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NCERT Solutions for Exercise 6.4 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT Solutions for Exercise 6.4 Class 12 Maths Chapter 10 - Application of Derivatives

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:46 PM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 01 Jan' 2025 - 14 Feb' 2025

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4

NCERT Solutions for Exercise 6.4 Class 12 Maths Chapter 6 Application of Derivatives are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 6.4 Class 12 Maths chapter 6 discuss problems related to the approximation of certain quantities using differentiation. Prior to exercise 6.4 Class 12 Maths, there are three exercises in which the concepts of rate of change of quantities, increasing and decreasing function and tangents and normals are discussed. Till the NCERT solutions for Class 12 Maths chapter 6 exercise 6.4 NCERT Book presents 25 solved examples. These NCERT syllabus solved examples give an insight into the topics covered in the chapter. The Class 12 Maths chapter 6 exercise 6.4 gives a detailed explanation of numerical related to the approximation of quantities. As mentioned, other than Class 12th Maths chapter 6 exercise 6.4 there are 5 exercises including the miscellaneous exercise.

12th class Maths exercise 6.4 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Application of Derivatives Class 12 Chapter 6 Exercise 6.4

Question:1(i) Using differentials, find the approximate value of each of the following up to 3
places of decimal. \sqrt {25.3 }

Answer:

Lets suppose y = \sqrt x and let x = 25 and \Delta x = 0.3
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{25+0.3} - \sqrt 25
\Delta y = \sqrt{25.3} - 5
\sqrt{25.3} = \Delta y +5
Now, we can say that \Delta y is approximate equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.3) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.3)\\ dy = \frac{1}{2\sqrt 25}.(0.3)\\ dy = \frac{1}{10}.(0.3)\\ dy = 0.03
Now,
\sqrt{25.3} = \Delta y +5\\ \sqrt {25.3} = 0.03 + 5\\ \sqrt{25.3} = 5.03
Hence, \sqrt{25.3} is approximately equals to 5.03

Question:1(ii) Using differentials, find the approximate value of each of the following up to 3 places of decimal.

\sqrt { 49.5 }

Answer:

Lets suppose y = \sqrt x and let x = 49 and \Delta x = 0.5
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{49+0.5} - \sqrt 49
\Delta y = \sqrt{49.5} - 7
\sqrt{49.5} = \Delta y +7
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.5)\\ dy = \frac{1}{2\sqrt 49}.(0.5)\\ dy = \frac{1}{14}.(0.5)\\ dy = 0.035
Now,
\sqrt{49.5} = \Delta y +7\\ \sqrt {49.5} = 0.035 + 7\\ \sqrt{49.5} = 7.035
Hence, \sqrt{49.5} is approximately equal to 7.035

Question:1(iii) Using differentials, find the approximate value of each of the following up to 3 places of decimal.

\sqrt {0.6}

Answer:

Lets suppose y = \sqrt x and let x = 1 and \Delta x = -0.4
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{1+(-0.4)} - \sqrt 1
\Delta y = \sqrt{0.6} - 1
\sqrt{0.6} = \Delta y +1
Now, we cam say that \Delta y is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(-0.4) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = -0.4)\\ dy = \frac{1}{2\sqrt 1}.(-0.4)\\ dy = \frac{1}{2}.(-0.4)\\ dy = -0.2
Now,
\sqrt{0.6} = \Delta y +1\\ \sqrt {0.6} = (-0.2) + 1\\ \sqrt{0.6} = 0.8
Hence, \sqrt{0.6} is approximately equal to 0.8

Question:1(iv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 0.009 ) ^{1/3 }

Answer:

Lets suppose y = (x)^{\frac{1}{3}} and let x = 0.008 and \Delta x = 0.001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({0.008+ 0.001})^{\frac{1}{3}} - (0.008)^{\frac{1}{3}}
\Delta y = ({0.009})^{\frac{1}{3}} - 0.2
({0.009})^{\frac{1}{3}} = \Delta y + 0.2
Now, we cam say that \Delta y is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = 0.001)\\ dy = \frac{1}{3(0.008)^{\frac{2}{3}}}.(0.001)\\ dy = \frac{1}{0.12}.(0.001)\\ dy = 0.008
Now,
(0.009)^{\frac{1}{3}} = \Delta y +0.2\\ (0.009)^{\frac{1}{3}} = (0.008) + 0.2\\ (0.009)^{\frac{1}{3}} = 0.208
Hence, (0.009)^{\frac{1}{3}} is approximately equal to 0.208

Question:1(v) Using differentials, find the approximate value of each of the following up to 3
places of decimal.

( 0.999) ^{1/10 }

Answer:

Lets suppose y = (x)^{\frac{1}{10}} and let x = 1 and \Delta x = -0.001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{10}} - (x)^{\frac{1}{10}}
\Delta y = ({1 - 0.001})^{\frac{1}{10}} - (1)^{\frac{1}{10}}
\Delta y = ({0.999})^{\frac{1}{10}} - 1
({0.999})^{\frac{1}{10}} = \Delta y + 1
Now, we cam say that \Delta y is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{10 (x)^{\frac{9}{10}}}.(-0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{10}} \ and \ \Delta x = -0.001)\\ dy = \frac{1}{10(1)^{\frac{9}{10}}}.(-0.001)\\ dy = \frac{1}{10}.(-0.001)\\ dy = -0.0001
Now,
(0.999)^{\frac{1}{10}} = \Delta y +1\\ (0.999)^{\frac{1}{10}} = (-0.0001) + 1\\ (0.999)^{\frac{1}{10}} = 0.9999 = 0.999 \ upto \ three\ decimal \ place
Hence, (0.999)^{\frac{1}{10}} is approximately equal to 0.999 (because we need to answer up to three decimal place)

Question:1(vi) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(15 )^{1/4}

Answer:

Let's suppose y = (x)^{\frac{1}{4}} and let x = 16 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({16 - 1})^{\frac{1}{4}} - (16)^{\frac{1}{4}}
\Delta y = ({15})^{\frac{1}{4}} - 2
({15})^{\frac{1}{4}} = \Delta y + 2
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(16)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 8}.(-1)\\dy = \frac{1}{32}.(-1) \\dy = -0.031
Now,
(15)^{\frac{1}{4}} = \Delta y +2\\ (15)^{\frac{1}{4}} = (-0.031) + 2\\ (15)^{\frac{1}{4}} = 1.969
Hence, (15)^{\frac{1}{4}} is approximately equal to 1.969

Question:1(vii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(26)^{1/3 }

Answer:

Lets suppose y = (x)^{\frac{1}{3}} and let x = 27 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({27 - 1})^{\frac{1}{3}} - (27)^{\frac{1}{3}}
\Delta y = ({26})^{\frac{1}{3}} - 3
({26})^{\frac{1}{3}} = \Delta y + 3
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -1)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-1)\\ dy = \frac{1}{3\times 9}.(-1)\\dy = \frac{1}{27}.(-1) \\dy = -0.037
Now,
(27)^{\frac{1}{3}} = \Delta y +3\\ (27)^{\frac{1}{3}} = (-0.037) + 3\\ (27)^{\frac{1}{3}} = 2.963
Hence, (27)^{\frac{1}{3}} is approximately equal to 2.963

Question:1(viii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 255) ^{1/4}

Answer:

Let's suppose y = (x)^{\frac{1}{4}} and let x = 256 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({256 - 1})^{\frac{1}{4}} - (256)^{\frac{1}{4}}
\Delta y = ({255})^{\frac{1}{4}} - 4
({255})^{\frac{1}{4}} = \Delta y + 4
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(256)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 64}.(-1)\\dy = \frac{1}{256}.(-1) \\dy = -0.003
Now,
(255)^{\frac{1}{4}} = \Delta y +4\\ (255)^{\frac{1}{4}} = (-0.003) + 4\\ (255)^{\frac{1}{4}} = 3.997
Hence, (255)^{\frac{1}{4}} is approximately equal to 3.997

Question:1(ix) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 82) ^{1/4 }

Answer:

Let's suppose y = (x)^{\frac{1}{4}} and let x = 81 and \Delta x = 1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({81 + 1})^{\frac{1}{4}} - (81)^{\frac{1}{4}}
\Delta y = ({82})^{\frac{1}{4}} - 3
({82})^{\frac{1}{4}} = \Delta y + 3
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 1)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(1)\\ dy = \frac{1}{4\times 27}.(1)\\dy = \frac{1}{108}.(1) \\dy = .009
Now,
(82)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.009) + 3\\ (82)^{\frac{1}{4}} = 3.009
Hence, (82)^{\frac{1}{4}} is approximately equal to 3.009

Question:1(x) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 401 ) ^{1/2 }

Answer:

Let's suppose y = (x)^{\frac{1}{2}} and let x = 400 and \Delta x = 1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}
\Delta y = ({400 + 1})^{\frac{1}{2}} - (400)^{\frac{1}{2}}
\Delta y = ({401})^{\frac{1}{2}} - 20
({401})^{\frac{1}{2}} = \Delta y + 20
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 1)\\ dy = \frac{1}{2(400)^{\frac{1}{2}}}.(1)\\ dy = \frac{1}{2\times 20}.(1)\\dy = \frac{1}{40}.(1) \\dy = 0.025
Now,
(401)^{\frac{1}{2}} = \Delta y +20\\ (401)^{\frac{1}{2}} = (0.025) + 20\\ (401)^{\frac{1}{2}} = 20.025
Hence, (401)^{\frac{1}{2}} is approximately equal to 20.025

Question:1(xi) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 0.0037 ) ^{1/2 }

Answer:

Lets suppose y = (x)^{\frac{1}{2}} and let x = 0.0036 and \Delta x = 0.0001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}
\Delta y = ({0.0036 + 0.0001})^{\frac{1}{2}} - (0.0036)^{\frac{1}{2}}
\Delta y = ({0.0037})^{\frac{1}{2}} - 0.06
({0.0037})^{\frac{1}{2}} = \Delta y + 0.06
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(0.0001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 0.0001)\\ dy = \frac{1}{2(0.0036)^{\frac{1}{2}}}.(0.0001)\\ dy = \frac{1}{2\times 0..06}.(0.0001)\\dy = \frac{1}{0.12}.(0.0001) \\dy = 0.0008
Now,
(0.0037)^{\frac{1}{2}} = \Delta y +0.06\\ (0.0037)^{\frac{1}{2}} = (0.0008) + 0.06\\ (0.0037)^{\frac{1}{2}} = 0.0608
Hence, (0.0037)^{\frac{1}{2}} is approximately equal to 0.060 (because we need to take up to three decimal places)

Question:1(xii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(26.57) ^ {1/3}

Answer:

Lets suppose y = (x)^{\frac{1}{3}} and let x = 27 and \Delta x = -0.43
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({27 - 0.43})^{\frac{1}{3}} - (27)^{\frac{1}{3}}
\Delta y = ({26.57})^{\frac{1}{3}} - 3
({26.57})^{\frac{1}{3}} = \Delta y + 3
Now, we cam say that \Delta y is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-0.43) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -0.43)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-0.43)\\ dy = \frac{1}{3\times 9}.(-0.43)\\dy = \frac{1}{27}.(-0.43) \\dy = -0.0159 = -0.016 (approx.)
Now,
(26.57)^{\frac{1}{3}} = \Delta y +3\\ (26.57)^{\frac{1}{3}} = (-0.016) + 3\\ (26.57)^{\frac{1}{3}} = 2.984
Hence, (0.0037)^{\frac{1}{2}} is approximately equal to 0.060 (because we need to take up to three decimal places)

Question:1(xiii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 81.5 ) ^{1/4 }

Answer:

Lets suppose y = (x)^{\frac{1}{4}} and let x = 81 and 0.5
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({81 + 0.5})^{\frac{1}{4}} - (81)^{\frac{1}{4}}
\Delta y = ({81.5})^{\frac{1}{4}} - 3
({81.5})^{\frac{1}{4}} = \Delta y + 3
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 0.5)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(0.5)\\ dy = \frac{1}{4\times 27}.(0.5)\\dy = \frac{1}{108}.(0.5) \\dy = .004
Now,
(81.5)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.004) + 3\\ (82)^{\frac{1}{4}} = 3.004
Hence, (81.5)^{\frac{1}{4}} is approximately equal to 3.004

Question:1(xiv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.

( 3.968) ^{3/2 }

Answer:

Let's suppose y = (x)^{\frac{3}{2}} and let x = 4 and \Delta x = -0.032
Then,
\Delta y = ({x+\Delta x})^{\frac{3}{2}} - (x)^{\frac{3}{2}}
\Delta y = ({4 - 0.032})^{\frac{3}{2}} - (4)^{\frac{3}{2}}
\Delta y = ({3.968})^{\frac{3}{2}} - 8
({3.968})^{\frac{3}{2}} = \Delta y + 8
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{3 (x)^{\frac{1}{2}}}{2}.(-0.032) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{3}{2}} \ and \ \Delta x = -0.032)\\ dy = \frac{3 (4)^{\frac{1}{2}}}{2}.(-0.032)\\ dy = \frac{3\times 2}{2}.(-0.032)\\\\dy = -0.096
Now,
(3.968)^{\frac{3}{2}} = \Delta y +8\\ (3.968)^{\frac{3}{2}} = (-0.096) + 8\\ (3.968)^{\frac{3}{2}} = 7.904
Hence, (3.968)^{\frac{3}{2}} is approximately equal to 7.904

Question:1(xv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
( 32.15 ) ^{1/5}

Answer:

Lets suppose y = (x)^{\frac{1}{5}} and let x = 32 and \Delta x = 0.15
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{5}} - (x)^{\frac{1}{5}}
\Delta y = ({32 + 0.15})^{\frac{1}{5}} - (32)^{\frac{1}{5}}
\Delta y = ({32.15})^{\frac{1}{5}} - 2
({32.15})^{\frac{1}{5}} = \Delta y + 2
Now, we can say that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{5 (x)^{\frac{4}{5}}}.(0.15) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{5}} \ and \ \Delta x = 0.15)\\ dy = \frac{1 }{5 (32)^{\frac{4}{5}}}.(0.15)\\ dy = \frac{1}{5\times16}.(0.15)\\\\dy = \frac{0.15}{80}\\ dy = 0.001
Now,
(32.15)^{\frac{1}{5}} = \Delta y +2\\ (32.15)^{\frac{1}{5}} = (0.001) + 2\\ (32.15)^{\frac{1}{5}} = 2.001
Hence, (32.15)^{\frac{1}{5}} is approximately equal to 2.001

Question:2 Find the approximate value of f (2.01), where f (x) = 4x^2 + 5x + 2.

Answer:

Let x = 2 and \Delta x = 0.01
f(x+\Delta x) = 4(x+\Delta x)^2 +5(x+\Delta x)+2
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (8x+5).(0.01) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 4x^2+5x+2 \ and \ \Delta x = 0.01)\\ dy = 0.08x+0.05
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.08x + 0.05 + 4x^2 + 5x +2\\ f(x+\Delta x) = 0.08(2)+0.05+4(2)^2+5(2)+2\\ f(x+\Delta x) = 0.16 + 0.05 + 16 + 10 + 2\\ f(x+\Delta x) = 28.21
Hence, the approximate value of f (2.01), where f (x) = 4x^2 + 5x + 2. is 28.21

Question:3 Find the approximate value of f (5.001), where f (x) = x^3 - 7x^2 + 15.

Answer:

Let x = 5 and \Delta x = 0.001
f(x+\Delta x) =(x+\Delta x)^3 - 7(x+\Delta x)^2 +15
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (3x^2 - 14x).(0.001) \ \ \ \ \ \ \ \ \ (\because y = f(x) = x^3-7x^2+15 \ and \ \Delta x = 0.001)\\ dy =0.003x^2 -0.014x
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.003x^2 - 0.014x + x^3 - 7x^2 +15\\ f(x+\Delta x) =0.003(5)^2-0.014(5)+(5)^3-7(5)^2+15\\ f(x+\Delta x) = 0.075-0.07+125-175+15\\ f(x+\Delta x) = -34.995
Hence, the approximate value of f (5.001), where f (x) = x^3 - 7x^2 + 15\ is \ -34.995

Question:4 Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

Answer:

Side of cube increased by 1% = 0.01x m
Volume of cube = x^3 \ m^3
we know that \Delta y is approximately equal to dy
So,
dy = \frac{dy}{dx}.\Delta x\\ dy =3x^2(0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = x^3 \ and \ \Delta x = 0.01x)\\ dy = 0.03x^3
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 1% is 0.03x^3 \ m^3

Question:5 Find the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%.

Answer:

Side of cube decreased by 1% (\Delta x) = -0.01x m
The surface area of cube = 6a^2 \ m^2
We know that, (\Delta y) is approximately equal to dy

dy = \frac{dy}{dx}.\Delta x\\ dy = 12a(-0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = 6a^2 \ and \ \Delta x = -0.01x)\\ dy = 12x(-0.01x)\\ dy=-0.12x^2 \ m^2
Hence, the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%. is -0.12x^2 \ m^2

Question:6 If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

Answer:

Error in radius of sphere (\Delta r) = 0.02 m
Volume of sphere = \frac{4}{3}\pi r^3
Error in volume (\Delta V)
dV = \frac{dV}{dr}.\Delta r\\ dV = 4\pi r^2 .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because V = \frac{4}{3}\pi r^3, r =7 \ and \ \Delta r = 0.02 )\\ dV = 4\pi (7)^2 (0.02)\\ dV= 4\pi (49) (0.02)\\ dV = 3.92\pi
Hence, the approximate error in its volume is 3.92\pi \ m^3

Question:7 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

Answer:

Error in radius of sphere (\Delta r) = 0.03 m
The surface area of sphere = 4\pi r^2
Error in surface area (\Delta A)
dA = \frac{dA}{dr}.\Delta r\\ dA = 8\pi r .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because A = 4\pi r^2, r =9 \ and \ \Delta r = 0.03 )\\ dA = 8\pi (9) (0.03)\\ dA= 2.16\pi
Hence, the approximate error in its surface area is 2.16\pi \ m^2

Question:8 If f(x) = 3x ^2 + 15x + 5 , then the approximate value of f (3.02) is
(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

Answer:

Let x = 3 and \Delta x = 0.02
f(x+\Delta x) = 3(x+\Delta x)^2 +15(x+\Delta x)+5
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (6x+15).(0.02) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 3x^2+15x+5 \ and \ \Delta x = 0.02)\\ dy = 0.12x+0.3
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.12x + 0.3 + 3x^2 + 15x +5\\ f(x+\Delta x) = 0.12(3)+0.3+3(3)^2+15(3)+5\\ f(x+\Delta x) = 0.36+ 0.3 + 27 + 45 + 5\\ f(x+\Delta x) = 77.66
Hence, the approximate value of f (3.02) is 77.66
Hence, (D) is the correct answer

Question:9 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(A) 0.06 x^3 \ m^3 (B) 0.6 x^3 \ m^3 (C) 0.09 x^3 \ m^3 (D) 0.9 x^3 \ m^3

Answer:

Side of cube increased by 3% = 0.03x m
The volume of cube = x^3 \ m^3
we know that \Delta y is approximately equal to dy
So,
dy = \frac{dy}{dx}.\Delta x\\ dy =3x^2(0.03x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = x^3 \ and \ \Delta x = 0.03x)\\ dy = 0.09x^3
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 3% is 0.09x^3 \ m^3
Hence, (C) is the correct answer

More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

Nine questions and their explanations are given in exercise 6.4 Class 12 Maths solutions. There are two objectives questions in the NCERT solutions for Class 12 Maths chapter 6 exercise 6.4. The first question of Class 12 Maths chapter 6 exercise 6.4 have 15 sub questions. All these questions are detailed in the Class 12th Maths chapter 6 exercise 6.4

Also Read| Application of Derivatives Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

  • Exercise 6.4 Class 12 Maths gives more clarity about the approximation technique using differentiation for certain quantities
  • One question from Class 12 Maths chapter 6 exercise 6.4 may appear for the Class 12 CBSE Exam.
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Key Features Of NCERT Solutions for Exercise 6.4 Class 12 Maths Chapter 6

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 6.4 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 6.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 6.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 6.4 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 6.4 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 6.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. Which topic is covered in NCERT solutions for Class 12 Maths chapter 6 exercise 6.4

The concept of application of differentiation on the approximation of particular quantities are discussed in the Class 12 Maths chapter 6 exercise 6.4

2. The topic covered just before approximations is ………...

Tangents and normals and their examples and practice questions are covered before the concept of approximation. 

3. Can one solve exercise 6.4 before solving 6.3

Yes, the concepts discussed in topics 6.4 and 6.5 are different and prior knowledge of tangent and normal is not required to solve questions on approximations.

4. What is the next discussion after approximation in NCERT Class 12 Maths chapter 6?

Maxima and minima

5. Is the concept of approximation helpful in higher education?

Yes, in the field of Mathematics, Engineering and Science the concepts of approximation are important.

6. How many questions come from Class 12th Maths chapter 6 exercise 6.4 for the CBSE board exam.

One question may come from the topic approximation for the board examination. Overall 8 to 10 mark questions are expected from the chapter application of derivatives for the CBSE board exam

7. Give some tips to prepare for CBSE Class 12.

The preparation tip for CBSE Maths Class 12 is detailed in the link given. 

CBSE Class 12 Maths Preparation Tips 2021 to Score 90+ Marks

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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