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

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    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

    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.

    Access NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

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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 Question (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 ?

    hello,

    Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

    I hope this was helpful!

    Good Luck

    Hello dear,

    If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


    As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


    Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


    Believe in Yourself! You can make anything happen


    All the very best.

    Hello Student,

    I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

    You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

    All the best.

    If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

    Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

    View All

    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

    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

     

    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

    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

    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 .

    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

    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.

    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.

    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

    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

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    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

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    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

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    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    Team Lead

    A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

    2 Jobs Available
    Quality Systems Manager

    A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party. 

    2 Jobs Available
    Merchandiser

    A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying. 

    2 Jobs Available
    Procurement Manager

    The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

    2 Jobs Available
    Production Planner

    Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner. 

    2 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    .NET Developer

    .NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    DevOps Architect

    A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties. 

    2 Jobs Available
    Cloud Solution Architect

    Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

    2 Jobs Available
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