Page No 18.16:
Question 1:
f(x) = (x5)4
Answer:
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
Page No 18.16:
Question 2:
f(x) = x3 3x
Answer:
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
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
.
Page No 18.16:
Question 3:
f(x) = x3 (x1)2
Answer:
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
Since f '(x) changes from positive to negative when x increases through , x = is the point of local maxima.
The local minimum value of f (x) at x = is given by
Since f '(x) does not change from positive as x increases through 0, x = 0 is a point of inflexion.
Disclaimer: The solution in the book is incorrect. The solution here is created according to the question given in the book.
Page No 18.16:
Question 4:
f(x) = (x1) (x+2)2
Answer:
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
Since f '(x) changes sign from positive to negative when x increases through , x = is the point of local maxima.
The local maximum value of f (x) at x = is given by
Page No 18.16:
Answer:
Now, for values close to x = 0 and to the left of 0, .
Also, for values close to x = 0 and to the right of 0, .
Therefore, by first derivative test,
x = 0 is a point of local maxima and the local maximum value of
Page No 18.16:
Question 6:
f(x) = x3 6x2 + 9x + 15
Answer:
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
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
Page No 18.16:
Question 7:
f(x) = sin 2x, 0<x<
Answer:
Since f '(x) changes from positive to negative when x increases through , x = is the point of maxima.
The local maximum value of f (x) at x = is given by
Since f '(x) changes from negative to positive when x increases through , x = is the point of minima.
The local minimum value of f (x) at x = is given by
Page No 18.16:
Question 8:
f(x) = sin x cos x, 0 < x<2
Answer:
Since f '(x) changes from positive to negative when x increases through , x = is the point of local maxima.
The local maximum value of f (x) at x = is given by
Since f '(x) changes from negative to positive when x increases through , x = is the point of local minima.
The local minimum value of f (x) at x = is given by
Page No 18.16:
Question 9:
f(x) = cos x, 0<x<
Answer:
.
Since , none is in the interval .
Page No 18.16:
Question 10:
f(x) = sin 2xx,
Answer:
.
Since f '(x) changes from positive to negative when x increases through , x = is the point of local maxima.
The local maximum value of f (x) at x = is given by
Since f '(x) changes from negative to positive when x increases through , x = is the point of local minima.
The local minimum value of f (x) at x = is given by
Page No 18.16:
Question 11:
f(x) = 2sin xx,
Answer:
Since f '(x) changes from positive to negative when x increases through , x = is the point of local maxima.
The local maximum value of f (x) at x = is given by
Since f '(x) changes from negative to positive when x increases through , x = is the point of local minima.
The local minimum value of f (x) at x = is given by
Page No 18.16:
Question 12:
f(x) =
Answer:
Since, f '(x) changes from positive to negative when x increases through , x = is a point of maxima.
The local maximum value of f (x) at x = is given by
Page No 18.16:
Question 13:
f(x) = x3 (2x1)3
Answer:
Since f '(x) changes from negative to positive when x increases through , x = is a point of local minima.
The local minimum value of f (x) at x = is given by
Page No 18.16:
Question 14:
f(x) =
Answer:
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
Page No 18.31:
Question 1:
(i) f(x) = x4 62x2 + 120x + 9
(ii) f(x) = x3 6x2 + 9x + 15
(iii) f(x) = (x1) (x+2)2
(iv) f(x) = 2/x2/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,
(viii) f(x) =
(ix) f(x) =
(x) f(x) = x ≠ 0
(xi) f(x) =
(xii) f(x) =
Answer:
Page No 18.31:
Question 2:
(i) f(x) = (x1) (x2)2
(ii) f(x) =
(iii) f(x) =
Answer:
Page No 18.31:
Question 3:
The function y = a log x+bx2 + x has extreme values at x=1 and x=2. Find a and b
Answer:
Page No 18.31:
Question 4:
Show that has a maximum value at x = e.
Answer:
Page No 18.31:
Question 5:
Find the maximum and minimum values of the function f(x) =
Answer:
Page No 18.31:
Question 6:
Find the maximum and minimum values of y = tan .
Answer:
Page No 18.31:
Question 7:
If f(x) = x3 + ax2 + bx + c has a maximum at x = 1 and minimum at x = 3. Determine a, b and c.
Answer:
Page No 18.31:
Question 8:
Prove that f(x) = sinx + cosx has maximum value at x = . [NCERT EXEMPLAR]
Answer:
Page No 18.37:
Question 1:
(i) f(x) = 4x in [2,4,5]
(ii) f(x) = (x1)2 + 3 in [3,1]
(iii) f(x) = 3x4 8x3 + 12x2 48x + 25 in [0,3]
(iv) f(x) = (x2)
Answer:
Page No 18.37:
Question 2:
Find the maximum value of 2x3 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [3, 1].
Answer:
Page No 18.37:
Question 3:
Find the absolute maximum and minimum values of the function of given by
Answer:
Page No 18.37:
Question 4:
Find the absolute maximum and minimum values of a function f given by
Answer:
Page No 18.37:
Question 5:
Find the absolute maximum and minimum values of a function f given by
Answer:
Page No 18.7:
Question 1:
f()=4-4+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 =
Thus, the minimum value of f (x) at x = is 3.
Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, function f does not have a maximum value.
Page No 18.7:
Question 2:
f(x)=(x-1)2+2 on R
Answer:
Given: f(x) = − (x − 1)2 + 2
Now,
(x − 1)2 0 for all x R
f(x) = − (x − 1)2 + 2 2 for all x R
The maximum value of f(x) is attained when (x − 1) = 0.
(x − 1) = 0
⇒ x = 1
Therefore, the maximum value of f (x) = 2
Since f(x) can be reduced, the minimum value does not exist, which is evident in the graph also.
Hence, function f does not have a minimum value.
Page No 18.7:
Question 3:
f(x)=| x+2 | on R
Answer:
Given: f(x) =
Now,
for all x R
Thus, f(x) 0 for all x R
Therefore, the minimum value of f at x = 2 is 0.
Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, the function f does not have a maximum value.
Page No 18.7:
Question 4:
f(x)=sin 2x+5 on R
Answer:
Given: f(x) = sin 2x + 5
We know that − 1 ≤ sin 2x ≤ 1.
⇒ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
⇒ 4 ≤ f(x) ≤ 6
Hence, the maximum and minimum values of f are 6 and 4, respectively.
Page No 18.7:
Question 5:
f(x) = | sin 4x+3 | on R
Answer:
Given: f(x) =
We know that −1 sin 4x 1.
⇒ 2 sin 4x + 3 4
⇒ 2 4
⇒ 2 f(x) 4
Hence, the maximum and minimum values of f are 4 and 2, respectively.
Page No 18.7:
Question 6:
f(x)=2x3 +5 on R
Answer:
We can observe that f(x) increases when the values of x are increased and f(x) decreases when the values of x are decreased. Also, f(x) can be reduced by giving small values of x.
Similarly, f(x) can be enlarged by giving large values of x.
So, f(x) does not have a minimum or maximum value.
Page No 18.7:
Question 7:
f(x) =| x + 1 | + 3 on R
Answer:
Given: f(x) = + 3
Now,
for all x R
f(x) = + 3 3 for all x R
f(x) 3 for all x R
The maximum value of f is attained when
Therefore, the maximum value of f at x = -1 is 3.
Since f(x) can be reduced, the minimum value does not exist, which is evident in the graph also.
Hence, the function f does not have a minimum value.
Page No 18.7:
Question 8:
f(x) = 16x2 16x + 28 on R
Answer:
Given: f(x) = 16x2 − 16x + 28
f(x) = 4(4x2 - 4x + 1) + 24
f(x) = 4(2x − 1)2 + 24
Now,
4(2x − 1)2 0 for all x R
f(x) = 4(2x − 1)2 + 24 24 for all x R
f(x) 24 for all x R
The minimum value of f is attained when (2x − 1) = 0.
(2x − 1) = 0
⇒ x =
Therefore, the minimum value of f at x = is 24.
Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, the function f does not have a maximum value.
Page No 18.7:
Question 9:
f(x) = x3 1 on R
Answer:
We can observe that f(x) increases when the values of x increase and f(x) decreases when the values of x decrease. Also, f(x) can be reduced by giving smaller values of x.
Similarly, f(x) can be enlarged by giving larger values of x.
So, f(x) does not have a minimum or maximum value.
Page No 18.72:
Question 1:
Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum.
Answer:
Page No 18.72:
Question 2:
Divide 64 into two parts such that the sum of the cubes of two parts is minimum.
Answer:
.
Page No 18.72:
Question 3:
How should we choose two numbers, each greater than or equal to 2, whose sum______________ so that the sum of the first and the cube of the second is minimum?
Answer:
Page No 18.72:
Question 4:
Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.
Answer:
Page No 18.72:
Question 5:
Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?
Answer:
Page No 18.72:
Question 6:
A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
(i)
(ii)
Find the point at which M is maximum in each case.
Answer:
Page No 18.72:
Question 7:
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?
Answer:
Page No 18.72:
Question 8:
A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum?
Answer:
Page No 18.72:
Question 9:
Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.
Answer:
Page No 18.73:
Question 10:
Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long. [CBSE 2000]
Answer:
Page No 18.73:
Question 11:
Two sides of a triangle have lengths 'a' and 'b' and the angle between them is . What value of will maximize the area of the triangle? Find the maximum area of the triangle also. [CBSE 2002 C]
Answer:
Page No 18.73:
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
x = 9 or x = 3
If x = 9, then length and breadth will become 0.
x ≠ 9
x = 3
Now,
x = 3 is the point of maxima.
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.
Page No 18.73:
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:
Page No 18.73:
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
Area of the base = lb = 4 m2
Area of the 4 walls, A= 2h (l + b)
However, the length cannot be negative.
Thus,
l = 2 m
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.
Page No 18.73:
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:
Page No 18.73:
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:
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.
Page No 18.73:
Question 17:
Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is
Answer:
Page No 18.73:
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:
Page No 18.73:
Question 19:
Prove that a conical tent of given capacity will require the least amount of canavas when the height is times the radius of the base.
Answer:
Page No 18.73:
Question 20:
Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to of the diameter of the sphere.
Answer:
Page No 18.73:
Question 21:
Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is . [CBSE 2014]
Answer:
Let:
Radius of the base = r,
Height = h,
Slant height = l,
Volume = V,
Curved surface area = C
Page No 18.73:
Question 22:
An isosceles triangle of vertical angle 2 is inscribed in a circle of radius a. Show that the area of the triangle is maximum when = .
[NCERT EXEMPLAR]
Answer:
Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.
Page No 18.73:
Question 23:
Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is r.
Answer:
Page No 18.73:
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:
Page No 18.73:
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:
Page No 18.73:
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:
Page No 18.73:
Question 27:
Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius
Answer:
Page No 18.74:
Question 28:
Show that among all positive numbers x and y with x2 + y2 =r2, the sum x+y is largest when x=y=r.
Answer:
Page No 18.74:
Question 29:
Determine the points on the curve x2 = 4y which are nearest to the point (0,5).
Answer:
Page No 18.74:
Question 30:
Find the point on the curve y2=4x which is nearest to the point (2, 8).
Answer:
Page No 18.74:
Question 31:
Find the point on the curve x2=8y which is nearest to the point (2,4).
Answer:
Page No 18.74:
Question 32:
Find the point on the parabolas x2 = 2y which is closest to the point (0,5).
Answer:
Page No 18.74:
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:
Page No 18.74:
Question 34:
Find the point on the curvey y2=2x which is at a minimum distance from the point (1,4).
Answer:
Page No 18.74:
Question 35:
Find the maximum slope of the curve y=
Answer:
Page No 18.74:
Question 36:
The total cost of producing x radio sets per day is Rs and the price per set at which they may be sold is Rs. Find the daily output to maximum the total profit.
Answer:
Page No 18.74:
Question 37:
Manufacturer can sell x items at a price of rupees each. The cost price is Rs Find the number of items he should sell to earn maximum profit.
Answer:
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.
Page No 18.74:
Question 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:
Page No 18.74:
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:
Page No 18.74:
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,
...(1)
Sum of volumes, V=
V = [From eq. (1)]
For the minimum or maximum values of V, we must have
...(2)
Now,
So, volume is minimum when x = 2r.
Page No 18.74:
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 .
Answer:
Page No 18.74:
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:
Page No 18.74:
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:
Page No 18.74:
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:
Page No 18.74:
Question 45:
The space s described in time t by a particle moving in a straight line is given by S = Find the minimum value of acceleration.
Answer:
Page No 18.74:
Question 46:
A particle is moving in a straight line such that its distance at any time t is given by S = Find when its velocity is maximum and acceleration minimum.
Answer:
Page No 18.79:
Question 1:
Write necessary condition for a point x = c to be an extreme point of the function f(x).
Answer:
We know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f '(x) = 0 at x = c
f '(c) = 0
Page No 18.80:
Question 2:
If for all positive x where a,b,>0, then
(a)
(b)
(c)
Answer:
Page No 18.80:
Question 2:
Write sufficient conditions for a point x=c to be a point of local maximum.
Answer:
We know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f '(x) = 0 at x = c
f '(c) = 0
Also, at the point of local maximum, the second order derivative of the function at the given point must be less than zero, i.e.
f ''(c) < 0
Page No 18.80:
Question 3:
The minimum value of is
(a) e
(b) 1/e
(c) 1
(d) none of these
Answer:
Page No 18.80:
Question 3:
If f(x) attains a local minimum at x=c, then write the values of f' (c) and f'' (c).
Answer:
If f(x) attains a local minimum at x = c, then the first order derivative of the function at the given point must be equal to zero, i.e.
f '(x) = 0 at x = c
f '(c) = 0
The second order derivative of the function at the given point must be greater than zero, i.e.
f ''(c) > 0
Page No 18.80:
Question 4:
For the function f(x) =
(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:
Page No 18.80:
Question 4:
Write the minimum value of f(x) =
Answer:
Page No 18.80:
Question 5:
Let f(x) = x3+3x29x+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:
Page No 18.80:
Question 5:
Write the maximum value of f(x) =
Answer:
Page No 18.80:
Question 6:
The minimum value of f(x) = is
(a) 6
(b) 4
(c) 8
(d) none of these
Answer:
Page No 18.80:
Question 6:
Write the point where f(x) = x log, x attains minimum value.
Answer:
.
Page No 18.80:
Question 7:
The number which exceeds its square by the greatest possible quantity is
(a)
(b)
(c)
(d) none of these
Answer:
Page No 18.80:
Question 7:
Find the least value of f(x) = , where a>0, b>0 and x>0.
Answer:
Page No 18.80:
Question 8:
Let f(x) = (xa)2 + (xb)2 + (xc)2. Then, f(x) has a minimum at x =
(a)
(b)
(c)
(d) none of these
Answer:
Page No 18.80:
Question 8:
Write the minimum value of f(x) = xx .
Answer:
Page No 18.80:
Question 9:
Write the maximum value of f(x) = x1/x.
Answer:
Page No 18.80:
Question 10:
Write the maximum value of f(x) = , if it exists.
Answer:
.
Page No 18.80:
Question 1:
The maximum value of x1/x, x>0 is
(a) e1/e
(b)
(c) 1
(d) none of these
Answer:
(a)
Disclaimer: The answer given in the book is incorrect. The solution provided here is according to the question.
Page No 18.81:
Question 9:
The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is
(a)
(b)
(c)
(d) none of these
Answer:
Page No 18.81:
Question 10:
The function f(x) = (xr)2 assumes minimum value at x =
(a) 5
(b)
(c) 3
(d) 2
Answer:
Page No 18.81:
Question 11:
At x= , f(x) = 2 sin 3x + 3 cos 3x is
(a) 0
(b) maximum
(c) minimum
(d) none of these
Answer:
(d) none of these
Page No 18.81:
Question 12:
If x lies in the interval [0,1], then the least value of x2 + x + 1 is
(a) 3
(b)
(c) 1
(d) none of these
Answer:
Page No 18.81:
Question 13:
The least value of the function f(x) = in the interval [0,9] is
(a) 126
(b) 135
(c) 160
(d) 0
Answer:
Page No 18.81:
Question 14:
The maximum value of f(x) = on [1, 1] is
(a)
(b)
(c)
(d)
Answer:
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 18.81:
Question 15:
The point on the curve y2 = 4x which is nearest to, the point (2,1) is
(a)
(b) (1,2)
(c) (1,2)
(d) (2,1)
Answer:
Page No 18.81:
Question 16:
If x+y=8, then the maximum value of xy is
(a) 8
(b) 16
(c) 20
(d) 24
Answer:
Page No 18.81:
Question 17:
The least and greatest values of f(x) = x36x2+9x in [0,6], are
(a) 3,4
(b) 0,6
(c) 0,3
(d) 3,6
Answer:
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 18.81:
Question 18:
f(x) = is maximum when x =
(a)
(b)
(c)
(d) 0
Answer:
Page No 18.81:
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)
(b)
(c)
(d)
Answer:
(d)
Page No 18.81:
Question 20:
The minimum value of is
(a) 75
(b) 50
(c) 25
(d) 55
Answer:
(a) 75
Page No 18.81:
Question 21:
If(x) = x+, x > 0, then its greatest value is
(a) 2
(b) 0
(c) 3
(d) none of these
Answer:
(d) none of these
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Question 22:
If(x) = , then its maximum value is
(a)
(b)
(c) 1
(d)
Answer:
(a)
Page No 18.81:
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)
(d)
Answer:
(b) 2
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Question 24:
f(x) = 1+2 sin x+3 cos2x, is
(a) Minimum at x =
(b) Maximum at x = sin ()
(c) Minimum at x =
(d) Maximum at sin()
Answer:
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Question 25:
The function f(x) = is maximum at x =
(a) 3
(b) 0
(c) 4
(d) 2
Answer:
(d) 2
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Question 26:
The maximum value of f(x) = on [1,1] is
(a)
(b)
(c)
(d)
Answer:
(c)
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Question 27:
Let f(x) = 2x3 3x2 12x + 5 on [2, 4]. The relative maximum occurs at x=
(a) 2
(b) 1
(c) 2
(d) 4
Answer:
(c) 2
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Question 28:
The minimum value of x loge x is equal to
(a) e
(b) 1/e
(c) 1/e
(d) 2/e
(e) e
Answer:
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Question 29:
The minimum value of the function f(x) = 2x321x2+36x20 is
(a) 128
(b) 126
(c) 120
(d) none of these
Answer:
(a)128
View NCERT Solutions for all chapters of Class 15