) = 4x+2+x.

Answer:

Given: fx = 4x+2+xf'x = -4x+22+1For a local maxima or a local minima, we must have f'x=0-4x+22+1=0-4x+22=-1x+22=4x+2=±2x=0 and -4Thus, x=0 and x=-4 are the possible points of local maxima or local minima.Now,  f''x = 8x+23At x=0: f''0 =823 =1>0So, x=0 is a point of local minimum.The local minimum value is given byf0 = 40+2+0 =2At x=-4: f''-4 =8-43 =-18<0So, x=-4 is a point of local minimum.The local maximum value is given byf-4 = 4-4+2-4 =-6

Page No 18.31:

Question 6:

Find the maximum and minimum values of y = tan x-2x.

Answer:

Given: fx =y=tan x-2xf'x =sec2 x-2For a local maxima or local minima, we must have f'x=0sec2 x-2=0sec2 x=2sec x=±2x=π4 and 3π4Thus, x=π4 and x=3π4 are the possible points of local maxima or a local minima.Now,f''x = 2 sec2 x tan xAt x=π4: f''π4 =2 sec2 π4 tan π4 = 4>0So, x=π4 is a point of local minimum.The local minimum value is given byfπ4 =tanπ4-2×π4 =1-π2At x=3π4: f''3π4 =2 sec2 3π4 tan 3π4 =-4<0So, x=3π4 is a point of local maximum.The local maximum value is given byf3π4 =tan 3π4-2×3π4 =-1-3π2

Page No 18.31:

Question 7:

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,fx=x3+ax2+bx+cf'x=3x2+2ax+bAs, fx is maximum at x=-1 and minimum at x=3.So, f-1=0 and f3=03-12+2a-1+b=0 and 332+2a3+b=03-2a+b=0        .....iand 27+6a+b=0   .....iiii-i, we get27-3+6a+2a=08a=-24a=-3Substituting a=-3 in i, we get3-2-3+b=03+6+b=0b=-9And, cR

Page No 18.31:

Question 8:

Prove that f(x) = sinx + 3cosx has maximum value at x = π6.                                                                                     [NCERT EXEMPLAR]

Answer:

We have,fx=sinx+3cosxf'x=cosx+3-sinxf'x=cosx-3sinxFor fx to have maximum or minimum value, we must have f'x=0cosx-3sinx=0cosx=3sinxcotx=3x=π6Also, f''x=-sinx-3cosxf''π6=-sinπ6-3cosπ6=-12-332=-12-32=-2<0So, x=π6 is point of maxima.



Page No 18.37:

Question 1:

(i) f(x) = 4x - x22 in [-2,4,5]

(ii) f(x) = (x-1)2 + 3 in [-3,1]

(iii) f(x) = 3x4 - 8x3 + 12x2 - 48x + 25 in [0,3]

(iv) f(x) = (x-2) x-1 in [1, 9]

Answer:

i Given: fx=4x-x22f'x=4-xFor a local maximum or a local minimum, we must havef'x=04-x=0x=4Thus, the critical points of f are -2, 4 and 4.5.Now,f-2=4-2--222=-8-2=-10f4=44-422=16-8=8f4.5=44.5-4.522=18-10.125=7.875Hence, the absolute maximum value when x=4 is 8 and the absolute minimum value when x=-2  is -10.


ii Given:fx=x-12+3f'x=2x-1For a local maximum or a local minimum, we must have f'x=02x-1=0x=1Thus, the critical points of f are -3 and 1.Now,f-3=-3-12+3=16+3=19f1=1-12+3=3Hence, the absolute maximum value when x=-3 is 19 and the absolute minimum value when x=1 is 3.


iii Given:fx=3x4-8x3+12x2-48x+25f'x=12x3-24x2+24x-48For a local maximum or a local minimum, we must havef'x=012x3-24x2+24x-48=0x3-2x2+2x-4=0x2x-2+2x-2=0x-2x2+2=0x-2=0  or x2+2=0 x=2 No real root exists for x2+2=0. Thus, the critical points of f are 0, 2 and 3.Now,f0=304-803+1202-480+25=25f2=324-823+1222-482+25=-39f3=334-833+1232-483+25=16Hence, the absolute maximum value when x=0 is 25 and the absolute minimum value when x=2 is -39.


iv Given: fx=x-2x-1f'x=x-1+x-22x-1For a local maximum or a local minimum, we must havef'x=0x-1+x-22x-1=02x-1+x-2=02x-2+x-2=03x-4=03x=4  x=43 Thus, the critical points of f are 1, 43 and 9.Now, f1=1-21-1=0 f43=43-243-1=-23×13=-233 f9=9-29-1=142Hence, the absolute maximum value when x=9 is 142 and the absolute minimum value when x=43 is -233.Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Page No 18.37:

Question 2:

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:

Given: fx=2x3-24x+107f'x=6x2-24For a local maximum or a local minimum, we must havef'x=06x2-24=06x2=24x2=4x=±2Thus, the critical points of f in the interval 1, 3 are 1, 2 and 3. Now, f1=213-241+107=85f2=223-242+107=75f3=233-243+107=89Hence, the absolute maximum value when x=3 in the interval 1, 3 is 89.Again, the critical points of f in the interval -3, -1  are -1, -2 and-3. So, f-3=2-33-24-3+107=125f-2=2-23-24-2+107=139f-1=2-13-24-1+107=129Hence, the absolute maximum value when x=-2 is 139.

Page No 18.37:

Question 3:

Find the absolute maximum and minimum values of the function of given by

f(x) = cos2x + sin x, x  [0, π]

Answer:

Given: fx=cos2 x+sin xf'x=2 cos x-sin x+cos x=-2 sin x cos x+cos xFor a local maximum or a local minimum, we must have f'x=0-2 sin x cos x+cos x=0cos x 2 sin x-1=0sin x=12 or cos x=0x=π6 or π2                         x0, πThus, the critical points of f are 0, π6, π2 and π.Now,f0=cos2 0+sin 0=1fπ6=cos2 π6+sin π6=54fπ2=cos2 π2+sin π2=1fπ=cos2 π+sin π=1Hence, the absolute maximum value when x=π6 is 54 and the absolute minimum value when x=0, π2, π is 1.

Page No 18.37:

Question 4:

Find the absolute maximum and minimum values of a function f given by

f(x) = 12x4/3 - 6x1/3 , x[-1, 1]

Answer:

Given: fx=12x43-6x13f'x=16x13-2x-23=28x-1x23For a local maximum or a local minimum, we must havef'x=028x-1x23=08x-1=0x=18Thus, the critical points of f are -1, 18  and 1.Now,f-1=12-143-6-113=18f18=121843-61813=-94f1=12143-6113=6Hence, the absolute maximum value when x=-1 is 18 and the absolute minimum value when x=18is -94. 

Page No 18.37:

Question 5:

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: fx=2x3-15x2+36x+1f'x=6x2-30x+36For a local maximum or a local minimum, we have f'x=06x2-30x+36=0x2-5x+6=0x-3x-2=0x=2 and  x=3 Thus, the critical points of f are 1,  2, 3 and 5.Now,f1=213-1512+361+1=24f2=223-1522+362+1=29f3=233-1532+363+1=28f5=253-1552+365+1=56Hence, the absolute maximum value when x=5 is 56 and the absolute minimum value when x=1 is 24.



Page No 18.7:

Question 1:

f(x)=4x2-4x+4 on R.

Answer:

Given: f(x) = 4x2 − 4x + 4
f(x) = (4x2 − 4x + 1)+3
f(x) = (2x − 1)2 + 3
Now,
(2x − 1)2 0 for all x R
f(x) = (2x − 1)2 + 3 3 for all x R
f(x) 3 for all x R



The minimum value of f is attained when (x − 1) = 0.
(2x − 1) = 0
x = 12
Thus, the minimum value of f (x) at x = 12 is 3.

Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, function f does not have a maximum value.

Page No 18.7:

Question 2:

f(x)=(x-1)2+2 on R

Answer:

Given: f(x) = − (x − 1)2 + 2
Now,
(x − 1)2 0 for all x R
f(x) = − (x − 1)2 + 2 2 for all x R



The maximum value of f(x) is attained when (x − 1) = 0.
(x − 1) = 0
x = 1
Therefore, the maximum value of f (x) = 2
Since f(x) can be reduced, the minimum value does not exist, which is evident in the graph also.
Hence, function f does not have a minimum value.

Page No 18.7:

Question 3:

f(x)=| x+2 | on R

Answer:

Given: f(x) = x+2

Now,
x+20 for all x R

Thus, f(x) 0 for all x R

Therefore, 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.

Page No 18.7:

Question 4:

f(x)=sin 2x+5 on R

Answer:

Given: f(x) = sin 2x + 5

We know that − 1 ≤ sin 2x ≤ 1.

⇒ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5

⇒ 4 ≤ sin 2x + 5 ≤ 6

⇒ 4 ≤ f(x) ≤ 6

Hence, the maximum and minimum values of f are 6 and 4, respectively.

Page No 18.7:

Question 5:

f(x) = | sin 4x+3 | on R

Answer:

Given: f(x) =sin 4x+3

We know that −1 sin 4x 1.

⇒ 2 sin 4x + 3 4

⇒ 2 sin 4x+3 4

⇒ 2  f(x) 4

Hence, the maximum and minimum values of f are 4 and 2, respectively.

Page No 18.7:

Question 6:

 f(x)=2x3 +5 on R

Answer:

We can observe that f(x) increases when the values of x are increased and f(x) decreases when the values of x are decreased. Also, f(x) can be reduced by giving small values of x.
Similarly, f(x) can be enlarged by giving large values of x.
So, f(x) does not have a minimum or maximum value.

Page No 18.7:

Question 7:

f(x) =-| x + 1 | + 3 on R

Answer:

Given: f(x) =-x+1 + 3

Now,
-x+10 for all x R

f(x) =  -x+1 + 3 3 for all x R
f(x) 3 for all x R

The maximum value of f is attained when x+1=0.x=-1

Therefore, the maximum value of f  at x = -1 is 3.

Since f(x) can be reduced, the minimum value does not exist, which is evident in the graph also.
Hence, the function f does not have a minimum value.

Page No 18.7:

Question 8:

f(x) = 16x2 - 16x + 28 on R

Answer:

Given: f(x) = 16x2 − 16x + 28
f(x) = 4(4x2 - 4x + 1) + 24
f(x) = 4(2x − 1)2 + 24

Now,
4(2x − 1)2 0 for all x R

f(x) = 4(2x − 1)2 + 24 24 for all x R
f(x) 24 for all x R




The minimum value of f is attained when (2x − 1) = 0.

(2x − 1) = 0
x = 12

Therefore, the minimum value of f  at x = 12 is 24.

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.

Page No 18.7:

Question 9:

f(x) = x3 -1 on R

Answer:




We can observe that f(x) increases when the values of x increase and f(x) decreases when the values of x decrease. Also, f(x) can be reduced by giving smaller values of x.
Similarly, f(x) can be enlarged by giving larger values of x.
So, f(x) does not have a minimum or maximum value.



Page No 18.72:

Question 1:

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                               ...1Now,z=x2+y2z=x2+15-x2           From eq. 1z=x2+x2+225-30xz=2x2+225-30xdzdx=4x-30For maximum or minimum values of z, we must havedzdx=04x-30=0x=152d2zdx2= 4 > 0Substituting x=152 in 1, we get y=152Thus, z is minimum when x=y=152.

Page No 18.72:

Question 2:

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Answer:

Suppose 64 is divided into two parts x and 64-x. Then,z=x3+64-x3dzdx=3x2+364-x2For maximum or minimum values of z, we must havedzdx=03x2+364-x2=03x2=364-x2x2=x2+4096-128xx=4096128x=32Now, d2zdx2=6x+664-x d2zdx2=384>0Thus, z is minimum when 64 is divided into two equal parts, 32 and 32..

Page No 18.72:

Question 3:

How should we choose two numbers, each greater than or equal to -2, whose sum______________ so that the sum of the first and the cube of the second is minimum?

Answer:

Let the two numbers be and y. Then,x, y >-2 and x+y=12                 ...(1)Now,z=x +y3z=x+12-x3                       From eq. 1dzdx=1+312-x2For maximum or minimum values of z, we must havedzdx=01+312-x2=012-x2=1312-x=±13x=12±13d2zdx2=612-xd2zdx2=3-6xAt x=12±13:d2zdx2=3-612+13-63<0Thus, z is maximum when x= 12+13.At x=12-13:d2zdx2=3-612-1363>0Thus, z is minimum when x= 12-13.x+y=12Substituting the value of x in eq. 1, we gety=-12+13+12y=13So, the required two numbers are 12-13 and 13.

Page No 18.72:

Question 4:

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Answer:

Let the two numbers be x and y. Then,x+y=15                                    ...(1)Now, z=x2y3z=x215-x3                From eq. 1dzdx=2x15-x3-3x215-x2For maximum or minimum values of z, we must havedzdx=02x15-x3-3x215-x2=02x15-x=3x230x-2x2=3x230x=5x2x=6 and y =9d2zdx2=215-x3-6x15-x2-6x15-x2+6x215-xAt x=6:d2zdx2=293-3692-3692+6369d2zdx2=-2430<0Thus, z is maximum when x= 6 and y =9.So, the required two parts into which 15 should be divided are 6 and 9.

Page No 18.72:

Question 5:

Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?

Answer:

Let r and h be the radius and height of the cylinder, respectively. Then, VolumeV of the cylinder =πr2h100=πr2hh=100πr2Surface area S of the cylinder=2πr2+2πrh=2πr2+2πr×100πr2S=2πr2+200rdSdr=4πr-200r2 For the maximum or minimum, we must have dSdr=04πr-200r2=04πr3=200r=50π13Now, d2Sdr2=4π+400r3 d2Sdr2>0  when r=50π13Thus, the surface area is minimum when r=50π13.At r=50π13:h=100π50π23 =250π13

Page No 18.72:

Question 6:

A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
(i) M=WL2x - W2x2

(ii) M=Wx3x - W3x3L2

Find the point at which M is maximum in each case.

Answer:

i  Given:M=WL2x-W2x2dMdx=WL2-2×Wx2dMdx=WL2-WxFor maximum or minimum values of M, we must havedMdx=0WL2-Wx=0WL2=Wxx=L2Now,d2Mdx2=-W <0So, is maximum at x=L2.

ii Given:M=Wx3-Wx33L2dMdx=W3-3×Wx23L2dMdx=W3-Wx2L2For maximum or minimum values of M, we must havedMdx=0W3-Wx2L2=0W3=Wx2L2x=L3Now,d2Mdx2=-2WxL2 <0So, M is maximum at x=L3.

Page No 18.72:

Question 7:

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

Answer:

Suppose the wire, which is to be made into a square and a circle, is cut into two pieces of length x m and y m, respectively. Then,x+y=28                                 ...1Perimeter of square, 4side=xSide=x4Area of square=x42=x216Circumference of circle, 2πr=yr=y2πArea of circle=πr2=πy2π2=y24πNow,z=Area of square+Area of circle z=x216+y24πz=x216+28-x24πdzdx=2x16-228-x4πFor maximum or minimum values of z, we must havedzdx=02x16-228-x4π=0           From eq. 1x4=28-xπxπ4+x=28xπ4+1=28x=28π4+1x=112π+4y=28-112π+4              From eq. 1y=28ππ+4 d2zdx2=18+12π>0Thus, z is minimum when x=112π+4 and y =28ππ+4.Hence, the length of the two pieces of wire are 112π+4 m and 28ππ+4 m respectively.

Page No 18.72:

Question 8:

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum?

Answer:

Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y, respectively. Then,x+y=20                                     ...1Perimeter of square, 4Side=xSide=x4Area of square=x42=x216Perimeter of triangle, 3Side=ySide=y3Area of triangle=34×Side2=34×y32=3y236Now,z=Area of square+Area of trianglez=x216+3y236z=x216+320-x236                   From eq. 1dzdx=2x16-2320-x36For maximum or minimum values of z, we must havedzdx=02x16-320-x18=09x4=320-x9x4+x3=203x94+3=203x=20394+3x=8039+43y=20-8039+43        From eq. 1y=1809+43 d2zdx2=18+318>0Thus, z is minimum when x=8039+43 and y =1809+43.Hence, the wire of length 20 cm should be cut into two pieces of lengths 8039+43 m and 1809+43 m.Disclaimer: The solution given in the book is incorrrect. The solution here is created according to the question given in the book.

Page No 18.72:

Question 9:

Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.

Answer:

Let the length of a side of the square and radius of the circle be x and r, respectively. It is given that the sum of the perimeters of square and circle is constant.   4x+2πr=K          Where K is some constantx=K-2πr4                           ...1Now,A=x2+πr2A=K-2πr216+πr2                        From eq. 1dAdr=K-2πr216+πr2dAdr=2K-2πr-2π16+2πrdAdr=K-2πr-π4+2πrK-2πr-π4+2πr=0K-2πrπ4=2πrK-2πr=8r                                      ...2d2Adx2=π22+2π>0So, the sum of the areas, A is least when K-2πr=8r.From eqs. 1 and 2, we getx=K-2πr4x=8r4x=2r Side of the square=Diameter of the circle



Page No 18.73:

Question 10:

Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.                                                              [CBSE 2000]

Answer:

Let the base of the right angled triangle be x and its height be y. Then,x2+y2=52y2=25-x2y=25-x2As, the area of the triangle,A=12×x×yAx=12×x×25-x2Ax=x25-x22A'x=25-x22+x-2x425-x2A'x=25-x22-x2225-x2A'x=25-x2-x2225-x2A'x=25-2x2225-x2For maxima or minima, we must have f'x=0A'x=025-2x2225-x2=025-2x2=02x2=25x=52So, y=25-252=50-252=252=52Also, A''x=-4x25-x2-25-2x2-2x225-x225-x2=-4x25-x2+25x-2x325-x225-x2=-100x+4x3+25x-2x325-x225-x2=-75x+2x325-x225-x2A''52=-7552+252325-52232<0So, x=52 is point of maxima. The largest possible area of the triangle=12×52×52=254 square units

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Question 11:

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.                                                                                                                                          [CBSE 2002 C]

Answer:

As, the area of the triangle, A=12absinθAθ=12absinθA'θ=12abcosθFor maxima or minima, A'θ=012abcosθ=0cosθ=0θ=π2Also, A''θ=-12absinθor, A''π2=-12absinπ2=-12ab<0i.e. θ=π2 is point of maximaNow,The maximum area of the triangle=12absinπ2=ab2

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Question 12:

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 a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume.

Answer:

Let the side of the square to be cut off be x cm.
Then, the length and the breadth of the box will be (18 − 2x) cm each and height of the box will be x cm.

Volume of the box, V(x) = x(18 − 2x)2

 V'x=18-2x2-4x18-2x        =18-2x18-2x-4x        =18-2x18-6x        =129-x3-xV''x=12-9-x-3-x          =-129-x+3-x          =-246-x
 

For maximum and minimum values of V, we must have V'x=0
x = 9 or x = 3 

If x = 9, then length and breadth will become 0.

x ≠ 9

x = 3

Now, 
V''3=-246-3=-72<0

x = 3 is the point of maxima.

Vx=318-62=3×144=432 cm3

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box so obtained would be the largest, i.e. 432 cm3.  

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Question 13:

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, in cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Answer:

Suppose square of side measuring x cm is cut off.Then, the length, breadth and height of the box will be 45-x, 24-2x and x, respectively. Volume of the box, V=45-2x24-2xxdVdx= 45-2x24-2x- 2x45-2x-2x24-2xFor maximum or minimum values of V, we must havedVdx=045-2x24-2x- 2x45-2x-2x24-2x=04x2+1080-138x-48x+4x2+4x2-90x=012x2-276x+1080=0x2-23x+90=0x2-18x-5x+90=0xx-18-5x-18=0x-18=0 or x-5=0x=18 or x=5Now,d2Vdx2=24x-276d2Vdx2x=5=120-276=-156<0     d2Vdx2x=18=432-276=156>0Thus, volume of the box is maximum when x=5 cm.Hence, the side of the square to be cut off measures 5 cm.                                 

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Question 14:

A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is 2 m and volume is 8 m3. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank?

Answer:

Let l, b and h be the length, breadth and height of the tank, respectively.

Height, h = 2 m

Volume of the tank = 8 m3

Volume of the tank = l × b × h

∴  l × b × 2 = 8

  lb=4b=4l

Area of the base = lb = 4 m2

Area of the 4 walls, A= 2h (l + b)

A=4l+4ldAdl=41-4l2For maximum or minimum values of A, we must havedAdl=041-4l2=0l=±2

However, the length cannot be negative.

Thus,
l = 2 m

 b=42=2 mNow, d2Adl2=32l3At l=2:d2Adl2=328=4>0

Thus, the area is the minimum when l = 2 m

We have
l = b = h = 2 m

Cost of building the base = Rs 70 × (lb) = Rs 70 × 4 = Rs 280

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2)= Rs 8 (90) = Rs 720

Total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

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Question 15:

A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.

Answer:

Let the dimensions of the rectangular part be x and y.  Radius of semi-circle =x2Total perimeter=10x+2y+πx2=102y=10-x-πx2y=1210-x1+π2                                         ...1Now, Area, A=π2x22+xyA=πx28+x210-x1+π2                       From eq. 1A=πx28+10x2-x221+π2dAdx=πx4+102-2x21+π2For maximum or minimum values of A, we must havedAdx=0πx4+102-2x21+π2=0xπ4-1-π2=-5x=-5-4-π4x=20π+4Substituting the value of x in  eq. 1, we gety=1210-20π+41+π2y=5-10π+22π+4y=5π+20-5π-10π+4y=10π+4 d2Adx2=π4-π2-1d2Adx2=π-2π-44d2Adx2=-π-44<0Thus, the area is maximum when x=20π+4 and y=10π+4So, the required dimensions are given below:Length=20π+4 m Breadth=10π+4 m

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Question 16:

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 rectangle be x and y.Perimeter of the window=x+y+x+x+y=123x+2y=12y=12-3x2                                       ...1Area of the window =xy +34x2A=x12-3x2 +34x2A=6x-3x22+34x2dAdx=6-6x2+234xdAdx=6-3x+32xdAdx=6-x3-32For maximum or a minimum values of A, we must havedAdx=06=x3-32x= 126-3Substituting the value of x in eq. 1, we gety=12-3126-32y=18-636-3Now, d2Adx2=-3+32<0Thus, the area is maximum when x=126-3 and y=18-636-3.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

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Question 17:

Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is 2R3.

Answer:

Let the height and radius of the base of the cylinder be h and r, respectively. Then,h24+r2=R2h=2R2-r2                                                         ...1Volume of cylinder, V=πr2hSquaring both sides, we getV2=π2r4h2V2=4π2r4R2-r2                                                   From eq. 1Now,Z=4π2r4R2-r6dZdr=4π24r3R2-6r5For maximum or minimum values of Z, we must havedZdr=04π24r3R2-6r5=04r3R2=6r56r2=4R2r2=4R26r=2R6Substituting the value of r in eq. 1, we get h=2R2-2R62h=26R2-4R26h=2R23h=2R3Now, d2Zdr2=4π212r2R2-30r4d2Zdr2=4π2122R62R2-302R64d2Zdr2=4π28R4-80R46d2Zdr2=4π248R4-80R46d2Zdr2=4π2-16R43<0So, volume of the cylinder is maximum when h=2R3.Hence proved.

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Question 18:

A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area.

Answer:

Let the dimensions of the rectangle be x and y. Then,x24+y2=r2x2+4y2=4r2x2=4r2-y2                                                   ...1Area of rectangle=xyA=xySquaring both sides, we getA2=x2y2Z=4y2r2-y2                          From eq. 1dZdy=8yr2-16y3For the maximum or minimum values of Z, we must havedZdy=08yr2-16y3=08r2=16y2y2=r22y=r2Substituting the value of y in eq. 1, we getx2=4r2-r22x2=4r2-r22x2=4r22x2=2r2x=r2Now, d2Zdy2=8r2-48y2d2Zdy2=8r2-48r22d2Zdy2=-16r2<0So, the area is maximum when x= r2 and y =r2.Area =xyA=r2×r2A=r2

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Question 19:

Prove that a conical tent of given capacity will require the least amount of  canavas when the height is 2 times the radius of the base.

Answer:

Let the surface area of conical tent be S = πrr2+h2Let the volume of the conical tent V = 13πr2hh=3Vπr2S=πrr2+3Vπr22S=1rπ2r6+9V6Now differentiating with respect to r we get,dSdr=ddr1rπ2r6+9V6        =1r6π2r52π2r6+9V6-π2r6+9V6r2For minima putting dSdr=0 we get,3π2r4π2r6+9V6=π2r6+9V6r23π2r6=π2r6+9V62π2r6= 9V6Substitutting the value of V we get,2π2r6= 913πr2h22π2r6= π2r4h22r2=h2h=2 r

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Question 20:

Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to 23 of the diameter of the sphere.

Answer:

Let h, r and R be the height, radius of base of the cone and radius of the sphere, respectively. Then,h=R+R2-r2h-R2=R2-r2h2+R2-2hr=R2-r2r2=2hR-h2                                                   ...1Volume of cone =13πr2hV=13πh2hR-h2                                From eq. 1V=13π2h2R-h3dVdh=π34hR-3h2For maximum or minimum values of V, we must havedVdh=0π34hR-3h2=04hR=3h2h=4R3Substituting the value of y in eq. 1, we get  x2=4r2-r22x2=4r2-r22x2=4r22x2=2r2x=r2Now, d2Vdh2=π34R-6hd2Vdh2=π34R-6×4R3d2Vdh2=-4πR3<0So, the volume is maximum when h= 4R3.h=23Diameter of sphereHence proved.

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Question 21:

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is cot-12.                     [CBSE 2014]

Answer:


Let:
Radius of the base = r,
Height = h,
Slant height = l,
Volume = V,
Curved surface area = C

As, Volume , V=13πr2hh=3Vπr2Also, the slant height, l=h2+r2=3Vπr22+r2=9V2π2r4+r2=9V2+π2r6π2r4l=9V2+π2r6πr2Now,CSA, C=πrlCr=πr9V2+π2r6πr2Cr=9V2+π2r6rC'r=r×6π2r529V2+π2r6-9V2+π2r6r2=3π2r6-9V2+π2r69V2+π2r6r2=3π2r6-9V2-π2r6r29V2+π2r6=2π2r6-9V2r29V2+π2r6For maxima or minima, C'r=02π2r6-9V2r29V2+π2r6=02π2r6-9V2=02π2r6=9V2V2=2π2r69V=2π2r69V=πr323 or r=3Vπ213So, h=3πr2×πr323h=r2hr=2cotθ=2 θ=cot-12Also,Since, for r<3Vπ213, C'r<0 and for r>3Vπ213, C'r>0So, the curved surface for r=3Vπ213 or V=πr323 is the least.

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Question 22:

An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when θ = π6.
                                                                                                                                                                                      [NCERT EXEMPLAR]

Answer:



Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.

AD=AO+OD=a+acos2θ=a1+cos2θandBC=2BD=2asin2θAs, area of the triangle AC, A=12BC×ADAθ=12×2asin2θ×a1+cos2θ=a2sin2θ1+cos2θ=a2sin2θ+a2sin2θcos2θAθ=a2sin2θ+a2sin4θ2A'θ=2a2cos2θ+4a2cos4θ2A'θ=2a2cos2θ+2a2cos4θA'θ=2a2cos2θ+cos4θFor maxima or minima, A'θ=02a2cos2θ+cos4θ=0cos2θ+cos4θ=0cos2θ=-cos4θcos2θ=cosπ-4θ2θ=π-4θ6θ=πθ=π6Also, A''θ=2a2-sin2θ-sin4θ=-2a2sin2θ+sin4θ<0 at θ=π6.So, the area of the triangle is maximum at θ=π6.

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Question 23:

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides.

Answer:

Let l, b and V be the length, breadth and volume of the rectangle, respectively. Then, 2l+b=36l=18-b                                                ...1Volume of the cylinder when revolved about the breadth, V=πl2bV=π18-b2b                                   From eq.1V=π324b+b3-36b2dVdb=π324+3b2-72bFor the maximum or minimum values of V, we must havedVdb=0π324+3b2-72b=0324+3b2-72b=0b2-24b+108=0b2-6b-18b+108=0b-6b-18=0b=6, 18Now, d2Vdb2=π6b-72At b=6:d2Vdb2=π6×6-72d2Vdb2=-36π<0At b=18:d2Vdb2=π6×18-72d2Vdb2=36π>0Substituting the value of b in eq. 1, we getl=18-6=12So, the volume is maximum when l=12 cm and b=6 cm.

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Question 24:

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

Answer:

Let the height, radius of base and volume of the cone be h, r and V, respectively. Then,h=R+R2-r2h-R=R2-r2Squaring both the sides, we geth2+R2-2hR=R2-r2r2=2hR-h2                                                         ...1Now, V=13πr2hV=π32h2R-h3                                       From eq. 1dVdh=π34hR-3h2For maximum or minimum values of V, we must havedVdh=0π34hR-3h2=04hR=3h2h=4R3Now, d2Vdh2=π34R-6hπ34R-8R=0-4πR3<0So, the volume is maximum when h=4R3.h=4×123=16 cm

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Question 25:

A closed cylinder has volume 2156 cm3. What will be the radius of its base so that its total surface area is minimum.

Answer:

Let the height, radius of the base and surface area of the cylinder be h, r and S, respectively. Then,Volume =πr2h2156=πr2h2156=227r2hh=2156×722r2h=686r2                                             ...1Surface area=2πrh+2πr2S=4312r+44r27                        From eq. 1dSdr=4312-r2+88r7For maximum or minimum values of S, we must havedSdr=04312-r2+88r7=04312r2=88r7r3=4312×788r3=343r=7 cmNow, d2sdr2=8624r3+887d2sdr2=8624343+887d2sdr2=1767>0So, the surface area is minimum when r=7 cm.

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Question 26:

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius 53 cm is 500 π cm3.

Answer:

Let the height, radius of base and volume of a cylinder be h, r and V, respectively. Then,h24+r2=R2h2=4R2-r2r2=R2-h24                                           ...1Now, V=πr2hV=πhR2-h34                                 From eq. 1dVdh=πR2-3h24For maximum or minimum values of V, we must havedVdh=0πR2-3h24=0R2-3h24=0R2=3h24h=2R3d2Vdh2=-3πh2d2Vdh2=-3π2×2R3d2Vdh2=-3πR3<0So, the volume is maximum when h=2R3.Maximum volume=πhR2-h24=π×2R3R2-4R212=2πR38R212=4πR333=4π53333=500π cm3



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Question 27:

Show that among all positive numbers x and y with x2 + y2 =r2, the sum x+y is largest when x=y=r2.

Answer:

Here,x2+y2=r2y=r2-x2                                                ...1Now,Z=x+yZ=x+r2-x2                                     From eq. 1dZdx=1 +-2x2r2-x2For maximum or minimum values of Z, we must havedZdx=01 +-2x2r2-x2=02x=2r2-x2x=r2-x2Squaring both the sides, we getx2=r2-x22x2=r2x=r2Substituting the value of x in eq. 1, we gety=r2-x2y=r2-r22y=r2d2zdx2=-r2-x2+x-xr2-x2r2-x2d2zdx2=-r2+x2-x2r2-x232d2zdx2=-r2r3×22d2zdx2=-22r<0So, z=x+y is maximum when x=y=r2.Hence proved.

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Question 28:

Determine the points on the curve x2 = 4y which are nearest to the point (0,5).

Answer:

Let the point x, y on the curve x2=4y be nearest to 0, 5. Then,x2=4yy=x24                                        ... 1Also,d2=x2+y-52              Using distance formulaNow,Z=d2=x2+y-52Z=x2+x24-52              Using eq. 1Z=x2+x416+25-5x22dZdy=2x+4x316-5xFor maximum or minimum values of Z, we must havedZdy=02x+4x316-5x=04x316=3xx3=12xx2=12x=±23Substituting the value of x in eq. 1, we gety=3Now,d2Zdy2=2+12x216-5d2Zdy2=9-3=6>0So, the required nearest point is ±23, 3 .

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Question 29:

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=4xx=y24                                              ... 1d2=x-22+y+82                    Using distance formulaNow,Z=d2=x-22+y+82Z=y24-22+y+82                   From eq. 1Z=y416+4-y2+y2+64+16ydZdy=4y316+16For maximum or minimum values of Z, we must havedZdy=04y316+16=04y316=-16y3=-64y=-4Substituting the value of x in eq. 1, we getx=4Now, d2Zdy2=12y216d2Zdy2=12>0So, the nearest point is 4,-4. 

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Question 30:

Find the point on the curve x2=8y which is nearest to the point (2,4).

Answer:

Let x, y be nearest to the point 2, 4. Then,x2=8yy=x28                                       ...1d2=x-22+y-42             Using distance formulaNow,Z=d2=x-22+y-42Z=x-22+x28-42               From eq. 1Z=x2+4-4x+x464+16-x2dZdy=-4+4x364For maximum or minimum values of Z, we must havedZdy=0-4+4x364=0x316=4x3=64x=4Substituting the value of x in eq. 1, we gety=2Now,d2Zdy2=12x264d2Zdy2=3>0So, the nearest point is 4, 2. 

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Question 31:

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=2yy=x22                                                   ...1The distance between points x, y and 0, 5 is given byd2=x2+y-52Now,d2=ZZ=x2+x22-52Z=x2+x44+25-5x2dZdy=2x+x3-10xFor maximum or a minimum values of Z, we must havedZdy=0x3-8x=0x2=8x=±22Substituting the value of x in eq. 1, we gety=4d2Zdy2=3x2-8d2Zdy2=24-8=16>0So, the nearest point is ±22, 4.

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Question 32:

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 the point on the parabola be x, y. Then,y=x2+7x+2                                 ... 1Let the distance of a point x, x2+7x+2  from the line y=3x-3 be S. Then,S=-3x+x2+7x+2 +310dSdt=-3+2x+710For maximum or minimum values of S, we must havedSdt=0-3+2x+710=02x=-4x=-2Now, d2Sdt2=210>0So, the nearest point is x, x2+7x+2 .-2, 4-14+2-2, -8

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Question 33:

Find the point on the curvey y2=2x which is at a minimum distance from the point (1,4).

Answer:

Suppose a point x, y on the curve y2=2x is nearest to the point 1, 4. Then,y2=2xx=y22                                        ...1d2=x-12+y-42              Using distance formulaNow,Z=d2=x-12+y-42Z=y22-12+y-42              From eq. 1Z=y44+1-y2+y2+16-8ydZdy=y3-8For maximum or minimum values of Z, we must havedZdy=0y3-8=0y3=8y=2Substituting the value of y in 1, we getx=2Now,d2Zdy2=3y2d2Zdy2=12>0So, the required nearest point is 2, 2. 

Page No 18.74:

Question 34:

Find the maximum slope of the curve y=-x3+3x2+2x-27.

Answer:

Given:y=-x3+3x2+2x-27        ...1Slope= dydx=-3x2+6x+2Now,M=-3x2+6x+2dMdx=-6x+6For maximum or minimum values of M, we must havedMdx=0-6x+6=06x=6x=1Substituing the value of x in eq. 1, we gety=-13+3×12+2×1-27=-23d2Mdx2=-6<0So, the slope is maximum when x=1 and y=-23. At 1, -23: Maximum slope=-312+61+2=-3+6+2=5 

Page No 18.74:

Question 35:

The total cost of producing x radio sets per  day is Rs x24+ 35x + 25 and the price per set  at which they may be sold is Rs. 50-x2 . Find the daily output to maximum the total profit.

Answer:

Profit =S.P. - C.P.P=x50-x2-x24+35x+25P=50x-x22-x24-35x-25dPdx=50-x-x2-35For maximum or minimum values of P, we must havedPdx=015-3x2=015=3x2x=303x=10Now, d2Pdx2=-32<0So, profit is maximum if daily output is 10 items. 

Page No 18.74:

Question 36:

Manufacturer can sell x items at a price of rupees 5-x100 each. The cost price is Rs  x5+500. Find the number of items he should sell to earn maximum profit.

Answer:

Profit =S.P. - C.P.P=x5-x100-500+x5P=5x-x2100-500-x5dPdx=5-x50-15For maximum or minimum values of P, we must havedPdx=05-x50-15=0245=x50x=24×505x=240Now, d2Pdx2=-150<0So, the profit is maximum if 240 items are sold. 

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Page No 18.74:

Question 37:

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.

Answer:

Let l, h, V and S be the length, height, volume and surface area of the tank to be constructed.Since volume, V is constant,l2h=Vh=Vl2                        ...1Surface area, S=l2+4lhS=l2+4Vl                          From eq. 1dSdl=2l-4Vl2For S to be maximum or minimum, we must havedSdl=02l-4Vl2=02l3-4V=02l3=4Vl3=2VNow, d2Sdl2=2+8Vl3d2Sdl2=2+8V2V=6>0Here, surface area is minimum.h=Vl2Substituting the value of V=l32 in eq. 1, we geth=l32l2h=l2Hence proved.

Page No 18.74:

Question 38:

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: l=2b                     ...1c=lbhc=2b2hh=c2b2                     ... 2Let cost of the material required for bottom be K/m2.Cost of the material required for 4 walls and top = Rs 3K/m2Total cost, T= Klb+3k2lh+2bh+lbT=2Kb2+3K4bc2b2+2bc2b2+2b2                                       From eqs. 1 and 2              dTdb=4Kb+3K-3cb2+4bFor maximum or minimum values of T, we must havedTdb=04kb+3K-3cb2+4b=04b=33cb2-4b4b=9cb2-12b4b=9c-12b3b24b3=9c-12b316b3=9cb=9c1613Now, d2Tdb2=4K+3K6cb3+4d2Tdb2=4K+3K6c9c×16+4K4+3×44348K>0 Cost is minimum when b=9c1613.Substituting 9c1613in eq. 1 and eq. 2l=29c1613h=c2b2h=c29c1623h=32c8113Thus, the most economic dimensions of the box are l=29c1613, b=9c1613 and h=32c8113.

Page No 18.74:

Question 39:

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:

Let r be the radius of the sphere, x be the side of the cube and S be the sum of the surface area of both. Then,
 S=4πr2+6x2
x=S-4πr2612                                            ...(1)

Sum of volumes, V43πr3+x3

V4πr33+S-4πr2632                        [From eq. (1)]

dVdr=4πr2-2πrS-4πr2612

For the minimum or maximum values of V, we must have
dVdr=0                                                          ...(2)
4πr2-2πrS-4πr2612=0                                      From eq. 24πr2=2πrS-4πr26124πr2=2πrx                                                                      From eq. 1          x=2r

Now,
d2Vdr2=8πr-2πS-4πr2612-2πr2S-4πr26-12-8πr6d2Vdr2=8πr-2πS-4πr2612+43π2r26S-4πr212d2Vdr2=8πr-2πx+43π2r21x=8πr-4πr+23π2rd2Vdr2=4πr+23π2r>0

So, volume is minimum when x = 2r.

Page No 18.74:

Question 40:

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi-circular ends is π : (π+2).

Answer:

Volume, V=12πlD22V=πD2l8l=8VπD2                                                                ...1Total surface area=πD24+lD+πDl2S=πD24+8VπD+8V2D                                   From eq. 1dSdD=πD2-8VπD2-8V2D2For maximum or minimum values of S, we must havedSdD=0πD2-8VπD2-8V2D2=0πD2=8VD21π+12D3=16Vπ1π+12Now, d2SdD2=π2+16VD31π+12d2SdD2=π2+π>0l=8VπD2l=8πD2πD3162ππ+2l=Dππ+2lD=ππ+2Hence proved.

Page No 18.74:

Question 41:

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.a2=h2+b244a2-b2=h2                                          ...1Here,Strength of beam, S=Kbh2                Where K is some constantS=kb4R2-b2                                 From eq. 1S=kb4a2-b3dSdb=k4a2-3b2For maximum or minimum values of S, we must havedSdb=0k4a2-3b2=04a2-3b2=04a2=3b2b=2a3Substituting the value of b in eq. 1, we get4a2-2a32=h212a2-4a23=h2h=223aNow, d2Sdb2=-6Kbd2Sdb2=-6K2a3d2Sdb2=-12Ka3<0So, the strength of beam is maximum when b=2a3 and h=223a.

Page No 18.74:

Question 42:

A straight line is drawn through a given point P(1,4). Determine the least value of the sum of the intercepts on the coordinate axes.

Answer:

The equation of line passing through 1, 4 with slope m is given by y-4=mx-1                                           ...1Substituting y=0, we get0-4=mx-1-4m=x-1x=m-4mSubstituting x=0, we get y-4=m0-1y=-m+4x=-m-4So, the intercepts on coordinate axes are m-4m and -m-4.Let S be the sum of the intercepts. Then,S=m-4m-m-4dSdm=4m2-1For maximum or minimum values of S, we must have dSdm=04m2-1=04m2=1m2=4m=±2Now, d2Sdm2=-8m3d2Sdm2m=2=-823=-1<0So, the sum is minimum at m=2.d2Sdm2m=-2=-8-23=1>0So, the sum is maximum at m=-2.Thus, the minimum value is given byS=-2-4-2--2-4=3+6=9

Page No 18.74:

Question 43:

The total area of a page is 150 cm2. The combined width of the margin at the top and bottom is 3 cm and the side 2 cm. 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 =150xy=150y=150x                                                             ...1Area of the printed matter= x-3y-2A=xy-2x-3y+6A=150-2x-450x+6dAdx=-2+450x2For maximum or minimum values of A, we must havedAdx=0-2+450x2=02x2=450x=15Substituting the value of x in 1, we gety=10Now,d2Adx2=-900x3d2Adx2=-900153d2Adx2=-9003375<0So, area of the printed matter is maximum when x=15 and y=10.

Page No 18.74:

Question 44:

The space s described in time t by a particle moving in a straight line is given by S = t5-40t3 + 30t2 + 80t - 250. Find the minimum value of acceleration.

Answer:

Given:s=t5-40t3+30t2+80t-250dsdt=5t4-120t2+60t+80Acceleration, a=d2sdt2=20t3-240t+60dadt=60t2-240For maximum or minimum values of a, we must havedadt=060t2-240=060t2=240t=2Now,d2adt2=120td2adt2=240>0So, acceleration is minimum at t= 2.amin=2023-2402+60=160-480+60=-260 At t=2:a=-260 

Page No 18.74:

Question 45:

A particle is moving in a straight line such that its distance at any time t is given by  S = t44-2t3  + 4t2 -7. Find when its velocity is maximum and acceleration minimum.

Answer:

Given:s=t44-2t3+4t2-7v=dsdt=t3-6t2+8ta=dvdt=3t2-12t+8For maximum or minimum values of v, we must havedvdt=03t2-12t+8=0On solving the equation, we gett=2±23Now,d2vdt2=6t-12At t=2-23:d2vdt2=62-23-12-123<0So, velocity is maximum at t=2-23.Again,dadt=6t-12For maximum or minimum values of a, we must havedadt=06t-12=0t=2Now,d2adt2=6>0So, acceleration is minimum at t=2.



Page No 18.80:

Question 1:

Write necessary condition for a point x = c to be an extreme point of the function f(x).

Answer:

We know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f '(x) = 0 at x = c
f '(c) = 0

Page No 18.80:

Question 2:

Write sufficient conditions for a point x=c to be a point of local maximum.

Answer:

We know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f '(x) = 0 at x = c
f '(c) = 0

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 ''(c) < 0

Page No 18.80:

Question 3:

If f(x) attains a local minimum at x=c, then write the values of f' (c) and f'' (c).

Answer:

If f(x) attains a local minimum at x = c, then the first order derivative of the function at the given point must be equal to zero, i.e.
f '(x) = 0 at x = c
f '(c) = 0

The second order derivative of the function at the given point must be greater than zero, i.e.
f ''(c) > 0

Page No 18.80:

Question 4:

Write the minimum value of f(x) = x+1x, x>0.

Answer:

Given:fx = x+1xf'x = 1-1x2For a local maxima or a local minima, we must have f'x=01-1x2=0x2=1x=1, -1But x>0x=1Now,f''x = 1x3At x=1: f''1 =213 =2>0So, x=1 is a point of local minimum.Thus, the local minimum value is given byf1 = 1+11 =1+1=2

Page No 18.80:

Question 5:

Write the maximum value of f(x) = x+1x, x>0.

Answer:

Given: fx = x+1xf'x = 1-1x2For a local maxima or a local minima, we must have f'x=01-1x2=0x2=1x=1, -1But x<0x=-1Now,f''x = 1x3At x=-1: f''-1 =2-13 =-2<0So, x=-1 is a point of local maximum.Thus, the local maximum value is given byf-1 = -1+1-1 =-1-1=-2

Page No 18.80:

Question 6:

Write the point where f(x) = x log, x attains minimum value.

Answer:

Given:fx = x loge xf'x = loge x+1For a local maxima or a local minima, we must havef'x=0loge x+1=0loge x=-1x=1ef1e = 1e loge 1e=-1eNow,f''x = 1xAt x=1e:f''1e =11e =e>0So, 1e, -1e is a point of local minimum..

Page No 18.80:

Question 7:

Find the least value of f(x) = ax+bx, where a>0, b>0 and x>0.

Answer:

We have,fx = ax+bxf'x =a-bx2For a local maxima or a local minima, we must havef'x=0a-bx2=0x2=bax=ba, -baBut, x>0 x=baNow,f''x = 2bx3At x=ba f''ba =2bba3 =2a32b12>0       a>0 and b>0So, x=ba is a point of local minimum.Hence, the least value isfba = aba+bba =ab+ab=2ab

Page No 18.80:

Question 8:

Write the minimum value of f(x) = xx .

Answer:

Given: fx = xxTaking log on both sides, we getlog fx=x log xDifferentiating w.r.t. x, we get1fx f'x=log x+1f'x=fx log x+1f'x=xx log x+1           ....1For a local maxima or a local minima, we must have f'x=0xx log x+1=0log x=-1x=1eNow,f''x =xx log x+12+xx×1x=xx log x+12+xx-1At x=1e:f''1e =1e1elog1e+12+1e1e-1=1e1e-1>0So, x=1e is a point of local minimum.Thus, the minimum value is given byf1e = 1e1e=e-1e

Page No 18.80:

Question 9:

Write the maximum value of f(x) = x1/x.

Answer:

Given: fx = x1xTaking log on both sides, we getlog fx=1xlog xDifferentiating w.r.t. x, we get1fxf'x=-1x2log x+1x2f'x=fx1x21-log xf'x=x1x1x2-1x2log x                 ...1f'x=x1x-21-log x       For a local maxima or a local minima, we must havef'x=0x1x-21-log x=0log x=1x=eNow, f''x =x1x1x2-1x2log x2+x1x-2x3+2x3log x-1x3=x1x1x2-1x2log x2+x1x-3x3+2x3log xAt x=ef''e =e1e1e2-1e2log e2+e1e-3e3+2e3log e=-e1e1e3<0So, x=e  is a point of local maximum.Thus, the maximum value is given byfe = e1e

Page No 18.80:

Question 10:

Write the maximum value of f(x) = log xx, if it exists.

Answer:

Given:fx =log xxf'x = 1-log xx2For a local maxima or a local minima, we must have f'x=01-log xx2=01-log x=0log x=1log x=log ex=eNow, f''x = -x-2x1-log xx4=-3x-2x log xx4At x=e:f''e =-3e-2e log ee4 =-5e3<0So, x=e is a point of local maximum.Thus, the local maximum value is given byfe =log ee=1e.

Page No 18.80:

Question 1:

The maximum value of x1/x, x>0 is

(a) e1/e

(b) 1ee

(c) 1

(d) none of these

Answer:

(a)        e1e
Given: fx = x1xTaking log on both sides, we getlog fx=1xlog xDifferentiating w.r.t. x, we get1fxf'x=-1x2log x+1x2f'x=fx1x21-log xf'x=x1x1x2-1x2log x                 ...1f'x=x1x-21-log x       For a local maxima or a local minima, we must havef'x=0x1x-21-log x=0log x=1x=eNow, f''x =x1x1x2-1x2log x2+x1x-2x3+2x3log x-1x3=x1x1x2-1x2log x2+x1x-3x3+2x3log xAt x=e:f''e =e1e1e2-1e2log e2+e1e-3e3+2e3log e=-e1e1e3<0So, x=e is a point of local maxima. Maximum value=fe = e1e

Disclaimer: The answer given in the book is incorrect. The solution provided here is according to the question.



Page No 18.81:

Question 2:

If ax + bx > c for all positive x where a,b,>0, then

(a) ab < c24

(b) ab > c24

(c) ab > c4

Answer:

(b) abc24Given: ax+bxcMinimum value of ax+bx=cNow,fx=ax+bxf'x=a-bx2For a local maxima or a local minima, we must havef'x=0a-bx2=0ax2-b=0ax2=bx2=bax=±baf''x=2bx3f''x=2bba3f''x=2b a32b32>0So, x=ba is a local minima. fba=aba+bbac=aaba+bbbac=ab+abc2abcc2abc24ab

Page No 18.81:

Question 3:

The minimum value of xlogex is

(a) e

(b) 1/e

(c) 1

(d) none of these

Answer:

(a) e Given: fx= xloge xf'x=loge x-1loge x2For a local maxima or a local minima, we must have f'x=0loge x-1loge x2=0loge x-1=0loge x=1x=eNow, f''x=-1xloge x2+2xloge x3 f''e=-1e+2e=1e>0So, x=e is a local minima. Minimum value of fx=eloge e=e

Page No 18.81:

Question 4:

For the function f(x) = x+ 1x

(a) x = 1 is a point of maximum
(b) x = -1 is a point of minimum
(c) maximum value > minimum value
(d) maximum value< minimum value

Answer:

(d) maximum value < minimum valueGiven: fx= x+1xf'x=1-1x2For a local maxima or a local minima, we must have f'x=01-1x2=0x2-1=0x2=1x=±1Now, f''x=2x3f''1=21=2>0So, x=1 is a local minima.Also,f''-1=-2<0So, x=-1 is a local maxima.The local minimum value is given byf1=2The local maximum value is given byf-1=-2 Maximum value<Minimum value

Page No 18.81:

Question 5:

Let f(x) = x3+3x2-9x+2. Then, f(x) has

(a) a maximum at x = 1
(b) a minimum at x = 1
(c) neither a maximum nor a minimum at x = -3
(d) none of these

Answer:

(b) a minimum at x=1Given: fx= x3+3x2-9x+2f'x=3x2+6x-9For a local maxima or a local minima, we must have f'x=03x2+6x-9=0x2+2x-3=0x+3x-1=0x=-3, 1Now, f''x=6x+6f''1=6+6=12>0So, x=1 is a local minima.Also, f''-3=-18+6=-12<0So, x=-3 is a local maxima.

Page No 18.81:

Question 6:

The minimum value of f(x) = x4-x2-2x+6 is
(a) 6
(b) 4
(c) 8
(d) none of these

Answer:

(b) 4Given: fx= x4-x2-2x+6f'x=4x3-2x-2f'x=x-14x2+4x+2For a local maxima or a local minima, we must have f'x=0x-14x2+4x+2=0x-1=0x=1Now, f''x=12x2-2f''1=12-2=10>0So, x=1 is a local minima.The local minimum value is given byf1=1-1-2+6=4

Page No 18.81:

Question 7:

The number which exceeds its square by the greatest possible quantity is
(a) 12

(b) 14

(c) 34

(d) none of these

Answer:

(a) 12Let the required number be x. Then,fx=x-x2       f'x=1-2xFor a local maxima or a local minima, we must have f'x=01-2x=02x=1x=12Now, f''x=-2<0So, x=12 is a local maxima.Hence, the required number is 12.

Page No 18.81:

Question 8:

Let f(x) = (x-a)2 + (x-b)2 + (x-c)2. Then, f(x) has a minimum at x =

(a) a+b+c3

(b) abc3

(c) 31a+1b+1c

(d) none of these

Answer:

(a) a+b+c3Given: fx= x-a2+x-b2+x-c2f'x= 2x-a+2x-b+2x-cFor a local maxima or a local minima, we must have f'x=02x-a+2x-b+2x-c=02x-2a+2x-2b+2x-2c=06x=2a+b+cx=a+b+c3Now, f''x=2+2+2=6>0So, x=a+b+c3 is a local minima.

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Question 9:

The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is

(a) 14

(b) 12

(c) 18

(d) none of these

Answer:

(b) 12Let the two non-zero numbers be x and y. Then,x+y=8y=8-x                      ... 1Now,fx=1x+1yfx=1x+18-x                  From eq. 1f'x=-1x2+18-x2For a local minima or a local maxima, we must have f'x=0-1x2+18-x2=0-8-x2+x2x28-x2=0-64-x2+16x+x2=016x-64=0x=4f''x=2x3-28-x3f''4=243-28-43f''4=264-264=0 Minimum value=14+14=12

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Question 10:

The function f(x) = r=15 (x-r)2 assumes minimum value at x =
(a) 5

(b) 52

(c) 3

(d) 2

Answer:

(c) 3Given: fx= r=15x-r2fx=x-12+x-22+x-32+x-42+x-52f'x=2x-1+x-2+x-3+x-4+x-5f'x=25x-15For a local maxima and a local minima, we must havef'x=025x-15=05x-15=05x=15x=3Now, f''x=10f''x=10>0So, x=3 is a local minima.

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Question 11:

At x= 5π6, f(x) = 2 sin 3x + 3 cos 3x is

(a) 0

(b) maximum

(c) minimum

(d) none of these

Answer:

(d) none of these

Given: fx=2 sin 3x+3 cos 3xf'x=6 cos 3x-9 sin 3xFor a local minima or a local maxima, we must have f'x=06 cos 3x-9 sin 3x=06 cos 3x=9 sin 3xsin 3xcos 3x=23tan 3x=23                         ...1At x=5π6: tan 3x=tan 5π2tan 3x=tan π2So, tan 3x is not defined.        tan 3x23 is not satisfying eq. 1Thus, x=5π6 is not a critical point.

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Question 12:

If x lies in the interval [0,1], then the least value of x2 + x + 1 is
(a) 3

(b)34

(c) 1

(d) none of these

Answer:

(c) 1Given: fx= x2+x+1f'x=2x+1For a local maxima or a local minima, we must have f'x=02x+1=02x=-1x=-120, 1At extreme points: f0=0f1=1+1+1=3>0So, x=1 is a local minima.

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Question 13:

The least value of the function f(x) = x3-18x2+96x in the interval [0,9] is
(a) 126
(b) 135
(c) 160
(d) 0

Answer:

(d) 0Given: fx= x3-18x2+96xf'x=3x2-36x+96For a local maxima or a local minima, we must have f'x=03x2-36x+96=0x2-12x+32=0x-4x-8=0x=4, 8So,f8=83-1882+968=512-1152+768=128f4=43-1842+964=64-288+384=160f0=03-1802+960=0f9=93-1892+969=729-1458+864=135Hence, 0 is the minimum value in the range 0, 9.

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Question 14:

The maximum value of f(x) = x4-x+x2 on [-1, 1] is

(a) -14

(b) -13

(c)  16

(d) 15

Answer:

Given: fx= x4-x+x2f'x=4-x+x2-x-1+2x4-x+x22For a local maxima or a local minima, we must have f'x=04-x+x2-x-1+2x4-x+x22=04-x+x2-x-1+2x=04-x+x2+x-2x2=0x2=4x=±2 -1, 1So,f-1=-14--1+-12=-16f1=14-1+12=14Hence, the maximum value is 14.

Disclaimer: The question in the book has some error. So, none of the options are matching with the solution. The solution here is according to the question given in the book.

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Question 15:

The point on the curve y2 = 4x which is nearest to, the point (2,1) is
(a) 1,22

(b) (1,2)

(c) (1,-2)

(d) (-2,1)

Answer:

(b) 1, 2Let the required point be x, y. Then, y2=4xx=y24                      ... 1Now, d=x-22+y-12Squaring both sides, we getd2=x-22+y-12d2=y24-22+y-12d2=y416+4-y2+y2+1-2y                     From eq. 1Now,Z=d2=y416+4-y2+y2+1-2ydZdy=y34-2y+2y-2dZdy=y34-2y34-2=0y3=8y=2Substituting the value of in 1, we getx=1Now, d2Zdy2=3y24d2Zdy2=3224=3>0So, the nearest point is 1, 2.



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Question 16:

If x+y=8, then the maximum value of xy is

(a) 8

(b) 16

(c) 20

(d) 24

Answer:

(b) 16Given: x+y=8y=8-x                           ... 1Let fx be xy. fx= x8-x             From eq. 1f'x=8-2xFor a local maxima or a local minima, we must have f'x=08-2x=08=2xx=4y=8-4=4             From eq. 1Now, f''x=-2f''4=-2<0So, x=4 is a local maxima.Hence, the local maximum value is given byf4=4×4=16

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Question 17:

The least and greatest values of f(x) = x3-6x2+9x in [0,6], are

(a) 3,4

(b) 0,6

(c) 0,3

(d) 3,6

Answer:

Given: fx= x3-6x2+9xf'x=3x2-12x+9For a local maxima or a local minima, we must have f'x=03x2-12x+9=0x2-4x+3=0x-1x-3=0x=1, 3Now,f0= 03-602+90=0f1= 13-612+91=1-6+9=4f3= 33-632+93=27-54+27=0f6= 63-662+96=216-216+54=54

The least and greatest values of f(x) = x3- 6x2+9x in [0, 6] are 0 and 54, respectively.

Disclaimer: The question in the book has some error. So, none of the options are matching with the solution. The solution here is according to the question given in the book.

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Question 18:

f(x) = sin + 3 cos x is maximum when x =

(a) π3

(b) π4

(c) π6

(d) 0

Answer:

(c) π6Given: fx= sin x +3 cos xf'x=cosx-3 sin xFor a local maxima or a local minima, we must have f'x=0cos x-3 sin x=0cos x=3 sin xtan x=13x=π6Now, f''x=-sin x-3 cos xf''π2=-sinπ2-3 cosπ2-12-32=-2<0So, x= π2  is a local maxima.

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Question 19:

If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is

(a) 34

(b) 13

(c) 14

(d) 23

Answer:

(d) 23


Let h, r, V and R be the height, radius of the base, volume of the cone and the radius of the sphere, respectively. Given: h=R+R2-r2h-R=R2-r2Squaring both side, we geth2+R2-2hR=R2-r2r2=2hr-h2                        ... 1Now, Volume =13πr2hV=π32h2R-h3                 From eq. 1dVdh=π34hR-3h2For maximum or minimum values of V, we must havedVdh=0π34hR-3h2=04hR-3h2=04hR=3h2h=4R3Now, d2Vdh2=π34R-6h=π34R-6×4R3=-4πR3<0So, volume is maximum when h=4R3.h=22R3h2R=23 HeightDiameter of sphere=23

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Question 20:

The minimum value of x2+250x is
(a) 75

(b) 50

(c) 25

(d) 55

Answer:

(a) 75


Given: fx= x2+250xf'x=2x-250x2For a local maxima or a local minima, we must have f'x=02x-250x2=02x3-250=0x3=125x=5Now, f''x=2+500x3f''5=2+50053=750125=6>0So, x=5 is a local minima. f'xmin= 52+2505=3755=75

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Question 21:

If(x) = x+1x, x > 0, then its greatest value is

(a) -2

(b) 0

(c) 3

(d) none of these

Answer:

(d) none of these

Given: fx= x+1xf'x=1-1x2For a local maxima or a local minima, we must have f'x=01-1x2=0x2-1=0x2=1x=±1x=1                                                Given: x>0Now, f''x=2x3f''1=2>0So, x=1 is a local minima.

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Question 22:

If(x) = 14x2+2x+1, then its maximum value is

(a) 43

(b) 23

(c) 1

(d) 34

Answer:

(a) 43

Maximum value of 14x2+2x+1= Minimum value of 4x2+2x+1 
Now, fx= 4x2+2x+1f'x=8x+2For a local maxima or a local minima, we must have f'x=08x+2=08x=-2x=-14Now, f''x=8f''1=8>0So, x=-14 is a local minima.Thus, 14x2+2x+1is maximum at x=-14.Maximum value of 14x2+2x+1 =14-142+2-14+1                                                                 =1416-12+1=1612=43

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Question 23:

Let x, y be two variables and x>0, xy=1, then minimum value of x+y is
(a) 1

(b) 2

(c) 212

(d) 313

Answer:

(b) 2

Given: xy=1y=1xfx= x+1xf'x= 1-1x2For a local maxima or a local minima, we must have f'x=01-1x2=0x2-1=0x2=1x=±1x=1                                                  Given: x>1y=1Now, f''x=2x3f''1=2>0So, x=1 is a local minima. Minimum value of fx=f1=1+1=2

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Question 24:

f(x) = 1+2 sin x+3 cos2x, 0<x<2π3 is

(a) Minimum at x = π2

(b) Maximum at x = sin-1 (13)

(c) Minimum at x = π6

(d) Maximum at sin-1(16)

Answer:

(a) Minimum at x=π2Given: fx=1+2 sin x+3 cos2 xf'x=2 cos x-6 cos x sin xf'x=2 cos x1-3 sin xFor a local maxima or a local minima, we must have f'x=02 cos x1-3 sin x=02 cos x=0  or 1-3 sin x=0cos x=0 or sin x=13x=π2 or x=sin-113Now, f''x=-2 sin x-6 cos 2xf''π2=-2 sin π2-6 cos 2×π2=-2+6=4>0So, x=π2 is a local minima.Also, f''sin-113=-2 sin sin-113-6 cos sin-113=-23-6×223=-23+42<0So, x=sin-113 is a local maxima.

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Question 25:

The function f(x) =2x3-15x2+36x+4 is maximum at x =

(a) 3

(b) 0

(c) 4

(d) 2

Answer:

(d) 2

Given: fx= 2x3-15x2+36x+4f'x=6x2-30x+36For a local maxima or a local minima, we must have f'x=06x2-30x+36=0x2-5x+6=0x-2x-3=0x=2, 3Now, f''x=12x-30f''2=24-30=-6<0So, x=1 is a local maxima.Also, f''3=36-30=6>0So, x=2 is a local maxima.

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Question 26:

The maximum value of f(x) = x4+x+x2 on [-1,1] is

(a) -14

(b) -13

(c) 16

(d) 15

Answer:

(c) 16


Given: fx= x4+x+x2f'x=4+x+x2-x1+2x4+x+x22For a local maxima or a local minima, we must have f'x=04+x+x2-x1+2x4+x+x22=04+x+x2-x1+2x=04-x2=0x=±2-1,1The values of fx at extreme points are given byf1=14+1+12=16f-1=-14-1+-12=-14Thus, 16 is the maximum value.

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Question 27:

Let f(x) = 2x3 - 3x2 - 12x + 5 on [-2, 4]. The relative maximum occurs at x=

(a) -2

(b) -1

(c) 2

(d) 4

Answer:

(c) 2

Given: fx= 2x3-3x2-12x+5f'x=6x2-6x-12For a local maxima or a local minima, we must have f'x=06x2-6x-12=0x2-x-2=0x-2x+1=0x=2, -1Now, f''x=12x-6f''-1=-12-6=-18<0So, x=1 is a local maxima.Also, f''2=24-6=18>0So, x=2 is a local minima.

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Question 28:

The minimum value of x loge x is equal to

(a) e

(b) 1/e

(c) -1/e

(d) 2/e

(e) -e

Answer:

(c) -1eHere, fx= x loge xf'x=loge x +1For a local maxima or a local minima, we must have f'x=0loge x +1=0loge x =-1x=e-1Now, f''x=1xf''e-1=e>0So, x=e-1  is a local minima.Hence, the minimum value of fx=fe-1.e-1 loge e-1 =-e-1=-1e

Page No 18.82:

Question 29:

The minimum value of the function f(x) = 2x3-21x2+36x-20 is

(a) -128

(b) -126

(c) -120

(d) none of these

Answer:

(a)-128

Given:fx= 2x3-21x2+36x-20f'x=6x2-42x+36For a local maxima or a local minima, we must have f'x=06x2-42x+36=0x2-7x+6=0x-1x-6=0x=1, 6Now, f''x=12x-42f''1=12-42=-30<0So, x=1 is a local maxima.Also, f''6=72-42=30>0So, x=6 is a local miniima.The local minimum value is given byf6==263-2162+366-20=-128



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