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

#CBSE Class 12th

Updated on 03 Dec 2023, 08:46 PM IST

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

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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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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.

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Frequently Asked Questions (FAQs)

Q: 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

Q: The topic covered just before approximations is ………...

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

Q: 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.

Q: What is the next discussion after approximation in NCERT Class 12 Maths chapter 6?

Maxima and minima

Q: Is the concept of approximation helpful in higher education?

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

Q: 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

Q: 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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Can i improve my 12 result this year in PCM

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If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

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Rahul Mandal

8 Sep'25

how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

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gopi

4 Sep'25

Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

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Prachi Kumari

4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

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CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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gopi

3 Sep'25

12 arts previous years board paper 2025

To find Class 12 Arts board papers, go to the official website of your education board, then click on the Sample Papers, Previous Years Question Papers(PYQ) or Model Papers section, and select the Arts stream. You will find papers for the various academic year. You can then select the year of which you want to solve and do your practice. There are many other educational websites that post pyqs on their website you can also visit that.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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

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