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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Maths Chapter 17 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 Maths 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 2020 Book of Class 12 Science Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 17.16:

Question 1:


f(x) = (x-5)4

Answer:

Given: fx=x-54⇒f'x=4x-53For a local maximum or a local minimum, we must havef'x=0⇒4x-53=0⇒x=5
 

Since f '(x) changes from negative to positive when x increases through 5, x = 5 is the point of local minima.
The local minimum value of  f (x) at x = 5 is given by
5-54=0

Page No 17.16:

Question 2:

f(x) = x3 - 3x

Answer:

Given: fx=x3-3x⇒f'x=3x2-3For a local maximum or a local minimum, we must havef'x=0⇒3x2-3=0⇒x2-1=0⇒x=±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 17.16:

Question 3:

f(x) = x3  (x-1)2

Answer:

Given: fx=x3x-12⇒f'x=3x2x-12+2x3x-1For a local maximum or a local minimum, we must have f'x=0⇒3x2x-12+2x3x-1=0⇒x2x-13x-3+2x=0⇒x2x-15x-3=0⇒x=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 17.16:

Question 4:

f(x) =  (x-1) (x+2)2

Answer:

Given: fx=x-1x+22⇒f'x=x+22+2x+2x-1For a local maximum or a local minimum, we must havef'x=0⇒x+22+2x+2x-1=0⇒x+2x+2+2x-2=0⇒x+23x=0⇒x=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 17.16:

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=0⇒x=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 17.16:

Question 6:

f(x) =  x3 - 6x2 + 9x + 15

Answer:

Given: fx=x3-6x2+9x+15⇒f'x=3x2-12x+9For a local maximum or a local minimum, we must havef'x=0⇒3x2-12x+9=0⇒x2-4x+3=0⇒x-1x-3=0⇒x=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 17.16:

Question 7:

f(x) = sin 2x, 0<x<π

Answer:

Given: fx=sin 2x⇒f'x=2 cos 2xFor a local maximum or a local minimum, we must havef'x=0⇒2 cos 2x=0⇒cos 2x=0⇒x=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 17.16:

Question 8:

f(x) =  sin x- cos x, 0 < x<2π

Answer:

Given: fx=sin x-cos x⇒f'x=cos x+sin xFor a local maximum or a local minimum, we must havef'x=0⇒cos x+sin x=0⇒cos x=-sin x⇒tan x=-1⇒x=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 17.16:

Question 9:

f(x) =  cos x, 0<x<π

Answer:

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

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

Page No 17.16:

Question 10:

f(x) = sin 2x-x, -π2<<x<π2

Answer:

Given: fx=sin 2x-x⇒f'x=2 cos 2x-1For a local maximum or a local minimum, we must havef'x=0⇒2 cos 2x-1=0⇒cos 2x=12⇒x=-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 17.16:

Question 11:

f(x) = 2sin x-x, -π2<<x<π2

Answer:

    Given: fx=2 sin x-x⇒f'x=2 cos x-1For a local maximum or a local minimum, we must have f'x=0⇒2 cos x-1=0⇒cos x=12⇒x=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 17.16:

Question 12:

f(x) = x1-x , x>0

Answer:

Given: fx=x1-x⇒f'x=1-x-x21-x=2-3x21-xFor the local maxima or minima, we must havef'x=0⇒2-3x21-x=0⇒x=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 17.16:

Question 13:

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

Answer:

Given: fx=x32x-13⇒f'x=3x22x-13+6x32x-12For the local maxima or minima, we must have f'x=0⇒3x22x-13+6x32x-12=0⇒3x22x-122x-1+2x=0⇒x22x-124x-1=0⇒x=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 17.16:

Question 14:

f(x) = x2+2x , x>0

Answer:

    Given: fx=x2+2x⇒f'x=12-2x2For the local maxima or minima, we must havef'x=0⇒12-2x2=0⇒12=2x2⇒x2=±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 17.31:

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 ≤x≤2

(xii) f(x) = x + 1-x, x≤1

Answer:

iGiven: fx = x4-62x2+120x+9⇒f'x = 4x3-124x+120For the local maxima or minima, we must have f'x=0⇒4x3-124x+120=0⇒x3-31x+30=0⇒x-1x2+x-30=0⇒x-1x+6x-5=0⇒x=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+15⇒f'x = 3x2-12x+9For the local maxima or minima, we must have f'x=0⇒ 3x2-12x+9=0⇒ x2-4x+3=0⇒x-1x-3=0⇒x=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+3x2-4⇒f'x = 3x2+6xFor the local maxima or minima, we must have f'x=0⇒  3x2+6x=0⇒3xx+2=0⇒x=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-2⇒f'x =-2x-2+4x-3=4x3-2x2For the local maxima or minima, we must have f'x=0⇒4x3-2x2=0⇒4-2x=0⇒x=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 = xex⇒f'x = ex+xexFor the local maxima or minima, we must have f'x=0⇒ex+xex=0⇒ex1+x=0⇒ex≠0 , x=-1⇒x=-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+2x⇒f'x = 12-2x2For the local maxima or minima, we must have f'x=0⇒12-2x2=0⇒x2=4⇒x=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+213⇒f'x =x+213+13x+1x+2-23For the local maxima or minima, we must have f'x=0⇒ x+213+13x+1x+2-23=0⇒13x+1=-x+213×x+223⇒13x+1=-x+2⇒x+1=-3x-6⇒x=-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-x2⇒f'x = 32-x2-x232-x2For the local maxima or minima, we must have f'x=0⇒32-x2-x232-x2=0⇒32-x2=x32-x2⇒32-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+a2x⇒f'x = 3x2-4ax+a2For the local maxima or minima, we must have f'x=0⇒3x2-4ax+a2=0⇒ 3x2-3ax-ax+a2=0⇒3xx-a-ax-a=0⇒3x-ax-a=0⇒x=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+a2x⇒f'x = 1-a2x2For the local maxima or minima, we must havef'x=0⇒ 1-a2x2=0⇒x2=a2⇒x=±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-x2⇒f'x = 2-x2-x22-x2For the local maxima or minima, we must have f'x=0⇒2-x2-x22-x2=0⇒2-x2=x2-x2⇒2-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-x⇒f'x = 1-121-xFor the local maxima or minima, we must have f'x=0⇒1-121-x=0⇒1-x=12⇒1-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 17.31:

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-4⇒f'x= 3x2-10x+8For the local maxima or minima, we must have f'x=0⇒  3x2-10x+8=0⇒3x2-6x-4x+8=0⇒x-23x-4=0⇒x=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-x⇒f'x = 1-x-x21-xFor the local maxima or minima, we must have f'x=0⇒1-x-x21-x=0⇒1-x=x21-x⇒2-2x=x⇒3x=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+12⇒f'x =- 3x-12x+12+2x+1x-13For the local maxima or minima, we must have f'x=0⇒ -3x-12x+12-2x+1x-13=0⇒x-12x+1-3x+1-2x-1=0⇒x-12x+1-3x-3-2x+2=0⇒x-12x+1-5x-1=0⇒x=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 17.31:

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+x⇒f'x=ax+2bx+1Since, f'x has extreme values at x=1 and x=2, f'1=0.⇒a1+2b1+1=0⇒a=-1-2b             ...1f'2=0⇒a2+2b2+1=0⇒a+8b=-2   ⇒a=-2-8b              ...2From eqs. 1 and 2, we get-2-8b=-1-2b⇒6b=-1⇒b=-16Substituting b=-16 in eq. 1, we geta=-1+13=-23

Page No 17.31:

Question 4:

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

Answer:

Here,fx = log xx⇒f'x = 1-log xx2For the local maxima or minima, we must have f'x=0⇒1-log xx2=0⇒1=log x⇒log e=log x⇒x=eNow, f''x = x2-1x-2x1-log xx4=-3+2 log xx3⇒f''e = -3+2 log ee3=-1e3<0So, x=e is the point of local maximum.

Page No 17.31:

Question 5:

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

Answer:

Given: fx = 4x+2+x⇒f'x = -4x+22+1For a local maxima or a local minima, we must have f'x=0⇒-4x+22+1=0⇒-4x+22=-1⇒x+22=4⇒x+2=±2⇒x=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 17.31:

Question 6:

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

Answer:

Given: fx =y=tan x-2x⇒f'x =sec2 x-2For a local maxima or local minima, we must have f'x=0⇒sec2 x-2=0⇒sec2 x=2⇒sec x=±2⇒x=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 17.31:

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+c⇒f'x=3x2+2ax+bAs, fx is maximum at x=-1 and minimum at x=3.So, f-1=0 and f3=0⇒3-12+2a-1+b=0 and 332+2a3+b=0⇒3-2a+b=0        .....iand 27+6a+b=0   .....iiii-i, we get27-3+6a+2a=0⇒8a=-24⇒a=-3Substituting a=-3 in i, we get3-2-3+b=0⇒3+6+b=0⇒b=-9And, c∈R

Page No 17.31:

Question 8:

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

Answer:

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



Page No 17.37:

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-x22⇒f'x=4-xFor a local maximum or a local minimum, we must havef'x=0⇒4-x=0⇒x=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+3⇒f'x=2x-1For a local maximum or a local minimum, we must have f'x=0⇒2x-1=0⇒x=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+25⇒f'x=12x3-24x2+24x-48For a local maximum or a local minimum, we must havef'x=0⇒12x3-24x2+24x-48=0⇒x3-2x2+2x-4=0⇒x2x-2+2x-2=0⇒x-2x2+2=0⇒x-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-1⇒f'x=x-1+x-22x-1For a local maximum or a local minimum, we must havef'x=0⇒x-1+x-22x-1=0⇒2x-1+x-2=0⇒2x-2+x-2=0⇒3x-4=0⇒3x=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 17.37:

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+107⇒f'x=6x2-24For a local maximum or a local minimum, we must havef'x=0⇒6x2-24=0⇒6x2=24⇒x2=4⇒x=±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 17.37:

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 x⇒f'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=0⇒cos x 2 sin x-1=0⇒sin x=12 or cos x=0⇒x=π6 or π2                        ∵ x∈0, 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 17.37:

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=12x43-6x13⇒f'x=16x13-2x-23=28x-1x23For a local maximum or a local minimum, we must havef'x=0⇒28x-1x23=0⇒8x-1=0⇒x=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 17.37:

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+1⇒f'x=6x2-30x+36For a local maximum or a local minimum, we have f'x=0⇒6x2-30x+36=0⇒x2-5x+6=0⇒x-3x-2=0⇒x=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 17.7:

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 17.7:

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 17.7:

Question 3:

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

Answer:

Given: f(x) = x+2

Now,
x+2≥0 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 17.7:

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 17.7:

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 17.7:

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 17.7:

Question 7:

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

Answer:

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

Now,
-x+1≤0 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 17.7:

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 17.7:

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 17.72:

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+y2⇒z=x2+15-x2           From eq. 1⇒z=x2+x2+225-30x⇒z=2x2+225-30x⇒dzdx=4x-30For maximum or minimum values of z, we must havedzdx=0⇒4x-30=0⇒x=152d2zdx2= 4 > 0Substituting x=152 in 1, we get y=152Thus, z is minimum when x=y=152.

Page No 17.72:

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

Page No 17.72:

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 x and y. Then,x, y >-2 and x+y=12                 ...(1)Now,z=x +y3⇒z=x+12-x3                       From eq. 1⇒dzdx=1+312-x2For maximum or minimum values of z, we must havedzdx=0⇒1+312-x2=0⇒12-x2=13⇒12-x=±13⇒x=12±13d2zdx2=612-x⇒d2zdx2=3-6xAt x=12±13:d2zdx2=3-612+13⇒-63<0Thus, z is maximum when x= 12+13.At x=12-13:d2zdx2=3-612-13⇒63>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 17.72:

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=x2y3⇒z=x215-x3                From eq. 1⇒dzdx=2x15-x3-3x215-x2For maximum or minimum values of z, we must havedzdx=0⇒2x15-x3-3x215-x2=0⇒2x15-x=3x2⇒30x-2x2=3x2⇒30x=5x2⇒x=6 and y =9d2zdx2=215-x3-6x15-x2-6x15-x2+6x215-xAt x=6:d2zdx2=293-3692-3692+6369⇒d2zdx2=-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 17.72:

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 =πr2h⇒100=πr2h⇒h=100πr2Surface area S of the cylinder=2πr2+2πrh=2πr2+2πr×100πr2⇒S=2πr2+200r∴dSdr=4πr-200r2 For the maximum or minimum, we must have dSdr=0⇒4πr-200r2=0⇒4πr3=200⇒r=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 17.72:

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-W2x2⇒dMdx=WL2-2×Wx2⇒dMdx=WL2-WxFor maximum or minimum values of M, we must havedMdx=0⇒WL2-Wx=0⇒WL2=Wx⇒x=L2Now,d2Mdx2=-W <0So, M is maximum at x=L2.

ii Given: M=Wx3-Wx33L2⇒dMdx=W3-3×Wx23L2⇒dMdx=W3-Wx2L2For maximum or minimum values of M, we must havedMdx=0⇒W3-Wx2L2=0⇒W3=Wx2L2⇒x=L3Now,d2Mdx2=-2WxL2 <0So, M is maximum at x=L3.

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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=x⇒Side=x4⇒Area of square=x42=x216Circumference of circle, 2πr=y⇒r=y2πArea of circle=πr2=πy2π2=y24πNow,z=Area of square+Area of circle ⇒z=x216+y24π⇒z=x216+28-x24π⇒dzdx=2x16-228-x4πFor maximum or minimum values of z, we must havedzdx=0⇒2x16-228-x4π=0           From eq. 1⇒x4=28-xπ⇒xπ4+x=28⇒xπ4+1=28⇒x=28π4+1⇒x=112π+4⇒y=28-112π+4              From eq. 1⇒y=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.

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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=x⇒Side=x4Area of square=x42=x216Perimeter of triangle, 3Side=y⇒Side=y3Area of triangle=34×Side2=34×y32=3y236Now,z=Area of square+Area of triangle⇒z=x216+3y236⇒z=x216+320-x236                   From eq. 1⇒dzdx=2x16-2320-x36For maximum or minimum values of z, we must havedzdx=0⇒2x16-320-x18=0⇒9x4=320-x⇒9x4+x3=203⇒x94+3=203⇒x=20394+3⇒x=8039+43⇒y=20-8039+43        From eq. 1⇒y=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.

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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 constant⇒x=K-2πr4                           ...1Now,A=x2+πr2⇒A=K-2πr216+πr2                        From eq. 1⇒dAdr=K-2πr216+πr2⇒dAdr=2K-2πr-2π16+2πr⇒dAdr=K-2πr-π4+2πr⇒K-2πr-π4+2πr=0⇒K-2πrπ4=2πr⇒K-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-2πr4⇒x=8r4⇒x=2r∴ Side of the square=Diameter of the circle



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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=52⇒y2=25-x2⇒y=25-x2As, the area of the triangle,A=12×x×y⇒Ax=12×x×25-x2⇒Ax=x25-x22⇒A'x=25-x22+x-2x425-x2⇒A'x=25-x22-x2225-x2⇒A'x=25-x2-x2225-x2⇒A'x=25-2x2225-x2For maxima or minima, we must have f'x=0A'x=0⇒25-2x2225-x2=0⇒25-2x2=0⇒2x2=25⇒x=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-x2⇒A''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=12absinθ⇒Aθ=12absinθ⇒A'θ=12abcosθFor maxima or minima, A'θ=0⇒12abcosθ=0⇒cosθ=0⇒θ=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=45-2x24-2xx⇒dVdx= 45-2x24-2x- 2x45-2x-2x24-2xFor maximum or minimum values of V, we must havedVdx=0⇒45-2x24-2x- 2x45-2x-2x24-2x=0⇒4x2+1080-138x-48x+4x2+4x2-90x=0⇒12x2-276x+1080=0⇒x2-23x+90=0⇒x2-18x-5x+90=0⇒xx-18-5x-18=0⇒x-18=0 or x-5=0⇒x=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=4⇒b=4l

Area of the base = lb = 4 m2

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

∴A=4l+4l⇒dAdl=41-4l2For maximum or minimum values of A, we must havedAdl=0⇒41-4l2=0⇒l=±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=10⇒x+2y+πx2=10⇒2y=10-x-πx2⇒y=1210-x1+π2                                         ...1Now, Area, A=π2x22+xy⇒A=πx28+x210-x1+π2                       From eq. 1⇒A=πx28+10x2-x221+π2⇒dAdx=πx4+102-2x21+π2For maximum or minimum values of A, we must havedAdx=0⇒πx4+102-2x21+π2=0⇒xπ4-1-π2=-5⇒x=-5-4-π4⇒x=20π+4Substituting the value of x in  eq. 1, we gety=1210-20π+41+π2⇒y=5-10π+22π+4⇒y=5π+20-5π-10π+4⇒y=10π+4 d2Adx2=π4-π2-1⇒d2Adx2=π-2π-44⇒d2Adx2=-π-44<0Thus, the area is maximum when x=20π+4 and y=10π+4. So, 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=12⇒3x+2y=12⇒y=12-3x2                                       ...1Area of the window =xy +34x2⇒A=x12-3x2 +34x2⇒A=6x-3x22+34x2⇒dAdx=6-6x2+234x⇒dAdx=6-3x+32x⇒dAdx=6-x3-32For maximum or a minimum values of A, we must havedAdx=0⇒6=x3-32⇒x= 126-3Substituting the value of x in eq. 1, we gety=12-3126-32⇒y=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=R2⇒h=2R2-r2                                                         ...1Volume of cylinder, V=πr2hSquaring both sides, we get⇒V2=π2r4h2⇒V2=4π2r4R2-r2                                                   From eq. 1Now,Z=4π2r4R2-r6⇒dZdr=4π24r3R2-6r5For maximum or minimum values of Z, we must havedZdr=0⇒4π24r3R2-6r5=0⇒4r3R2=6r5⇒6r2=4R2⇒r2=4R26⇒r=2R6Substituting the value of r in eq. 1, we get ⇒h=2R2-2R62⇒h=26R2-4R26⇒h=2R23⇒h=2R3Now, d2Zdr2=4π212r2R2-30r4⇒d2Zdr2=4π2122R62R2-302R64⇒d2Zdr2=4π28R4-80R46⇒d2Zdr2=4π248R4-80R46⇒d2Zdr2=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=r2⇒x2+4y2=4r2⇒x2=4r2-y2                                                   ...1Area of rectangle=xy⇒A=xySquaring both sides, we get⇒A2=x2y2⇒Z=4y2r2-y2                          From eq. 1⇒dZdy=8yr2-16y3For the maximum or minimum values of Z, we must havedZdy=0⇒8yr2-16y3=0⇒8r2=16y2⇒y2=r22⇒y=r2Substituting the value of y in eq. 1, we get⇒x2=4r2-r22⇒x2=4r2-r22⇒x2=4r22⇒x2=2r2⇒x=r2Now, d2Zdy2=8r2-48y2⇒d2Zdy2=8r2-48r22⇒d2Zdy2=-16r2<0So, the area is maximum when x= r2 and y =r2.Area =xy⇒A=r2×r2⇒A=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 = 13πr2h⇒h=3Vπr2∴S=πrr2+3Vπr22⇒S=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+9V6r2⇒3π2r6=π2r6+9V6⇒2π2r6= 9V6Substitutting the value of V we get,2π2r6= 913πr2h2⇒2π2r6= π2r4h2⇒2r2=h2∴h=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-r2⇒h-R2=R2-r2⇒h2+R2-2hr=R2-r2⇒r2=2hR-h2                                                   ...1Volume of cone =13πr2h⇒V=13πh2hR-h2                                From eq. 1⇒V=13π2h2R-h3⇒dVdh=π34hR-3h2For maximum or minimum values of V, we must havedVdh=0⇒π34hR-3h2=0⇒4hR=3h2⇒h=4R3Substituting the value of y in eq. 1, we get  x2=4r2-r22⇒x2=4r2-r22⇒x2=4r22⇒x2=2r2⇒x=r2Now, d2Vdh2=π34R-6h⇒d2Vdh2=π34R-6×4R3⇒d2Vdh2=-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=13πr2h⇒h=3Vπr2Also, the slant height, l=h2+r2=3Vπr22+r2=9V2π2r4+r2=9V2+π2r6π2r4⇒l=9V2+π2r6πr2Now,CSA, C=πrl⇒Cr=πr9V2+π2r6πr2⇒Cr=9V2+π2r6r⇒C'r=r×6π2r529V2+π2r6-9V2+π2r6r2=3π2r6-9V2+π2r69V2+π2r6r2=3π2r6-9V2-π2r6r29V2+π2r6=2π2r6-9V2r29V2+π2r6For maxima or minima, C'r=0⇒2π2r6-9V2r29V2+π2r6=0⇒2π2r6-9V2=0⇒2π2r6=9V2⇒V2=2π2r69⇒V=2π2r69⇒V=πr323 or r=3Vπ213So, h=3πr2×πr323⇒h=r2⇒hr=2⇒cotθ=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×AD⇒Aθ=12×2asin2θ×a1+cos2θ=a2sin2θ1+cos2θ=a2sin2θ+a2sin2θcos2θ⇒Aθ=a2sin2θ+a2sin4θ2⇒A'θ=2a2cos2θ+4a2cos4θ2⇒A'θ=2a2cos2θ+2a2cos4θ⇒A'θ=2a2cos2θ+cos4θFor maxima or minima, A'θ=0⇒2a2cos2θ+cos4θ=0⇒cos2θ+cos4θ=0⇒cos2θ=-cos4θ⇒cos2θ=cosπ-4θ⇒2θ=π-4θ⇒6θ=π⇒θ=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:

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 63r.

Answer:


Let ABC is an isosceles triangle with AB=AC=x and a circle with centre O and radius r is inscribed in the triangle. O,A and O,E and O,D are joined.From ΔABF, AF2+BF2=AB2⇒(3r)2+(y2)2=x2      .....(1)Again,From ΔADO, (2r)2=r2+AD2⇒3r2=AD2⇒AD=3rNow, BD=BF and EC=FC (Since tangents drawn from an external point are equal)Now, AD+DB=x⇒(3r)+(y2)=x⇒y2=x−3     .....(2)∴(3r)2+(x−3r)2=x2⇒9r2+x2−23rx+3r2=x2⇒12r2=23rx⇒6r=3x⇒x=6r3Now, From (2),y2=63r−3r⇒y2=633r−3r⇒y2=(63−33)r3⇒y2=33r3⇒y=23rPerimeter=2x+y=263r+23r=123r+23r=12r+6r3=183r=18×33×3r=63r

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Question 24:

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=36⇒l=18-b                                                ...1Volume of the cylinder when revolved about the breadth, V=πl2b⇒V=π18-b2b                                   From eq.1⇒V=π324b+b3-36b2⇒dVdb=π324+3b2-72bFor the maximum or minimum values of V, we must havedVdb=0⇒π324+3b2-72b=0⇒324+3b2-72b=0⇒b2-24b+108=0⇒b2-6b-18b+108=0⇒b-6b-18=0⇒b=6, 18Now, d2Vdb2=π6b-72At b=6:d2Vdb2=π6×6-72⇒d2Vdb2=-36π<0At b=18:d2Vdb2=π6×18-72⇒d2Vdb2=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 25:

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-r2⇒h-R=R2-r2Squaring both the sides, we geth2+R2-2hR=R2-r2⇒r2=2hR-h2                                                         ...1Now, V=13πr2h⇒V=π32h2R-h3                                       From eq. 1⇒dVdh=π34hR-3h2For maximum or minimum values of V, we must havedVdh=0⇒π34hR-3h2=0⇒4hR=3h2⇒h=4R3Now, d2Vdh2=π34R-6h⇒π34R-8R=0⇒-4πR3<0So, the volume is maximum when h=4R3.⇒h=4×123=16 cm

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Question 26:

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 =πr2h⇒2156=πr2h⇒2156=227r2h⇒h=2156×722r2⇒h=686r2                                             ...1Surface area=2πrh+2πr2⇒S=4312r+44r27                        From eq. 1⇒dSdr=4312-r2+88r7For maximum or minimum values of S, we must havedSdr=0⇒4312-r2+88r7=0⇒4312r2=88r7⇒r3=4312×788⇒r3=343⇒r=7 cmNow, d2sdr2=8624r3+887⇒d2sdr2=8624343+887⇒d2sdr2=1767>0So, the surface area is minimum when r=7 cm.

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Question 27:

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=R2⇒h2=4R2-r2⇒r2=R2-h24                                           ...1Now, V=πr2h⇒V=πhR2-h34                                 From eq. 1⇒dVdh=πR2-3h24For maximum or minimum values of V, we must havedVdh=0⇒πR2-3h24=0⇒R2-3h24=0⇒R2=3h24⇒h=2R3d2Vdh2=-3πh2d2Vdh2=-3π2×2R3⇒d2Vdh2=-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 28:

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=r2⇒y=r2-x2                                                ...1Now,Z=x+y⇒Z=x+r2-x2                                     From eq. 1⇒dZdx=1 +-2x2r2-x2For maximum or minimum values of Z, we must havedZdx=0⇒1 +-2x2r2-x2=0⇒2x=2r2-x2⇒x=r2-x2Squaring both the sides, we get⇒x2=r2-x2⇒2x2=r2⇒x=r2Substituting the value of x in eq. 1, we gety=r2-x2⇒y=r2-r22⇒y=r2d2zdx2=-r2-x2+x-xr2-x2r2-x2⇒d2zdx2=-r2+x2-x2r2-x232⇒d2zdx2=-r2r3×22⇒d2zdx2=-22r<0So, z=x+y is maximum when x=y=r2.Hence proved.

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Question 29:

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=4y⇒y=x24                                        ... 1Also,d2=x2+y-52              Using distance formulaNow,Z=d2=x2+y-52⇒Z=x2+x24-52              Using eq. 1⇒Z=x2+x416+25-5x22⇒dZdy=2x+4x316-5xFor maximum or minimum values of Z, we must havedZdy=0⇒2x+4x316-5x=0⇒4x316=3x⇒x3=12x⇒x2=12⇒x=±23Substituting the value of x in eq. 1, we gety=3Now,d2Zdy2=2+12x216-5⇒d2Zdy2=9-3=6>0So, the required nearest point is ±23, 3 .

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Question 30:

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⇒x=y24                                              ... 1d2=x-22+y+82                    Using distance formulaNow,Z=d2=x-22+y+82⇒Z=y24-22+y+82                   From eq. 1⇒Z=y416+4-y2+y2+64+16y⇒dZdy=4y316+16For maximum or minimum values of Z, we must havedZdy=0⇒4y316+16=0⇒4y316=-16⇒y3=-64⇒y=-4Substituting the value of x in eq. 1, we getx=4Now, d2Zdy2=12y216⇒d2Zdy2=12>0So, the nearest point is 4,-4. 

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Question 31:

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=8y⇒y=x28                                       ...1d2=x-22+y-42             Using distance formulaNow,Z=d2=x-22+y-42⇒Z=x-22+x28-42               From eq. 1⇒Z=x2+4-4x+x464+16-x2⇒dZdy=-4+4x364For maximum or minimum values of Z, we must havedZdy=0⇒-4+4x364=0⇒x316=4⇒x3=64⇒x=4Substituting the value of x in eq. 1, we gety=2Now,d2Zdy2=12x264⇒d2Zdy2=3>0So, the nearest point is 4, 2. 

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Question 32:

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

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Question 33:

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 +310⇒dSdt=-3+2x+710For maximum or minimum values of S, we must havedSdt=0⇒-3+2x+710=0⇒2x=-4⇒x=-2Now, d2Sdt2=210>0So, the nearest point is x, x2+7x+2 .⇒-2, 4-14+2⇒-2, -8

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Question 34:

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

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Question 35:

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+2⇒dMdx=-6x+6For maximum or minimum values of M, we must havedMdx=0⇒-6x+6=0⇒6x=6⇒x=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 

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Question 36:

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+25⇒P=50x-x22-x24-35x-25⇒dPdx=50-x-x2-35For maximum or minimum values of P, we must havedPdx=0⇒15-3x2=0⇒15=3x2⇒x=303⇒x=10Now, d2Pdx2=-32<0So, profit is maximum if daily output is 10 items. 

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Question 37:

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+x5⇒P=5x-x2100-500-x5⇒dPdx=5-x50-15For maximum or minimum values of P, we must havedPdx=0⇒5-x50-15=0⇒245=x50⇒x=24×505⇒x=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.

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Question 38:

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=V⇒h=Vl2                        ...1Surface area, S=l2+4lh⇒S=l2+4Vl                          From eq. 1⇒dSdl=2l-4Vl2For S to be maximum or minimum, we must havedSdl=0⇒2l-4Vl2=0⇒2l3-4V=0⇒2l3=4V⇒l3=2VNow, d2Sdl2=2+8Vl3⇒d2Sdl2=2+8V2V=6>0Here, surface area is minimum.h=Vl2Substituting the value of V=l32 in eq. 1, we geth=l32l2⇒h=l2Hence proved.

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Question 39:

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                     ...1⇒c=lbh⇒c=2b2h⇒h=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+lb⇒T=2Kb2+3K4bc2b2+2bc2b2+2b2                                       From eqs. 1 and 2              ⇒dTdb=4Kb+3K-3cb2+4bFor maximum or minimum values of T, we must havedTdb=0⇒4kb+3K-3cb2+4b=0⇒4b=33cb2-4b⇒4b=9cb2-12b⇒4b=9c-12b3b2⇒4b3=9c-12b3⇒16b3=9c⇒b=9c1613Now, d2Tdb2=4K+3K6cb3+4⇒d2Tdb2=4K+3K6c9c×16+4⇒K4+3×443⇒48K>0∴ Cost is minimum when b=9c1613.Substituting b = 9c1613in eq. 1 and eq. 2⇒l=29c1613h=c2b2⇒h=c29c1623⇒h=32c8113Thus, the most economic dimensions of the box are l=29c1613, b=9c1613 and h=32c8113.

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Question 40:

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, V=  43πr3+x3

⇒V =  4πr33+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. 2⇒4πr2=2πrS-4πr2612⇒4πr2=2πrx                                                                      From eq. 1          ⇒x=2r

Now,
d2Vdr2=8πr-2πS-4πr2612-2πr2S-4πr26-12-8πr6⇒d2Vdr2=8πr-2πS-4πr2612+43π2r26S-4πr212⇒d2Vdr2=8πr-2πx+43π2r21x=8πr-4πr+23π2r⇒d2Vdr2=4πr+23π2r>0

So, volume is minimum when x = 2r.

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Question 41:

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=12πlD22⇒V=πD2l8⇒l=8VπD2                                                                ...1Total surface area=πD24+lD+πDl2⇒S=πD24+8VπD+8V2D                                   From eq. 1⇒dSdD=πD2-8VπD2-8V2D2For maximum or minimum values of S, we must havedSdD=0⇒πD2-8VπD2-8V2D2=0⇒πD2=8VD21π+12⇒D3=16Vπ1π+12Now, d2SdD2=π2+16VD31π+12⇒d2SdD2=π2+π>0l=8VπD2⇒l=8πD2πD3162ππ+2⇒l=Dππ+2⇒lD=ππ+2Hence proved.

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Question 42:

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+b24⇒4a2-b2=h2                                          ...1Here,Strength of beam, S=Kbh2                Where K is some constant⇒S=kb4R2-b2                                 From eq. 1⇒S=kb4a2-b3⇒dSdb=k4a2-3b2For maximum or minimum values of S, we must havedSdb=0⇒k4a2-3b2=0⇒4a2-3b2=0⇒4a2=3b2⇒b=2a3Substituting the value of b in eq. 1, we get⇒4a2-2a32=h2⇒12a2-4a23=h2⇒h=223aNow, d2Sdb2=-6Kb⇒d2Sdb2=-6K2a3⇒d2Sdb2=-12Ka3<0So, the strength of beam is maximum when b=2a3 and h=223a.

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Question 43:

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-1⇒x=m-4mSubstituting x=0, we get y-4=m0-1⇒y=-m+4⇒x=-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-4⇒dSdm=4m2-1For maximum or minimum values of S, we must have dSdm=0⇒4m2-1=0⇒4m2=1⇒m2=4⇒m=±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

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Question 44:

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 =150⇒xy=150⇒y=150x                                                             ...1Area of the printed matter= x-3y-2⇒A=xy-2x-3y+6⇒A=150-2x-450x+6⇒dAdx=-2+450x2For maximum or minimum values of A, we must havedAdx=0⇒-2+450x2=0⇒2x2=450⇒x=15Substituting the value of x in 1, we gety=10Now,d2Adx2=-900x3⇒d2Adx2=-900153⇒d2Adx2=-9003375<0So, area of the printed matter is maximum when x=15 and y=10.

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Question 45:

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

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Question 46:

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+4t2-7⇒v=dsdt=t3-6t2+8t⇒a=dvdt=3t2-12t+8For maximum or minimum values of v, we must havedvdt=0⇒3t2-12t+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=0⇒6t-12=0⇒t=2Now,d2adt2=6>0So, acceleration is minimum at t=2.



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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+1x2⇒f'x=fx1x21-log x⇒f'x=x1x1x2-1x2log x                 ...1⇒f'x=x1x-21-log x       For a local maxima or a local minima, we must havef'x=0⇒x1x-21-log x=0⇒log x=1⇒x=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.



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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) ab≥c24Given: ax+bx≥cMinimum value of ax+bx=cNow,fx=ax+bx⇒f'x=a-bx2For a local maxima or a local minima, we must havef'x=0⇒a-bx2=0⇒ax2-b=0⇒ax2=b⇒x2=ba⇒x=±baf''x=2bx3⇒f''x=2bba3⇒f''x=2b a32b32>0So, x=ba is a local minima.∴ fba=aba+bba≥c=aaba+bbba≥c=ab+ab≥c⇒2ab≥c⇒c2≤ab⇒c24≤ab

Page No 17.80:

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 x⇒f'x=loge x-1loge x2For a local maxima or a local minima, we must have f'x=0⇒loge x-1loge x2=0⇒loge x-1=0⇒loge x=1⇒x=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

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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+1x⇒f'x=1-1x2For a local maxima or a local minima, we must have f'x=0⇒1-1x2=0⇒x2-1=0⇒x2=1⇒x=±1Now, f''x=2x3⇒f''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 17.80:

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+2⇒f'x=3x2+6x-9For a local maxima or a local minima, we must have f'x=0⇒3x2+6x-9=0⇒x2+2x-3=0⇒x+3x-1=0⇒x=-3, 1Now, f''x=6x+6⇒f''1=6+6=12>0So, x=1 is a local minima.Also, f''-3=-18+6=-12<0So, x=-3 is a local maxima.

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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+6⇒f'x=4x3-2x-2⇒f'x=x-14x2+4x+2For a local maxima or a local minima, we must have f'x=0⇒x-14x2+4x+2=0⇒x-1=0⇒x=1Now, f''x=12x2-2⇒f''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 17.80:

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=0⇒1-2x=0⇒2x=1⇒x=12Now, f''x=-2<0So, x=12 is a local maxima.Hence, the required number is 12.

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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-c2⇒f'x= 2x-a+2x-b+2x-cFor a local maxima or a local minima, we must have f'x=0⇒2x-a+2x-b+2x-c=0⇒2x-2a+2x-2b+2x-2c=0⇒6x=2a+b+c⇒x=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=8⇒y=8-x                      ... 1Now,fx=1x+1y⇒fx=1x+18-x                  From eq. 1⇒f'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=0⇒16x-64=0⇒x=4f''x=2x3-28-x3⇒f''4=243-28-43⇒f''4=264-264=0∴ Minimum value=14+14=12

Page No 17.80:

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-r2⇒fx=x-12+x-22+x-32+x-42+x-52⇒f'x=2x-1+x-2+x-3+x-4+x-5⇒f'x=25x-15For a local maxima and a local minima, we must havef'x=0⇒25x-15=0⇒5x-15=0⇒5x=15⇒x=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 3x⇒f'x=6 cos 3x-9 sin 3xFor a local minima or a local maxima, we must have f'x=0⇒6 cos 3x-9 sin 3x=0⇒6 cos 3x=9 sin 3x⇒sin 3xcos 3x=23⇒tan 3x=23                         ...1At x=5π6: tan 3x=tan 5π2⇒tan 3x=tan π2So, tan 3x is not defined.        tan 3x≠23 is not satisfying eq. 1Thus, x=5π6 is not a critical point.

Page No 17.80:

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+1⇒f'x=2x+1For a local maxima or a local minima, we must have f'x=0⇒2x+1=0⇒2x=-1⇒x=-12∉0, 1At extreme points: f0=0f1=1+1+1=3>0So, x=1 is a local minima.

Page No 17.80:

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+96x⇒f'x=3x2-36x+96For a local maxima or a local minima, we must have f'x=0⇒3x2-36x+96=0⇒x2-12x+32=0⇒x-4x-8=0⇒x=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.

Page No 17.80:

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+x2⇒f'x=4-x+x2-x-1+2x4-x+x22For a local maxima or a local minima, we must have f'x=0⇒4-x+x2-x-1+2x4-x+x22=0⇒4-x+x2-x-1+2x=0⇒4-x+x2+x-2x2=0⇒x2=4⇒x=±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.

Page No 17.80:

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=4x⇒x=y24                      ... 1Now, d=x-22+y-12Squaring both sides, we get⇒d2=x-22+y-12⇒d2=y24-22+y-12⇒d2=y416+4-y2+y2+1-2y                     From eq. 1Now,Z=d2=y416+4-y2+y2+1-2y⇒dZdy=y34-2y+2y-2⇒dZdy=y34-2⇒y34-2=0⇒y3=8⇒y=2Substituting the value of y in 1, we getx=1Now, d2Zdy2=3y24⇒d2Zdy2=3224=3>0So, the nearest point is 1, 2.



Page No 17.81:

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=8⇒y=8-x                           ... 1Let fx be xy.⇒ fx= x8-x             From eq. 1⇒f'x=8-2xFor a local maxima or a local minima, we must have f'x=0⇒8-2x=0⇒8=2x⇒x=4⇒y=8-4=4             From eq. 1Now, f''x=-2⇒f''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+9x⇒f'x=3x2-12x+9For a local maxima or a local minima, we must have f'x=0⇒3x2-12x+9=0⇒x2-4x+3=0⇒x-1x-3=0⇒x=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.

Page No 17.81:

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 x⇒f'x=cosx-3 sin xFor a local maxima or a local minima, we must have f'x=0⇒cos x-3 sin x=0⇒cos x=3 sin x⇒tan x=13⇒x=π6Now, f''x=-sin x-3 cos x⇒⇒f''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-r2⇒h-R=R2-r2Squaring both side, we geth2+R2-2hR=R2-r2⇒r2=2hr-h2                        ... 1Now, Volume =13πr2h⇒V=π32h2R-h3                 From eq. 1⇒dVdh=π34hR-3h2For maximum or minimum values of V, we must havedVdh=0⇒π34hR-3h2=0⇒4hR-3h2=0⇒4hR=3h2⇒h=4R3Now, d2Vdh2=π34R-6h=π34R-6×4R3=-4πR3<0So, volume is maximum when h=4R3.⇒h=22R3⇒h2R=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+250x⇒f'x=2x-250x2For a local maxima or a local minima, we must have f'x=0⇒2x-250x2=0⇒2x3-250=0⇒x3=125⇒x=5Now, f''x=2+500x3⇒f''5=2+50053=750125=6>0So, x=5 is a local minima.∴ f'xmin= 52+2505=3755=75

Page No 17.81:

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+1x⇒f'x=1-1x2For a local maxima or a local minima, we must have f'x=0⇒1-1x2=0⇒x2-1=0⇒x2=1⇒x=±1⇒x=1                                                Given: x>0Now, f''x=2x3⇒f''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+1⇒f'x=8x+2For a local maxima or a local minima, we must have f'x=0⇒8x+2=0⇒8x=-2⇒x=-14Now, f''x=8⇒f''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=1⇒y=1xfx= x+1x⇒f'x= 1-1x2For a local maxima or a local minima, we must have f'x=0⇒1-1x2=0⇒x2-1=0⇒x2=1⇒x=±1⇒x=1                                                  Given: x>1⇒y=1Now, f''x=2x3⇒f''1=2>0So, x=1 is a local minima.∴ Minimum value of fx=f1=1+1=2

Page No 17.81:

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 x⇒f'x=2 cos x-6 cos x sin x⇒f'x=2 cos x1-3 sin xFor a local maxima or a local minima, we must have f'x=0⇒2 cos x1-3 sin x=0⇒2 cos x=0  or 1-3 sin x=0⇒cos x=0 or sin x=13⇒x=π2 or x=sin-113Now, f''x=-2 sin x-6 cos 2x⇒f''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.

Page No 17.81:

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

Page No 17.81:

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+x2⇒f'x=4+x+x2-x1+2x4+x+x22For a local maxima or a local minima, we must have f'x=0⇒4+x+x2-x1+2x4+x+x22=0⇒4+x+x2-x1+2x=0⇒4-x2=0⇒x=±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.

Page No 17.81:

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

Page No 17.81:

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 x⇒f'x=loge x +1For a local maxima or a local minima, we must have f'x=0⇒loge x +1=0⇒loge x =-1⇒x=e-1Now, f''x=1x⇒f''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 17.81:

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-20⇒f'x=6x2-42x+36For a local maxima or a local minima, we must have f'x=0⇒6x2-42x+36=0⇒x2-7x+6=0⇒x-1x-6=0⇒x=1, 6Now, f''x=12x-42⇒f''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



Page No 17.82:

Question 30:

Let the function f : R → R be defined by f(x) = 2x + cos x, then f(x) 
(a) has a minimum at x = π                         (b) has a maximum at x = 0
(c) is a decreasing function                        (d) is an increasing function

Answer:


The given function is f(x) = 2x + cosx.

f(x) = 2x + cosx

Differentiating both sides with respect to x, we get

f'x=2-sinx

We know

−1 ≤ sinx ≤ 1

∴f'x=2-sinx>0 ∀ x ∈ R

⇒ f(x) is an increasing function for all x ∈ R

Thus, the function f : R → R defined by f(x) = 2x + cosx is an increasing function.

Hence, the correct answer is option (d).

Page No 17.82:

Question 31:

If x is real, the minimum value x2 - 8x + 17 is 
(a) -1          (b) 0              (c) 1              (d) 2

Answer:


Let f(x) = x2 − 8x + 17

Differentiating both sides with respect to x, we get

f'x=2x-8

For maxima or minima,

f'x=0

⇒2x-8=0

⇒x=4

Now,

f''x=2>0

So, x = 4 is the point of local minimum of f(x).

∴ Minimum value of f(x) = f(4) = (4)2 − 8 × 4 + 17 = 16 − 32 + 17 = 1

Thus, the minimum value of x2 − 8x + 17 is 1.

Hence, the correct answer is option (c).

Page No 17.82:

Question 32:

The maximum value of sin x cos x is 
(a) 14              (b) 12                (c) 2                    (d) 22

Answer:


Let fx=sinxcosx

⇒fx=12sin2x

Differentiating both sides with respect to x, we get

f'x=12cos2x×2=cos2x

For maxima or minima,

f'x=0

⇒cos2x=0

⇒2x=π2

⇒x=π4

Now,

f''x=-2sin2x

⇒f''π4=-2sin2×π4=-2sinπ2=-2<0

So, x=π4 is the point of local maximum.

∴ Maximum value of f(x)

 =fπ4

=12sin2×π4

=12sinπ2

=12×1

=12

Thus, the maximum value of sinx cosx is 12.

Hence, the correct answer is option (b).

Page No 17.82:

Question 33:

The maximum value of 1xx is
(a) e               (b) ee             (c) e1/e              (d) 1e1/e

Answer:


Let fx=1xx.

fx=1xx

⇒logfx=log1xx

⇒logfx=xlog1x

⇒logfx=-xlogx

Differentiating both sides with respect to x, we get

1fx×f'x=-x×1x+logx×1

⇒f'x=-1xx1+logx          .....(1)

For maxima or minima,

f'x=0

⇒-1xx1+logx=0

⇒1+logx=0        1xx>0

⇒logx=-1

⇒x=e-1=1e

Now,

f''x=-1xx×1x+1+logx×-1xx1+logx                          [Using (1)]

⇒f''1e=-e1e×e+1+log1e×-e1e1+log1e                    log1e=loge-1=-1

⇒f''1e=-e1e+1-0=-e1e+1

⇒f''1e<0

So, x=1e is the point of local maximum of f(x).

∴ Maximum value of f(x) = 11e1e=e1e

Thus, the maximum value of 1xx is e1e.

Hence, the correct answer is option (c).

Page No 17.82:

Question 34:

 The function f(x) = xx has a stationary point at
(a) x = e             (b) x = 1e             (c) x = 1                   (d) x = e

Answer:


The values of x for which f'x=0 are called stationary values.

Let fx=xx.

fx=xx

⇒logfx=logxx

⇒logfx=xlogx

Differentiating both sides with respect to x, we get

1fx×f'x=x×1x+logx×1

⇒f'x=xx1+logx   

For stationary point, we have

f'x=0

⇒xx1+logx=0

⇒1+logx=0        xx>0

⇒logx=-1

⇒x=e-1=1e

Thus, the function f(x) = xx has a stationary point at x=1e.

Hence, the correct answer is option (b).

Page No 17.82:

Question 35:

Maximum slope of the curve y = -x3 + 3x2 + 9x - 27 is 
(a) 0              (b) 12             (c) 16               (d) 32

Answer:


The given curve is y = −x3 + 3x2 + 9x − 27.

Slope of the curve, m = dydx

∴ m = dydx=-3x2+6x+9

⇒ dmdx=-6x+6

For maxima or minima,

dmdx=0

⇒-6x+6=0

⇒x=1

Now,

d2mdx2=-6 < 0

⇒ x  = 1 is the point of local maximum

So, the slope of the given curve is maximum when x = 1.

∴ Maximum value of the slope, m

=-3×12+6×1+9       (m = −3x2 + 6x + 9)

=-3+6+9

=12

Thus, the maximum slope of the curve y = −x3 + 3x2 + 9x − 27 is 12.

Hence, the correct answer is option (b).

Page No 17.82:

Question 36:

The function f(x) = 2x3 - 3x2 - 12x + 4, has
(a) two points of local maximum              (b) two points of local minimum
(c) one maximum and one minimum        (d) no maximum no minimum

Answer:


The given function is f(x) = 2x3 − 3x2 − 12x + 4.

f(x) = 2x3 − 3x2 − 12x + 4

Differentiating both sides with respect to x, we get

f'x=6x2-6x-12

⇒f'x=6x2-x-2

⇒f'x=6x+1x-2

For maxima or minima,

f'x=0

⇒ 6(x + 1)(x − 2) = 0

⇒ x + 1 = 0 or x − 2 = 0

⇒ x = −1 or x = 2

Now,

f''x=12x-6

At x = −1, we have

f''-1=12×-1-6=-12-6=-18 < 0

So, x = −1 is the point of local maximum.

At x = 2, we have

f''2=12×2-6=24-6=18 > 0

So, x = 2 is the point of local minimum.

Thus, the given function f(x) = 2x3 − 3x2 − 12x + 4 has one point of local maximum and one point of local minimum.

Hence, the correct answer is option (c).

Page No 17.82:

Question 1:

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

Answer:


The given function is fx=14x2+2x+1.

The function f(x) would attain its maximum value, when the value of 4x2 + 2x + 1 is minimum.

Let g(x) = 4x2 + 2x + 1

∴g'x=8x+2

For maxima or minima,

g'x=0

⇒8x+2=0

⇒x=-14

Now,

g''x=8>0

So, x=-14 is the point of local minimum of g(x)

Minimum value of function g(x)

=g-14

=4×-142+2×-14+1

=14-12+1

=34

∴ Maximum value of f(x) =1Minimum value of gx=134=43

Thus, the maximum value of the function fx=14x2+2x+1 is 43.


If f(x) = 14x2+2x+1, then its maximum value is      43     .

Page No 17.82:

Question 2:

The minimum value of f(x) = sin x in -π2,π2 is ____________________.

Answer:


The given function is fx=sinx, x∈-π2,π2.

fx=sinx

Differentiating both sides with respect to x, we get

f'x=cosx

For maxima or minima,

f'x=0

⇒cosx=0

⇒x=-π2 or x=π2, for all x∈-π2,π2

Now,

f''x=-sinx

At x=π2, we have

f''π2=-sinπ2=-1<0

So, x=π2 is the point of local maximum of f(x).

At x=-π2, we have

f''-π2=-sin-π2=sinπ2=1>0              [sin(−θ) = −sinθ]

So, x=-π2 is the point of local minimum of f(x).

∴ Minimum value of f(x) = f-π2=sin-π2=-sinπ2=-1

Thus, the minimum value of f(x) = sinx in -π2,π2 is −1.


The minimum value of f(x) = sin x in -π2,π2 is ___−1___.

Page No 17.82:

Question 3:

The maximum value of f(x) = sin x + cos x is _______________.

Answer:


The given function is fx=sinx+cosx.

fx=sinx+cosx

Differentiating both sides with respect to x, we get

f'x=cosx-sinx

For maxima or minima,

f'x=0

⇒cosx-sinx=0

⇒cosx=sinx

⇒tanx=1

⇒x=π4,5π4          (Let us only consider the values of x ∈ [0, 2π])

Now,

f''x=-sinx-cosx

At x=5π4, we have

f''5π4 =-sin5π4-cos5π4 =--12--12 =12+12 =22 =2 > 0

So, x=5π4 is the point of local minimum of f(x).

At x=π4, we have

f''π4 =-sinπ4-cosπ4 =-12-12 =-22 =-2 < 0

So, x=π4 is the point of local maximum of f(x).

∴ Maximum value of f(x) =fπ4 =sinπ4+cosπ4 =12+12 =22 =2

Thus, the maximum value of fx=sinx+cosx is 2.


The maximum value of f(x) = sinx + cosx is      2     .

Page No 17.82:

Question 4:

If f(x) has the second order derivative at x = c such that f '(c) = 0 and f ''(c) > 0, then c is a point of _______________.

Answer:


Second derivative test: Let f(x) be a function defined on an interval I and c ∈ I. Suppose f(x) be twice differentiable at x = c. Then, x = c is a point of local minima if f '(c) = 0 and f ''(c) > 0. In this case, f(c) is then the local minimum value of f(x).

So, if f(x) has the second order derivative at x = c such that f '(c) = 0 and f ''(c) > 0, then c is a point of local minima.


If f(x) has the second order derivative at x = c such that f '(c) = 0 and f ''(c) > 0, then c is a point of ___local minima___.

Page No 17.82:

Question 5:

If f '(x) changes its sign from positive to negative as x increases through c in the interval (c − h, c + h), then x = c is a point of ______________.

Answer:


First derivative test states that if f '(x) changes sign from positive to negative as x increases through c, then c is a point of local maxima, and f(c) is local maximum value.

Thus, if f '(x) changes its sign from positive to negative as x increases through c in the interval (c − h, c + h), then x = c is a point of local maximum.


If f '(x) changes its sign from positive to negative as x increases through c in the interval (c − h, c + h), then x = c is a point of ___local maximum___.



Page No 17.83:

Question 6:

If f '(x) changes its sign from negative to positive as x increases through c in the interval (c − h, c + h), then x = c is a point of ______________.

Answer:


First derivative test states that if f '(x) changes sign from negative to positive as x increases through c, then c is a point of local minima, and f(c) is local minimum value.

Thus, if f '(x) changes its sign from negative to positive as x increases through c in the interval (c − h, c + h), then x = c is a point of local minimum.


If f '(x) changes its sign from negative to positive as x increases through c in the interval (c − h, c + h), then x = c is a point of ___local minimum___.

Page No 17.83:

Question 7:

The positive real number x when added to its reciprocal gives the minimum value of the sum when, x = __________________.

Answer:


Let S(x) be the sum of the positive real number x and its reciprocal.

∴Sx=x+1x

Differentiating both sides with respect to x, we get

S'x=1-1x2

For maxima or minima,

S'x=0

⇒1-1x2=0

⇒x2=1

⇒ x = −1 or x = 1

Now,

S''x=2x3

At x = −1, we have

S''1=2-13=-2<0

So, x = −1 is the point of local maximum of S(x).

At x = 1, we have

S''1=213=2>0

So, x = 1 is the point of local minimum of S(x).

Therefore, S(x) is minimum when x =1.

Thus, the sum of positive real number x and its reciprocal is minimum when x = 1.


The positive real number x when added to its reciprocal gives the minimum value of the sum when, x = ___1___.

Page No 17.83:

Question 8:

The real number which must exceeds its cube is _______________.

Answer:


Let the real number be x.

The cube of the number is x3.

Let f(x) = x − x3

Differentiating both sides with respect to x, we get

f'x=1-3x2

For maxima or minima,

f'x=0

⇒1-3x2=0

⇒x2=13

⇒x=±13

Now,

f''x=-6x

At x=-13, we have

f''-13=-6×-13=23>0

So, x=-13 is the point of local minimum of f(x).

At x=13, we have

f''13=-6×13=-23<0

So, x=13 is the point of local maximum of f(x).

Thus, the real number which must exceeds its cube is x=13.


The real number which must exceeds its cube is      13     .

Page No 17.83:

Question 9:

The function f(x) = ax + bx , a, b, x > 0 takes on the least value at x equal to __________________.

Answer:


The given function is fx=ax+bx, a, b, x > 0.

fx=ax+bx

Differentiating both sides with respect to x, we get

f'x=a-bx2

For maxima or minima,

f'x=0

⇒a-bx2=0

⇒x2=ba

⇒ x=ba    (x > 0)

Now,

f''x=2bx3

At x=ba, we have

f''ba=2bba3=2aab>0

So, x=ba is the point of local minimum of f(x).

Thus, the function takes the least value at x=ba.


The function f(x) = ax + bx , a, b, x > 0 takes on the least value at x equal to       ba      .

Page No 17.83:

Question 10:

If y = a log x + bx2 + x has its extreme values at x = 1 and x = 2, then (a, b) = ____________________.

Answer:


It is given that, y=alogx+bx2+x has its extreme values at x = 1 and x = 2.

∴dydx=0 at x = 1 and x = 2

y=alogx+bx2+x

Differentiating both sides with respect to x, we get

dydx=ax+2bx+1

Now,

dydxx=1=0

⇒a+2b+1=0

⇒a+2b=-1        .....1

Also,

dydxx=2=0

⇒a2+4b+1=0

⇒a+8b=-2        .....2

Subtracting (1) from (2), we get

6b = −1

⇒b=-16

Putting b=-16 in (1), we get

a+2×-16=-1

⇒a=-1+13=-23

Thus, the values of a and b are -23 and -16, respectively.

Hence, the ordered pair (a, b) is -23,-16.


If y = alogx + bx2 + x has its extreme values at x = 1 and x = 2, then (a, b) =      -23,-16     .

Page No 17.83:

Question 11:

If the function f(x) = a sin x + 13sin 3x has an extremum at x = π3 then a = _________________.

Answer:


It is given that, the function fx=asinx+13sin3x has an extremum at x = π3.

∴f'x=0 at x = π3

fx=asinx+13sin3x

Differentiating both sides with respect to x, we get

f'x=acosx+13×3cos3x

⇒f'x=acosx+cos3x

Now,

f'π3=0

⇒acosπ3+cos3π3=0

⇒a×12+cosπ=0

⇒a2-1=0

⇒a=2

Thus, the value of a is 2.


If the function fx=asinx+13sin3x has an extremum at x = π3 then a = ___2___.

Page No 17.83:

Question 12:

The maximum value of f(x) = x e−x is _______________.

Answer:


The given function is f(x) = xe−x.

fx=xe-x

Differentiating both sides with respect to x, we get

f'x=x×e-x×-1+e-x×1

⇒f'x=e-x-x+1

For maxima or minima,

f'x=0

⇒e-x-x+1=0

⇒-x+1=0        e-x>0∀x∈R

⇒x=1

Now,

f''x=e-x×-1+-x+1×e-x×-1

⇒f''x=e-xx-2

At x = 1, we have

f''1=e-11-2=-1e<0

So, x = 1 is the point of local maximum of f(x).

∴ Maximum value of f(x) = f(1) = 1 × e−1 = 1e

Thus, the maximum value of f(x) = xe−x is 1e.


The maximum value of f(x) = xe−x is      1e     .

Page No 17.83:

Question 13:

If the function f(x) = x4 - 62x2 + ax + 9 attains a local maximum at x = 1, then a = _________________.

Answer:


It is given that, the function f(x) = x4 − 62x2 + ax + 9 attains a local maximum at x = 1.

∴f'x=0 at x = 1

f(x) = x4 − 62x2 + ax + 9

Differentiating both sides with respect to x, we get

f'x=4x3-124x+a

Now,

f'1=0    

⇒4×13-124×1+a=0

⇒a=124-4=120

Thus, the value of a is 120.

Also,

f''x=12x2-124

At x = 1, we have

f''1=12×12-124=12-124=-112<0

So, x = 1 is the point of local maximum of f(x).


If the function f(x) = x4 − 62x2 + ax + 9 attains a local maximum at x = 1, then a = ___120____.

Page No 17.83:

Question 14:

If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is _______________.

Answer:


Let the two numbers be x and 4 − x (x ≠ 0, 4).

Suppose S be the sum of their reciprocals.

∴S=1x+14-x, x≠0, 4

Differentiating both sides with respect to x, we get

dSdx=-1x2-14-x2×-1

⇒dSdx=-1x2+14-x2

For maxima or minima,

dSdx=0

⇒-1x2+14-x2=0

⇒-16-8x+x2+x2=0

⇒8x=16

⇒x=2

Now, 

d2Sdx2=2x3+24-x2

At x = 2, we have

d2Sdx2x=2= 223+24-22= 14+12= 34>0

So, x = 2 is the point of local minimum.

Thus, S is minimum when x = 2.

∴ Minimum value of S = 12+14-2 = 12+12 = 1          S=1x+14-x 

Thus, if the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is 1.


If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is ___1___.

Page No 17.83:

Question 15:

If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ________________.

Answer:


It is given that, x and y are two real numbers such that x > 0 and xy = 1.

Let S = x + y.

Now, xy=1⇒y=1x

∴S=x+y=x+1x      

Differentiating both sides with respect to x, we get

dSdx=1-1x2

For maxima or minima,

dSdx=0

⇒1-1x2=0

⇒x2=1

⇒ x = 1    (x > 0)

Now,

d2Sdx2=2x3

At x = 1, we have

d2Sdx2x=1=213=2>0

So, x = 1 is the point of local minimum.

Thus, S is minimum when x = 1.

When x = 1, y=1x=1

∴ Minimum value of S = x + y = 1 + 1 = 2           

Thus, the minimum value of x + y is 2.


If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ___2___.

Page No 17.83:

Question 16:

The number that exceeds its square by the greatest amount is _______________. 

Answer:


Let the number be x.

The square of the number is x2.

Let f(x) = x − x2. Now, we need to find the value of x for which f(x) is maximum.

f(x) = x − x2

Differentiating both sides with respect to x, we get

f'x=1-2x

For maxima or minima,

f'x=0

⇒1-2x=0

⇒x=12

Now,

f''x=-2<0

So, x=12 is the point of local maximum of f(x). Therefore, f(x) is maximum when x=12.

Thus, the number that exceeds its square by the greatest amount is x=12.


The number that exceeds its square by the greatest amount is      12     .

Page No 17.83:

Question 17:

If m and M respectively denote the minimum and maximum values of f(x) = (x + 1)2 + 3 in the interval [−3, 1], then the ordered pair (m, M) = _________.

Answer:


The given function is f(x) = (x + 1)2 + 3, x ∈ [−3, 1].

f(x) = (x + 1)2 + 3

Differentiating both sides with respect to x, we get

f'x=2x+1

For maxima or minima,

f'x=0

⇒2x+1=0

⇒x+1=0

⇒x=-1

Now, 

f''x=2>0

So, x = −1 is the point of local minimum of f(x).

At x = −1, we have

f(−1) = (−1 + 1)2 + 3 = 0 + 3 = 3

At x = −3, we have

f(−3) = (−3 + 1)2 + 3 = 4 + 3 = 7

At x = 1, we have

f(1) = (1 + 1)2 + 3 = 4 + 3 = 7

Thus, the minimum value of f(x) is 3 and the maximum value of f(x) is 7.

∴ m = 3 and M = 7

Thus, the ordered pair (m, M) is (3, 7).


If m and M respectively denote the minimum and maximum values of f(x) = (x + 1)2 + 3 in the interval [−3, 1], then the ordered pair (m, M) = ___(3, 7)___.

Page No 17.83:

Question 18:

The minimum value of f(x) = x2 + 250x is _______________.

Answer:


The given function is fx=x2+250x, x ≠ 0.

fx=x2+250x

Differentiating both sides with respect to x, we get

f'x=2x-250x2

For maxima or minima,

f'x=0

⇒2x-250x2=0

⇒x3=2502=125

⇒x=5

Now,

f''x=2+500x3

At x = 5, we have

f''5=2+50053=2+500125=2 + 4 = 6 > 0

So, x = 5 is the point of local minimum.

∴ Minimum value of f(x) = f5=52+2505 = 25 + 50 = 75                  fx=x2+250x

Thus, the minimum value of the given function is 75.


The minimum value of fx=x2+250x is ____75____.

Page No 17.83:

Question 19:

The maximum slope of the curve y = −x3 + 3x2 + 9x − 27 is _________________.

Answer:


The given curve is y = −x3 + 3x2 + 9x − 27.

Slope of the curve, m = dydx

∴ m = dydx=-3x2+6x+9

Differentiating both sides with respect to x, we get

dmdx=-6x+6

For maxima or minima,

dmdx=0

⇒-6x+6=0

⇒x=1

Now,

d2mdx2=-6<0

So, x = 1 is the point of local maximum.

Thus, the slope of the given curve is maximum when x = 1.

∴ Maximum value of the slope

=-3×12+6×1+9           (Slope, m = -3x2+6x+9)

= −3 + 6 + 9

= 12

Hence, the maximum slope of the curve y = −x3 + 3x2 + 9x − 27 is 12.


The maximum slope of the curve y = −x3 + 3x2 + 9x − 27 is ___12___.

Page No 17.83:

Question 20:

The function f(x) = x2+2x has a local minimum at x = __________________.

Answer:


The given function is fx=x2+2x, x ≠ 0.

fx=x2+2x

Differentiating both sides with respect to x, we get

f'x=12-2x2

For maxima or minima,

f'x=0

⇒12-2x2=0

⇒x2=4

⇒ x = −2 or x = 2

Now,

f''x=4x3

At x = −2, we have

f''-2=4-23=-12<0

So, x = −2 is the point of local maximum of f(x).

At x = 2, we have

f''2=423=12>0

So, x = 2 is the point of local minimum of f(x).

Thus, the given function f(x) = x2+2x has a local minimum at x = 2.


The function f(x) = x2+2x has a local minimum at x = ____2____.

Page No 17.83:

Question 21:

The least value of the function f(x) = ax + bx(a > 0, b > 0, x > 0) is ________________.

Answer:


The given function is fx=ax+bx (a > 0, b > 0, x > 0).

fx=ax+bx

Differentiating both sides with respect to x, we get

f'x=a-bx2

For maxima or minima,

f'x=0

⇒a-bx2=0

⇒x2=ba

⇒x=ba        (x > 0)

Now,

f''x=2bx3

At x=ba, we have

f''ba=2bba3=2aab>0           (a > 0, b > 0)

So, x=ba is the point of local minimum of f(x). Thus, the function takes the least value at x=ba.

∴ Least value of the given function

=fba

=a×ba+bba              fx=ax+bx

=ab+ab

=2ab

Thus, the least value of the function fx=ax+bx (a > 0, b > 0, x > 0) is 2ab.


The least value of the function fx=ax+bx (a > 0, b > 0, x > 0) is       2ab      .

Page No 17.83:

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 17.83:

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 17.83:

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 17.84:

Question 4:

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

Answer:

Given: fx = x+1x⇒f'x = 1-1x2For a local maxima or a local minima, we must have f'x=0⇒1-1x2=0⇒x2=1⇒x=1, -1But x>0⇒x=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 17.84:

Question 5:

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

Answer:

Given:  fx = x+1x⇒f'x = 1-1x2For a local maxima or a local minima, we must have f'x=0⇒1-1x2=0⇒x2=1⇒x=1, -1But x<0⇒x=-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 17.84:

Question 6:

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

Answer:

Given: fx = x loge x⇒f'x = loge x+1For a local maxima or a local minima, we must havef'x=0⇒loge x+1=0⇒loge x=-1⇒x=1e⇒f1e = 1e loge 1e=-1eNow,f''x = 1xAt x=1e:f''1e =11e =e>0So, 1e, -1e is a point of local minimum..

Page No 17.84:

Question 7:

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

Answer:

We have,fx = ax+bx⇒f'x =a-bx2For a local maxima or a local minima, we must havef'x=0⇒a-bx2=0⇒x2=ba⇒x=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 17.84:

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+1⇒f'x=fx log x+1⇒f'x=xx log x+1           ....1For a local maxima or a local minima, we must have f'x=0⇒xx log x+1=0⇒log x=-1⇒x=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 17.84:

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+1x2⇒f'x=fx1x21-log x⇒f'x=x1x1x2-1x2log x                 ...1⇒f'x=x1x-21-log x       For a local maxima or a local minima, we must havef'x=0⇒x1x-21-log x=0⇒log x=1⇒x=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 17.84:

Question 10:

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

Answer:

Given: fx =log xx⇒f'x = 1-log xx2For a local maxima or a local minima, we must have f'x=0⇒1-log xx2=0⇒1-log x=0⇒log x=1⇒log x=log e⇒x=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.



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