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

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 $\mathrm{\Delta }x=0.5$
Then,
$\mathrm{\Delta }y=\sqrt{x+\mathrm{\Delta }x}-\sqrt{x}$
$\mathrm{\Delta }y=\sqrt{49+0.5}-\sqrt{4}9$
$\mathrm{\Delta }y=\sqrt{49.5}-7$
$\sqrt{49.5}=\mathrm{\Delta }y+7$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{2\sqrt{x}}.(0.5)(\because y=\sqrt{x}and\mathrm{\Delta }x=0.5) \\ dy=\frac{1}{2\sqrt{4}9}.(0.5) \\ dy=\frac{1}{14}.(0.5) \\ dy=0.035 \end{gathered}$$
Now,
$$\begin{gathered} \sqrt{49.5}=\mathrm{\Delta }y+7 \\ \sqrt{49.5}=0.035+7 \\ \sqrt{49.5}=7.035 \end{gathered}$$
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 $\mathrm{\Delta }x=-0.4$
Then,
$\mathrm{\Delta }y=\sqrt{x+\mathrm{\Delta }x}-\sqrt{x}$
$\mathrm{\Delta }y=\sqrt{1+(-0.4)}-\sqrt{1}$
$\mathrm{\Delta }y=\sqrt{0.6}-1$
$\sqrt{0.6}=\mathrm{\Delta }y+1$
Now, we cam say that $\mathrm{\Delta }y$ is approximately equals to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{2\sqrt{x}}.(-0.4)(\because y=\sqrt{x}and\mathrm{\Delta }x=-0.4) \\ dy=\frac{1}{2\sqrt{1}}.(-0.4) \\ dy=\frac{1}{2}.(-0.4) \\ dy=-0.2 \end{gathered}$$
Now,
$$\begin{gathered} \sqrt{0.6}=\mathrm{\Delta }y+1 \\ \sqrt{0.6}=(-0.2)+1 \\ \sqrt{0.6}=0.8 \end{gathered}$$
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 $\mathrm{\Delta }x=0.001$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{3}}-(x{)}^{\frac{1}{3}}$
$\mathrm{\Delta }y=(0.008+0.001{)}^{\frac{1}{3}}-(0.008{)}^{\frac{1}{3}}$
$\mathrm{\Delta }y=(0.009{)}^{\frac{1}{3}}-0.2$
$(0.009{)}^{\frac{1}{3}}=\mathrm{\Delta }y+0.2$
Now, we cam say that $\mathrm{\Delta }y$ is approximately equals to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{3(x{)}^{\frac{2}{3}}}.(0.001)(\because y=(x{)}^{\frac{1}{3}}and\mathrm{\Delta }x=0.001) \\ dy=\frac{1}{3(0.008{)}^{\frac{2}{3}}}.(0.001) \\ dy=\frac{1}{0.12}.(0.001) \\ dy=0.008 \end{gathered}$$
Now,
$$\begin{gathered} (0.009{)}^{\frac{1}{3}}=\mathrm{\Delta }y+0.2 \\ (0.009{)}^{\frac{1}{3}}=(0.008)+0.2 \\ (0.009{)}^{\frac{1}{3}}=0.208 \end{gathered}$$
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 $\mathrm{\Delta }x=-0.001$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{10}}-(x{)}^{\frac{1}{10}}$
$\mathrm{\Delta }y=(1-0.001{)}^{\frac{1}{10}}-(1{)}^{\frac{1}{10}}$
$\mathrm{\Delta }y=(0.999{)}^{\frac{1}{10}}-1$
$(0.999{)}^{\frac{1}{10}}=\mathrm{\Delta }y+1$
Now, we cam say that $\mathrm{\Delta }y$ is approximately equals to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{10(x{)}^{\frac{9}{10}}}.(-0.001)(\because y=(x{)}^{\frac{1}{10}}and\mathrm{\Delta }x=-0.001) \\ dy=\frac{1}{10(1{)}^{\frac{9}{10}}}.(-0.001) \\ dy=\frac{1}{10}.(-0.001) \\ dy=-0.0001 \end{gathered}$$
Now,
$$\begin{gathered} (0.999{)}^{\frac{1}{10}}=\mathrm{\Delta }y+1 \\ (0.999{)}^{\frac{1}{10}}=(-0.0001)+1 \\ (0.999{)}^{\frac{1}{10}}=0.9999=0.999uptothreedecimalplace \end{gathered}$$
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 $\mathrm{\Delta }x=-1$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{4}}-(x{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(16-1{)}^{\frac{1}{4}}-(16{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(15{)}^{\frac{1}{4}}-2$
$(15{)}^{\frac{1}{4}}=\mathrm{\Delta }y+2$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{4(x{)}^{\frac{3}{4}}}.(-1)(\because y=(x{)}^{\frac{1}{4}}and\mathrm{\Delta }x=-1) \\ dy=\frac{1}{4(16{)}^{\frac{3}{4}}}.(-1) \\ dy=\frac{1}{4\times 8}.(-1) \\ dy=\frac{1}{32}.(-1) \\ dy=-0.031 \end{gathered}$$
Now,
$$\begin{gathered} (15{)}^{\frac{1}{4}}=\mathrm{\Delta }y+2 \\ (15{)}^{\frac{1}{4}}=(-0.031)+2 \\ (15{)}^{\frac{1}{4}}=1.969 \end{gathered}$$
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 $\mathrm{\Delta }x=-1$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{3}}-(x{)}^{\frac{1}{3}}$
$\mathrm{\Delta }y=(27-1{)}^{\frac{1}{3}}-(27{)}^{\frac{1}{3}}$
$\mathrm{\Delta }y=(26{)}^{\frac{1}{3}}-3$
$(26{)}^{\frac{1}{3}}=\mathrm{\Delta }y+3$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{3(x{)}^{\frac{2}{3}}}.(-1)(\because y=(x{)}^{\frac{1}{3}}and\mathrm{\Delta }x=-1) \\ dy=\frac{1}{3(27{)}^{\frac{2}{3}}}.(-1) \\ dy=\frac{1}{3\times 9}.(-1) \\ dy=\frac{1}{27}.(-1) \\ dy=-0.037 \end{gathered}$$
Now,
$$\begin{gathered} (27{)}^{\frac{1}{3}}=\mathrm{\Delta }y+3 \\ (27{)}^{\frac{1}{3}}=(-0.037)+3 \\ (27{)}^{\frac{1}{3}}=2.963 \end{gathered}$$
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 $\mathrm{\Delta }x=-1$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{4}}-(x{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(256-1{)}^{\frac{1}{4}}-(256{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(255{)}^{\frac{1}{4}}-4$
$(255{)}^{\frac{1}{4}}=\mathrm{\Delta }y+4$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{4(x{)}^{\frac{3}{4}}}.(-1)(\because y=(x{)}^{\frac{1}{4}}and\mathrm{\Delta }x=-1) \\ dy=\frac{1}{4(256{)}^{\frac{3}{4}}}.(-1) \\ dy=\frac{1}{4\times 64}.(-1) \\ dy=\frac{1}{256}.(-1) \\ dy=-0.003 \end{gathered}$$
Now,
$$\begin{gathered} (255{)}^{\frac{1}{4}}=\mathrm{\Delta }y+4 \\ (255{)}^{\frac{1}{4}}=(-0.003)+4 \\ (255{)}^{\frac{1}{4}}=3.997 \end{gathered}$$
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 $\mathrm{\Delta }x=1$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{4}}-(x{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(81+1{)}^{\frac{1}{4}}-(81{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(82{)}^{\frac{1}{4}}-3$
$(82{)}^{\frac{1}{4}}=\mathrm{\Delta }y+3$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{4(x{)}^{\frac{3}{4}}}.(1)(\because y=(x{)}^{\frac{1}{4}}and\mathrm{\Delta }x=1) \\ dy=\frac{1}{4(81{)}^{\frac{3}{4}}}.(1) \\ dy=\frac{1}{4\times 27}.(1) \\ dy=\frac{1}{108}.(1) \\ dy=.009 \end{gathered}$$
Now,
$$\begin{gathered} (82{)}^{\frac{1}{4}}=\mathrm{\Delta }y+3 \\ ({82}^{\frac{1}{4}}=(0.009)+3 \\ (82{)}^{\frac{1}{4}}=3.009 \end{gathered}$$
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 $\mathrm{\Delta }x=1$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{2}}-(x{)}^{\frac{1}{2}}$
$\mathrm{\Delta }y=(400+1{)}^{\frac{1}{2}}-(400{)}^{\frac{1}{2}}$
$\mathrm{\Delta }y=(401{)}^{\frac{1}{2}}-20$
$(401{)}^{\frac{1}{2}}=\mathrm{\Delta }y+20$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{2(x{)}^{\frac{1}{2}}}.(1)(\because y=(x{)}^{\frac{1}{2}}and\mathrm{\Delta }x=1) \\ dy=\frac{1}{2(400{)}^{\frac{1}{2}}}.(1) \\ dy=\frac{1}{2\times 20}.(1) \\ dy=\frac{1}{40}.(1) \\ dy=0.025 \end{gathered}$$
Now,
$$\begin{gathered} (401{)}^{\frac{1}{2}}=\mathrm{\Delta }y+20 \\ (401{)}^{\frac{1}{2}}=(0.025)+20 \\ (401{)}^{\frac{1}{2}}=20.025 \end{gathered}$$
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 $\mathrm{\Delta }x=0.0001$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{2}}-(x{)}^{\frac{1}{2}}$
$\mathrm{\Delta }y=(0.0036+0.0001{)}^{\frac{1}{2}}-(0.0036{)}^{\frac{1}{2}}$
$\mathrm{\Delta }y=(0.0037{)}^{\frac{1}{2}}-0.06$
$(0.0037{)}^{\frac{1}{2}}=\mathrm{\Delta }y+0.06$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{2(x{)}^{\frac{1}{2}}}.(0.0001)(\because y=(x{)}^{\frac{1}{2}}and\mathrm{\Delta }x=0.0001) \\ dy=\frac{1}{2(0.0036{)}^{\frac{1}{2}}}.(0.0001) \\ dy=\frac{1}{2\times 0..06}.(0.0001) \\ dy=\frac{1}{0.12}.(0.0001) \\ dy=0.0008 \end{gathered}$$
Now,
$$\begin{gathered} (0.0037{)}^{\frac{1}{2}}=\mathrm{\Delta }y+0.06 \\ (0.0037{)}^{\frac{1}{2}}=(0.0008)+0.06 \\ (0.0037{)}^{\frac{1}{2}}=0.0608 \end{gathered}$$
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 $\mathrm{\Delta }x=-0.43$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{3}}-(x{)}^{\frac{1}{3}}$
$\mathrm{\Delta }y=(27-0.43{)}^{\frac{1}{3}}-(27{)}^{\frac{1}{3}}$
$\mathrm{\Delta }y=(26.57{)}^{\frac{1}{3}}-3$
$(26.57{)}^{\frac{1}{3}}=\mathrm{\Delta }y+3$
Now, we cam say that $\mathrm{\Delta }y$ is approximately equals to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{3(x{)}^{\frac{2}{3}}}.(-0.43)(\because y=(x{)}^{\frac{1}{3}}and\mathrm{\Delta }x=-0.43) \\ dy=\frac{1}{3(27{)}^{\frac{2}{3}}}.(-0.43) \\ dy=\frac{1}{3\times 9}.(-0.43) \\ dy=\frac{1}{27}.(-0.43) \\ dy=-0.0159=-0.016(approx.) \end{gathered}$$
Now,
$$\begin{gathered} (26.57{)}^{\frac{1}{3}}=\mathrm{\Delta }y+3 \\ (26.57{)}^{\frac{1}{3}}=(-0.016)+3 \\ (26.57{)}^{\frac{1}{3}}=2.984 \end{gathered}$$
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,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{4}}-(x{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(81+0.5{)}^{\frac{1}{4}}-(81{)}^{\frac{1}{4}}$
$\mathrm{\Delta }y=(81.5{)}^{\frac{1}{4}}-3$
$(81.5{)}^{\frac{1}{4}}=\mathrm{\Delta }y+3$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{4(x{)}^{\frac{3}{4}}}.(0.5)(\because y=(x{)}^{\frac{1}{4}}and\mathrm{\Delta }x=0.5) \\ dy=\frac{1}{4(81{)}^{\frac{3}{4}}}.(0.5) \\ dy=\frac{1}{4\times 27}.(0.5) \\ dy=\frac{1}{108}.(0.5) \\ dy=.004 \end{gathered}$$
Now,
$$\begin{gathered} (81.5{)}^{\frac{1}{4}}=\mathrm{\Delta }y+3 \\ ({82}^{\frac{1}{4}}=(0.004)+3 \\ (82{)}^{\frac{1}{4}}=3.004 \end{gathered}$$
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 $\mathrm{\Delta }x=-0.032$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{3}{2}}-(x{)}^{\frac{3}{2}}$
$\mathrm{\Delta }y=(4-0.032{)}^{\frac{3}{2}}-(4{)}^{\frac{3}{2}}$
$\mathrm{\Delta }y=(3.968{)}^{\frac{3}{2}}-8$
$(3.968{)}^{\frac{3}{2}}=\mathrm{\Delta }y+8$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{3(x{)}^{\frac{1}{2}}}{2}.(-0.032)(\because y=(x{)}^{\frac{3}{2}}and\mathrm{\Delta }x=-0.032) \\ dy=\frac{3(4{)}^{\frac{1}{2}}}{2}.(-0.032) \\ dy=\frac{3\times 2}{2}.(-0.032) \\ dy=-0.096 \end{gathered}$$
Now,
$$\begin{gathered} (3.968{)}^{\frac{3}{2}}=\mathrm{\Delta }y+8 \\ (3.968{)}^{\frac{3}{2}}=(-0.096)+8 \\ (3.968{)}^{\frac{3}{2}}=7.904 \end{gathered}$$
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 $\mathrm{\Delta }x=0.15$
Then,
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{5}}-(x{)}^{\frac{1}{5}}$
$\mathrm{\Delta }y=(32+0.15{)}^{\frac{1}{5}}-(32{)}^{\frac{1}{5}}$
$\mathrm{\Delta }y=(32.15{)}^{\frac{1}{5}}-2$
$(32.15{)}^{\frac{1}{5}}=\mathrm{\Delta }y+2$
Now, we can say that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}\mathrm{\Delta }x \\ dy=\frac{1}{5(x{)}^{\frac{4}{5}}}.(0.15)(\because y=(x{)}^{\frac{1}{5}}and\mathrm{\Delta }x=0.15) \\ dy=\frac{1}{5(32{)}^{\frac{4}{5}}}.(0.15) \\ dy=\frac{1}{5\times 16}.(0.15) \\ dy=\frac{0.15}{80} \\ dy=0.001 \end{gathered}$$
Now,
$$\begin{gathered} (32.15{)}^{\frac{1}{5}}=\mathrm{\Delta }y+2 \\ (32.15{)}^{\frac{1}{5}}=(0.001)+2 \\ (32.15{)}^{\frac{1}{5}}=2.001 \end{gathered}$$
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 $\mathrm{\Delta }x=0.01$
$f(x+\mathrm{\Delta }x)=4(x+\mathrm{\Delta }x{)}^2+5(x+\mathrm{\Delta }x)+2$
$$\begin{gathered} \mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x) \\ f(x+\mathrm{\Delta }x)=\mathrm{\Delta }y+f(x) \end{gathered}$$
We know that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ dy=(8x+5).(0.01)(\because y=f(x)=4x^2+5x+2and\mathrm{\Delta }x=0.01) \\ dy=0.08x+0.05 \end{gathered}$$
$$\begin{gathered} f(x+\mathrm{\Delta }x)=\mathrm{\Delta }y+f(x) \\ f(x+\mathrm{\Delta }x)=0.08x+0.05+4x^2+5x+2 \\ f(x+\mathrm{\Delta }x)=0.08(2)+0.05+4(2{)}^2+5(2)+2 \\ f(x+\mathrm{\Delta }x)=0.16+0.05+16+10+2 \\ f(x+\mathrm{\Delta }x)=28.21 \end{gathered}$$
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 $\mathrm{\Delta }x=0.001$
$f(x+\mathrm{\Delta }x)=(x+\mathrm{\Delta }x{)}^3-7(x+\mathrm{\Delta }x{)}^2+15$
$$\begin{gathered} \mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x) \\ f(x+\mathrm{\Delta }x)=\mathrm{\Delta }y+f(x) \end{gathered}$$
We know that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ dy=(3x^2-14x).(0.001)(\because y=f(x)=x^3-7x^2+15and\mathrm{\Delta }x=0.001) \\ dy=0.003x^2-0.014x \end{gathered}$$
$$\begin{gathered} f(x+\mathrm{\Delta }x)=\mathrm{\Delta }y+f(x) \\ f(x+\mathrm{\Delta }x)=0.003x^2-0.014x+x^3-7x^2+15 \\ f(x+\mathrm{\Delta }x)=0.003(5{)}^2-0.014(5)+(5{)}^3-7(5{)}^2+15 \\ f(x+\mathrm{\Delta }x)=0.075-0.07+125-175+15 \\ f(x+\mathrm{\Delta }x)=-34.995 \end{gathered}$$
Hence, the approximate value of f (5.001), where $f(x)=x^3-7x^2+15is-34.995$

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 $\mathrm{\Delta }y$ is approximately equal to dy
So,
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ dy=3x^2(0.01x)(\because y=x^{3}and\mathrm{\Delta }x=0.01x) \\ dy=0.03x^3 \end{gathered}$$
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% $(\mathrm{\Delta }x)$ = -0.01x m
The surface area of cube = $6a^2{m}^2$
We know that, $(\mathrm{\Delta }y)$ is approximately equal to dy

$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ dy=12a(-0.01x)(\because y=6a^{2}and\mathrm{\Delta }x=-0.01x) \\ dy=12x(-0.01x) \\ dy=-0.12x^2{m}^2 \end{gathered}$$
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 $(\mathrm{\Delta }r)$ = 0.02 m
Volume of sphere = $\frac{4}{3}\pi r^3$
Error in volume $(\mathrm{\Delta }V)$
$$\begin{gathered} dV=\frac{dV}{dr}.\mathrm{\Delta }r \\ dV=4\pi r^2.\mathrm{\Delta }r(\because V=\frac{4}{3}\pi r^3,r=7and\mathrm{\Delta }r=0.02) \\ dV=4\pi (7{)}^2(0.02) \\ dV=4\pi (49)(0.02) \\ dV=3.92\pi \end{gathered}$$
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 $(\mathrm{\Delta }r)$ = 0.03 m
The surface area of sphere = $4\pi r^2$
Error in surface area $(\mathrm{\Delta }A)$
$$\begin{gathered} dA=\frac{dA}{dr}.\mathrm{\Delta }r \\ dA=8\pi r.\mathrm{\Delta }r(\because A=4\pi r^2,r=9and\mathrm{\Delta }r=0.03) \\ dA=8\pi (9)(0.03) \\ dA=2.16\pi \end{gathered}$$
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 $\mathrm{\Delta }x=0.02$
$f(x+\mathrm{\Delta }x)=3(x+\mathrm{\Delta }x{)}^2+15(x+\mathrm{\Delta }x)+5$
$$\begin{gathered} \mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x) \\ f(x+\mathrm{\Delta }x)=\mathrm{\Delta }y+f(x) \end{gathered}$$
We know that $\mathrm{\Delta }y$ is approximately equal to dy
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ dy=(6x+15).(0.02)(\because y=f(x)=3x^2+15x+5and\mathrm{\Delta }x=0.02) \\ dy=0.12x+0.3 \end{gathered}$$
$$\begin{gathered} f(x+\mathrm{\Delta }x)=\mathrm{\Delta }y+f(x) \\ f(x+\mathrm{\Delta }x)=0.12x+0.3+3x^2+15x+5 \\ f(x+\mathrm{\Delta }x)=0.12(3)+0.3+3(3{)}^2+15(3)+5 \\ f(x+\mathrm{\Delta }x)=0.36+0.3+27+45+5 \\ f(x+\mathrm{\Delta }x)=77.66 \end{gathered}$$
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 $\mathrm{\Delta }y$ is approximately equal to dy
So,
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ dy=3x^2(0.03x)(\because y=x^{3}and\mathrm{\Delta }x=0.03x) \\ dy=0.09x^3 \end{gathered}$$
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