# NCERT Solutions for Class 12 Maths Chapter 6

NCERT solutions for class 12 Maths chapter 6 Applications of Derivatives exercise 6.5, 6.4, 6.3, 6.2, 6.1 (rate of change, increasing decreasing, approximation, tangent normal and maxima minima) in PDF format for new academic session 2020-21.

12th NCERT solutions of other subjects, NCERT books, revisions books, assignments, chapter tests based on applications of derivatives class xii, Previous Year’s Board Papers questions are in PDF format. UP Board Intermediate students 2020-2021 can take help from these solutions. Download UP Board Solution for Class 12 Math chapter 6 in Hindi Medium or English Medium or in Videos format.

## NCERT solutions for class 12 Maths chapter 6

 Class: 12 Subject: Maths Chapter 6: Applications of Derivatives

### 12th Maths Chapter 6 Solutions

NCERT solutions for class 12 Maths chapter 6 AOD is given below to free download in PDF form. Download NCERT Books 2020-21 and Offline Apps based on latest CBSE Syllabus. Join the discussion forum to ask your doubts of NIOS and CBSE Board with our experts and other users.

• ### Chapter 6 Solutions in PDF

#### Class 12 Maths Chapter 6 Solutions in Videos

Class 12 Maths Chapter 6 Exercise 6.1 Solution
Class 12 Maths Chapter 6 Exercise 6.2 Solution
Class 12 Maths Chapter 6 Exercise 6.3 Solution
Class 12 Maths Chapter 6 Exercise 6.4 Solution
Class 12 Maths Chapter 6 Exercise 6.5 Solution
Class 12 Maths Chapter 6 Miscellaneous Exercise Solution

#### Class 12 Maths Chapter 6 Solution in Hindi Medium Videos

Class 12 Maths Exercise 6.1 Solution in Hindi
Class 12 Maths Exercise 6.2 Solution in Hindi
Class 12 Maths Exercise 6.3 Solution in Hindi
Class 12 Maths Exercise 6.4 Solution in Hindi
Class 12 Maths Exercise 6.5 Solution in Hindi
Class 12 Maths Miscellaneous Exercise Solution

#### Previous Years CBSE Questions

1. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which its area increases, when side is 10 cm long. [CBSE Sample Paper 2017]
2. The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm. [Delhi 2017]
3. The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing when the side of the triangle is 20 cm? [Delhi 2015]
4. Determine for what values of x, the function f(x) = x^3 + 1/x^3, where x ≠ 0, is strictly increasing or strictly decreasing. [CBSE Sample Paper 2017]
5. Show that the function f(x) = 4x^3 – 18x^2 + 27x – 7 is always increasing on R. [Delhi 2017]
6. Find the interval in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing. [Delhi 2016] 7. Find the point on the curve y = x^3 – 11x + 5 at which the tangent is y = x – 11. [CBSE Sample Paper 2017]

##### Questions from Board Papers

1. Find the equation of tangents to the curve y = cos(x + y), where x lies in [- 2π, 2π], that are parallel to the line x + 2y = 0. [Foreign 2016]
2. Find the shortest distance between the line x – y + 1 = 0 and the curve y^2 = x. [CBSE Sample Paper 2017]
3. If the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is π/3. [Delhi 2017]
4. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3. Also find maximum volume in terms of volume of the sphere. [Delhi 2016]
5. The sum of the surface areas of a cuboid with sides x, 2x and x/3 and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes. [Foreign 2016]
6. A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is 3 m and volume is 75 cubic meter. If building of tank costs ₹ 100 per square metre for the base and ₹ 50 per square meters for the sides, find the cost of least expensive tank. [Delhi 2015C]
7. A point on the hypotenuse of a right triangle is at distances ‘a’ and ‘b’ from the sides of the triangle. Show that the minimum length of the hypotenuse is (a^2/3 + b^2/3)^3/2. [Delhi 2015C]
8. Find the local maxima and local minima, of the function f(x) = sin x – cos x, 0 < x < 2π. Also find the local maximum and local minimum values. [Delhi 2015]

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### Important Questions on 12th Maths Chapter 6

Find the rate of change of the area of a circle with respect to its radius r when r = 3 cm.
Area of circle with radius r is given by A=πr^2
Therefore, the rate of change of A with respect to r = dA/dr = 2πr
When r=3 cm,
we have dA/dr=6π
Hence, the area of circle is changing at the rate of 6π cm^2/cm.
सिद्ध कीजिए R पर f(x)=3x+17 से प्रदत्त फलन निरंतर वर्धमान है।
f(x)=3x+17
⇒f^’ (x) = 3 > 0, सभी x∈R के लिए
इसलिए, फलन R पर निरंतर वर्धमान है।
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Let the radius of circle =r cm
Area of circle with radius r is given by A=πr^2
Therefore, the rate of change of A with respect to t
= dA/dt = 2πr.dr/dt
= 2πr.3
= 6πr
When r=10 cm, then dA/dt=6π.(10)=60π cm^2/s
Hence, the area of circle is increasing at the rate of 60π cm^2/s.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Let the radius of circle =r cm
Area of circle with radius r is given by A=πr^2
Therefore, the rate of change of A with respect to t
= dA/dt = 2πr.dr/dt
= 2πr.5
= 10πr
When r=8 cm, then dA/dt=10π.(8)=80π cm^2/s
Hence, the enclosed area is increasing at the rate of 80π cm^2/s.
सिद्ध कीजिए कि लघुगणकीय फलन (0,∞) में निरंतर वर्धमान फलन है।
f(x)=log⁡x
⇒f'(x) = 1/x > 0, सभी x∈(0,∞) के लिए
इसलिए, फलन (0,∞) में निरंतर वर्धमान है।
सिद्ध कीजिए कि (-1,1) में f(x) = x^2-x+1 से प्रदत्त फलन न तो वर्धमान है और न ही ह्रासमान है।
f(x)=x^2-x+1 ⇒f'(x)=2x-1
यदि f^’ (x)=0
⇒ 2x-1=0
⇒x=1/2
x=1/2 अंतराल (-1,1) को अंतरालों (-1,1/2 ) और (1/2,1) में विभक्त करता है।
f^’ (x)=2x-1<0, सभी x∈(-1,1/2 ) के लिए, इसलिए, फलन (-1,1/2 ) में निरंतर ह्रासमान है। f^' (x)=1/x>0, सभी x∈(1/2,1) के लिए,
इसलिए, फलन (1/2,1) में निरंतर वर्धमान है।
इसप्रकार, (-1,1) में फलन न तो वर्धमान है और न ही ह्रासमान है।
Find the slope of the tangent to the curve y = 3x^4 – 4x at x = 4.
The slope of tangent to the curve y=3x^4-4x is
dy/dx=12x^3-4
If x=4,
then slope =├ dy/dx]=12(4)^3-4
= 768-4
= 764
Find points at which the tangent to the curve y = x^3 – 3x^2 – 9x + 7 is parallel to the x-axis.
The slope of tangent to the curve y=x^3-3x^2-9x+7 is dy/dx=3x^2-6x-9
If the tangent is parallel to x – axis, then slope dy/dx=0
⇒3x^2-6x-9=0
⇒3(x^2-2x-3)=0
⇒3(x-3)(x+1)=0
⇒x=-1,3
If x=-1, then y=(-1)^3-〖3(-1)〗^2-9(-1)+7=-1-3+9+7=12, so, the points = (-1,12)
If x=3, then y=(3)^3-〖3(3)〗^2-9(3)+7=27-27-27+7=-20, so, the point = (3,-20)
Hence, on the points (-1,12) and (3,-20), the tangents area parallel to x – axis.
f(2.01) का सन्निकट मान ज्ञात कीजिए जहाँ f(x)=〖4x〗^2+5x+2 है।
मान लीजिए x=2 और ∆x=0.01 है।
f(2.01)= f(x+∆x)=〖4(x+∆x)〗^2+5(x+∆x)+2
यहाँ, ∆y=f(x+∆x)-f(x) है।
इसलिए, f(x+∆x)=f(x)+∆y ≈ f(x)+f'(x)∆x [क्योंकि dx=∆x]
≈(〖4x〗^2+5x+2)+(8x+5)∆x
⇒f(2.01)= [〖4(2)〗^2+5(2)+2]+[8(2)+5](0.01) [क्योंकि x=2,∆x=0.01]
= [16+10+2]+[16+5](0.01)
= 28+0.21
=28.31
अतः, f(2.01) का सन्निकट मान 28.31 है।
x m भुजा वाले घन की भुजा में 1% वृद्धि के कारण घन के आयतन में होने वाले सनन्निकट परिवर्तन ज्ञात कीजिए।
यहाँ, x m भुजा वाले घन का आयतन V=x^3 है।
इसलिए, dV=(dV/dx)∆x=(3x^2)∆x
=(3x^2 )(0.01x)
[क्योंकि x का 1% = 0.01x]
= 0.03x^3 m^3
अतः, आयतन में होने वाला सनन्निकट परिवर्तन 0.03x^3 m^3 है।
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.
The slope of tangent to the curve y=x^3-11x+5 is dy/dx=3x^2-11
Slope of the line y=x-11 is 1.
Given that:
The slope of tangent to the curve y=x^3-11x+5 = the slope of the line y=x-11
⇒2x^2-11=1
⇒3x^2=12
⇒x^2=4
⇒x=±2
If x=2, then y=(2)^3-11(2)+5=-9, therefore, the point = (2,-9)
If x=-2, then y=(-2)^3-11(-2)+5=18, therefore, the point = (-2,19)
Out of the points (2,-9) and (-2,19), only (2,-9) satisfy the equation of line y=x-11.
Hence, (2,-9) is the point on y=x^3-11x+5 at which tangent is y=x-11.
Prove that f(x)=e^x do not have maxima or minima.
We have,
f(x)=e^x
〖∴f〗^’ (x)=e^x
Now if f^’ (x)=0,then e^x=0
But, the exponential function can never assume 0 for any value of x.
Therefore, there does not exist c ∈ R such that f^’ (c)=0
Hence, function f does not have maxima or minima.
Find both the maximum value and the minimum value of 〖3x〗^4-8x^3+12x^2-48x+25 on the interval [0, 3].
Let f(x)=3x^4-8x^3+12x^2-48x+25
f^’ (x)=12x^3-24x^2+24x-48 =12(x^3-2x^2+2x-4) =12[x^2 (x-2)+2(x-2)]
=12(x-2)(x^2+2)
Now, f^’ (x)=0 gives x =2 or x^2+2=0 for which there are real roots
Therefore, we consider only x=2∈[0,3]
Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3].
f(2)=3〖(2)〗^4-8〖(2)〗^3+12〖(2)〗^2-48(2)+25
=3(16)-8(8)+12(4)-48(2)+25=48-64+48-96+25 =-39
f(0)=3〖(0)〗^4-8〖(0)〗^3+12〖(0)〗^2-48(0)+25=3(0)-8(0)+12(0)-48(0)+25=25
f(3)=3〖(3)〗^4-8〖(3)〗^3+12〖(3)〗^2-48(3)+25 F(3)=3(81)-8(27)+12(9)-48(3)+25
=243-216+108-144+25 =16
Hence, we can conclude that the absolute value of f on [0, 3] is 25 occurring at x=0 and the absolute minimum value of f at [0, 3] is – 39 occurring at x = 2.
यदि दिया है कि अंतराल [0,2] में x=1 पर फलन x^4-〖62x〗^2+ax+9 उच्चतम मान प्राप्त करता है तो a का मान ज्ञात कीजिए।
यहाँ, f(x)=x^4-〖62x〗^2+ax+9 है इसलिए, f'(x)=4x^3-124x+a है।
दिया है कि अंतराल [0,2] में x=1 पर फलन f उच्चतम मान प्राप्त करता है।
इसलिए,
f'(1)=0
⇒ 4(1)^3-124(1)+a = 0
⇒ a = 120,
अतः, a का मान 120 है।
ऐसी दो धन संख्याएँ x और y ज्ञात कीजिए ताकि x+y=60 और xy^3 उच्चतम हो।
प्रश्नानुसार, x + y = 60 ⇒y=60-x … (1)
माना P=xy^3
⇒P=x(60-x)^3 [समीकरण (1) से मान रखने पर]
इसलिए, P'(x)=-3x(60-x)^2 +(60-x)^3=(60-x)^2 (-3x+60-x)=(60-x)^2 (60-4x) है।
अब, P'(x)=0⇒(60-x)^2 (60-4x)=0 ⇒x=15
[x≠60, क्योंकि यदि x=60 तो y=60-60=0 होगा। परन्तु दिया गया है कि y एक धन संख्या है।]
अब, P”(x)=-4(60-x)^2-2(60-4x)(60-x)=-2(60-x)(180-6x)
x=15 के लिए, P”(15)=-2(60-15)(180-90)=-8100<0 यहाँ, P''(15)<0 है, इसलिए, द्वितीय अवकलज परिक्षण द्वारा, x=15 स्थानीय उच्चतम बिंदु है। समीकरण (1) से, y=60-15=45 अतः, संख्याओं x और y के मान क्रमशः 15 और 45 हैं।
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
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,
⇒S(x)=x^3+(16-x)^3
∴〖 S〗^’ (x)=3x^2-3(16-x)^2, S^” (x)=6x+6(16-x)
Now, S^’ (x)=0
⇒ 3x^2-3(16-x)^2 = 0
⇒ x^2-(16-x)^2=0
⇒ x^2-256-x^2+32x = 0
⇒ x = 256/32 = 8
Now S^” (8)=6(8)+6(16-8)=48+48=96>0
∴ 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.