Chapter 6-Application Of Derivatives Exercise 6.5

NCERT solutions for class 12 Maths chapter 6-Application Of Derivatives Exercise 6.5

NCERT Solutions for Class 12 Maths Chapter-6 Applications of Derivatives

NCERT solutions for class 12 maths Chapter-6 Applications of Derivatives is prepared by academic team of Physics Wallah. We have prepared NCERT Solutions for all exercise of chapter-6. Given below is step by step solutions of all questions given in NCERT textbook for Chapter-6.

NCERT Solutions for Class 12 Maths Exercise 6.5

Solve The Following Questions.

Question1. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x − 1)2 + 3

(ii) f(x) = 9×2 + 12x + 2

(iii) f(x) = −(x − 1)2 + 10

(iv) g(x) =x3 + 1

Solution :
(i) The given function is f(x) = (2x − 1)2 + 3.

It can be observed that (2x− 1)2 ≥ 0 for every x∈R.

Therefore, f(x) = (2x − 1)2 + 3 ≥ 3 for every x∈R.

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

2x − 1 = 0⇒ x = 1/2

∴Minimum value of f == 3

Hence, functionf does not have a maximum value.

(ii) The given function is f(x) = 9×2 + 12x + 2 = (3x + 2)2 − 2.

It can be observed that (3x + 2)2 ≥ 0 for every x∈R.

Therefore, f(x) = (3x + 2)2 − 2 ≥ −2 for every x∈R.

The minimum value of f is attained when 3x + 2 = 0.

3x + 2 = 0⇒ x = -2/3

∴Minimum value off =

Hence, functionf does not have a maximum value.

(iii) The given function is f(x) = − (x − 1)2 + 10.

It can be observed that (x − 1)2 ≥ 0 for every x∈R.

Therefore,f(x) = − (x − 1)2 + 10 ≤ 10 for every x∈R.

The maximum value off is attained when (x − 1) = 0.

(x − 1) = 0⇒x = 1

∴Maximum value off = f(1) = − (1 − 1)2 + 10 = 10

Hence, functionf does not have a minimum value.

(iv) The given function isg(x) =x3 + 1.

Hence, functiong neither has a maximum value nor a minimum value.

Question2. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = |x + 2| − 1

(ii) g(x) = − |x + 1| + 3

(iii) h(x) = sin(2x) + 5

(iv) f(x) = |sin 4x + 3|

(v) h(x) =x+ 1,x ∈ (−1, 1)

Solution :

.

Question3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i).f(x) =x2

(ii).g(x) =x3 − 3x

(iii) h(x) = sinx + cosx, 0 < x < π/2

(iv) f(x) = sinx − cosx, 0 <x < 2π

(v) f(x) =x3 − 6×2+ 9x + 15

(vi)

(vii)

(viii)

Solution :

Question4. Prove that the following functions do not have maxima or minima:

(i) f(x) =ex

(ii) g(x) = logx

(iii) h(x) =x3 +x2 + x + 1

Solution :
We have,

f(x) = ex

∴ f'(x) = ex

Now, if f'(x) = 0. But, the exponential function can never assume 0 for any value ofx.

Therefore, there does not existc∈R such that f'(c) = 0

Hence, functionf does not have maxima or minima.

We have,

g(x) = logx

∴ g'(x) = 1/x

Since log xnis defined for a positive number x, g'(x) > 0 for any x

Therefore, there does not existc∈R such that g'(c) = 0

Hence, functiong does not have maxima or minima.

We have,

h(x) =x3 +x2 +x + 1

h'(x) =3×2 +2x + 1

Now,

h(x) = 0 ⇒ 3×2 + 2x + 1 = 0⇒

Therefore, there does not exist c∈R such that h'(c) = 0

Hence, functionh does not have maxima or minima.

Question5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i)

(ii)

(iii)

(iv)

Solution :
(i) The given function isf(x) = x3.

f'(x) = 3×2

Now, f'(x) = 0 ⇒ x = 0

Then, we evaluate the value off at critical point x = 0 and at end points of the interval [−2, 2].

f(0) = 0

f(−2) = (−2)3 = −8

f(2) = (2)3 = 8

Hence, we can conclude that the absolute maximum value off on [−2, 2] is 8 occurring atx = 2. Also, the absolute minimum value off on [−2, 2] is −8 occurring at x = −2.

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

Then, we evaluate the value off at critical point π/4 and at the end points of the interval [0, π].

Hence, we can conclude that the absolute maximum value off on [0, π] is√2 occurring atx = π4 and the absolute minimum value off on [0, π] is −1 occurring atx = π.

Then, we evaluate the value off at critical pointx = 4 and at the end points of the interval.

Hence, we can conclude that the absolute maximum value off onis 8 occurring atx = 4 and the absolute minimum value off onis −10 occurring atx = −2.

Hence, we can conclude that the absolute maximum value off on [−3, 1] is 19 occurring atx = −3 and the minimum value off on [−3, 1] is 3 occurring atx = 1.

Question6. Find the maximum profit that a company can make, if the profit function is given by

p(x) = 41 − 72x − 18×2

Solution :
The profit function is given as p (x) = 41 − 72x− 18×2.

∴ p'(x)=−72−36x

⇒ x=−7236 =−2

Also, p”(−2)=−36 <0 By second derivative test, x=−2

is the point of local maxima of p.

∴ Maximum profit=p(−2)

=41−72(−2)−18(−2)2

=41+144−72=113

Hence, the maximum profit that the company can make is 113 units. The solution given in the book has some error. The solution is created according to the question given in the book.

Question7. Find both the maximum value and the minimum value of 3×4 − 8×3 + 12×2 − 48x + 25 on the interval [0, 3]

Solution :
Let f(x) = 3×4 − 8×3 + 12×2 − 48x + 25.

Now,f'(x) = 0 gives x = 2 or x2+ 2 = 0 for which there are no real roots.

Therefore, we consider only x = 2∈[0, 3].

Now, we evaluate the value off at critical pointx = 2 and at the end points of the interval [0, 3].

Hence, we can conclude that the absolute maximum value offon [0, 3] is 25 occurring atx= 0 and the absolute minimum value off at [0, 3] is − 39 occurring atx = 2.

Question8. At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

Solution :
Letf(x) = sin 2x.

Question9. What is the maximum value of the function sinx + cosx?

Solution :
Let f(x) = sinx + cosx.

Now,f”(x) will be negative when (sinx + cosx) is positive i.e., when sin xand cosxare both positive. Also, we know that sinx and cosx both are positive in the first quadrant. Then,f”(x) will be negative when x ∊ (0, π/2).

Thus, we consider x = π/4

∴By second derivative test,f will be the maximum at x = π/4 and the maximum value of f is.

Question10. Find the maximum value of 2×3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].

Solution :
Let f(x) = 2×3 − 24x + 107.

We first consider the interval [1, 3].

Then, we evaluate the value off at the critical pointx = 2∈ [1, 3] and at the end points of the interval [1, 3].

f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75

f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85

f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89

Hence, the absolute maximum value off(x) in the interval [1, 3] is 89 occurring atx = 3.

Next, we consider the interval [−3, −1].

Evaluate the value off at the critical pointx = −2∈ [−3, −1] and at the end points of the interval [1, 3].

f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125

f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129

f(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139

Hence, the absolute maximum value off(x) in the interval [−3, −1] is 139 occurring atx = −2.

Question11. It is given that atx = 1, the function x4− 62×2 +ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

Solution :
Let f(x) =x4 − 62×2 +ax + 9.

Hence, the value ofa is 120.

Question12. Find the maximum and minimum values ofx + sin 2x on [0, 2π].

Solution :
Let f(x) =x + sin 2x.

Hence, we can conclude that the absolute maximum value off(x) in the interval [0, 2π] is 2π occurring atx = 2π and the absolute minimum value off(x) in the interval [0, 2π] is 0 occurring atx = 0.

Question13. Find two numbers whose sum is 24 and whose product is as large as possible.

Solution :
Let one number be x. Then, the other number is (24 −x).

Let P(x) denote the product of the two numbers. Thus, we have:

∴By second derivative test,x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

Question14. Find two positive numbers x andy such that x +y = 60 and xy3 is maximum.

Solution :
The two numbers arex andy such that x +y = 60.

⇒y = 60 −x

Let f(x) =xy3.

∴By second derivative test,x= 15 is a point of local maxima off. Thus, functionxy3 is maximum whenx = 15 andy = 60 − 15 = 45.

Hence, the required numbers are 15 and 45.

Question15. Find two positive numbersxandysuch that their sum is 35 and the product x2y5 is a maximum

Solution :
Let one number bex. Then, the other number is y = (35 −x).

Let P(x) =x2y5. Then, we have:

∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 = 25.

Hence, the required numbers are 10 and 25.

Question16. Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution :
Let one number be x. Then, the other number is (16 −x).

Let the sum of the cubes of these numbers be denoted by S(x). Then,

∴ By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8.

Question17. A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Solution :
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 the height of the box is x cm.

Therefore, the volume V(x) of the box is given by,

∴ By second derivative test,x = 3 is the point of maxima of V.

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 obtained is the largest possible.

Question18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner 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 the maximum possible?

Solution :
Let the side of the square to be cut off be x cm. Then, the height of the box isx, the length is 45 − 2x, and the breadth is 24 − 2x.

Therefore, the volumeV(x) of the box is given by,

∴By second derivative test,x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

Question19. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution :
Let a rectangle of lengthl and breadthb be inscribed in the given circle of radiusa.

Then, the diagonal passes through the centre and is of length 2a cm.

∴By the second derivative test, when l = √2a , then the area of the rectangle is the maximum.

Since l = b = √2a, the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Question20. Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

Solution :
Letr andh be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by,

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

Question21. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Solution :
Letr andh be the radius and height of the cylinder respectively.

Then, volume (V) of the cylinder is given by,

Question22. 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 length of the two pieces so that the combined area of the square and the circle is minimum?

Solution :
Let a piece of lengthl be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length (28 −l) m.

Now, side of square = l/4

Letr be the radius of the circle. Then, 2πr = 28 – l ⇒ r = 1/2π(28 – l)

The combined areas of the square and the circle (A) is given by,

Question23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Solution :
Letr andh be the radius and height of the cone respectively inscribed in a sphere of radius R.

Question24. Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.

Solution :
Let randh be the radius and the height (altitude) of the cone respectively.

Then, the volume (V) of the cone is given as: V =13πr2h ⇒ h =3Vπr2

The surface area (S)of the cone is given by,

S = πrl (wherel is the slant height)

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to√2 times the radius of the base.

Question25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan-1 √2

Solution :
Letθ be the semi-vertical angle of the cone.

It is clear that

Let r,h, and l be the radius, height, and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now,r =l sinθ and h =l cosθ

The volume (V) of the cone is given by,

V=1/3πr2h =1/3π (l2sin2θ)(lcosθ)

=1/3πl3 sinθ cosθ

⇒dV/dθ =l3π/3 [sin2θ (−sinθ) + cosθ(2sinθ cosθ)]

=l3π / 3[−sin3θ+2sinθcos2θ]

⇒d2V / dθ2 =l3π / 3 [−3sin2θ cosθ + 2cos3θ− 4sin2θ cosθ ]

=l3π/3 [2cos3θ − 7sin2θ cosθ]

∴By second derivative test, the volume (V) is the maximum when θ = tan-1 √2

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan-1 √2.

Question26. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin-1 (1/3)

Solution :
Let r be the radius, l be the slant height and h be the height of the cone of given surface area,S.

Also, let α be the semi-vertical angle of the cone.

Question27. The point on the curve x2 = 2y which is nearest to the point (0, 5) is

(A) (2√2 , 4)

(B) (2√2 , 0)

(C) (0, 0)

(D) (2, 2)

Solution :
Equation of the curve is x2 = 2y

For each value of x, the position of the point will be(x, x2/2)

Let P (x, x2/2) be any point on the curve (i), then according to question,

Distance between given point (0, 5) and

Therefore, option (A) is correct.

Question28. For all real values ofx, the minimum value ofis

(A) 0

(B) 1

(C) 3

(D) 1/3

Solution :
Let f =

∴By second derivative test,f is the minimum atx= 1 and the minimum value is given by f(1) = 1-1+1/1+1+1 = 1/3

The correct answer is D.

Question29. The maximum value ofis

(A)(1/3)1/3

(B)1/2

(C) 1

(D) 0

Solution :

Then, we evaluate the value off at critical point x = 1/2 and at the end points of the interval [0, 1] {i.e., atx = 0 andx = 1}.

Hence, we can conclude that the maximum value off in the interval [0, 1] is 1.

The correct answer is C.

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