RD Sharma Class 12 Exercise 13.1 Differentials, Errors and Approxim Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 13.1 Differentials, Errors and Approxim Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 10:00 AM IST

The class 12 mathematics syllabus is vast and pretty difficult to finish. Students in class 12 suffer a lot because there is not enough time to clear their doubts in school and to make sure they have understood the concepts properly. Therefore, students are recommended to use RD Sharma class 12th exercise 13.1 solutions for their home practice as it would help them improve their performance. The 13th chapter of the NCERT Mathematics Textbook is titled Differentials, Errors, and Approximations. This section will explore concepts like Absolute error, Relative error, Percentage error, Geometrical meaning of differentials, etc. The RD Sharma Solutions contains 10 questions based on the entire chapter.

## Differentials, Errors and Approximations Excercise: 13.1

### Differentials, errors and approximations exercise 13.1 question 1

Answer: $\Delta y=0-$No change

Given: $y=sinx$ and $x$ changes from $\frac{\pi }{2}$ to $\frac{22}{14}$
Solution: Suppose $\chi=\frac{\pi}{2}$ Therefore $x+\Delta x=\frac{22}{14}$
\begin{aligned} &\rightarrow \frac{\pi}{2}+\Delta x=\frac{22}{14} \\\\ &\Delta x=\frac{22}{14}-\frac{\pi}{2} \end{aligned}

→ Differentiate y with respect to x

$\frac{d y}{d x}=\cos x$
⇒ As we know that $(\sin x)=\cos x$
$\therefore \frac{d y}{d x}=\cos x$
⇒ when,$x=\frac{\pi}{2}$ we have $\frac{d y}{d x}=\cos \left(\frac{\pi}{2}\right)$
\begin{aligned} &\Rightarrow\left(\frac{d y}{d x}\right) x=\frac{\pi}{2}=0 \\\\ &\Rightarrow \Delta y=\left(\frac{d y}{d x}\right) \Delta x \end{aligned}

Here $\frac{d y}{d x}=0$ and $\Delta x=\frac{22}{14}-\frac{\pi}{2}$

$\Delta y=(0)\left(\frac{22}{14}\right)-\left(\frac{\pi}{2}\right)$

$\Delta y=(0)\$

Differentials, errors and approximations exercise 13.1 question 2

Answer: decrease $80\pi \; cm^{2}$
Hint: As we know, volume of a sphere of radius $x$ is given by $v=\left(\frac{4}{3}\right) \pi x^{3}$
Given: the radius of a sphere changes from 10cm to 9.8cm
Solution: Suppose $x$ be the radius of the sphere and $\Delta x$ be the change in the value of $x$
$\Rightarrow$ Thus, we have $x=10$ and $x+ \Delta x=9.8$\begin{aligned} &\Rightarrow 10+\Delta x=9.8 \\\\ &\Delta x=9.8-10 \\\\ &=-0.2 \end{aligned}

Differentiate $v$ with respect to $x$

\begin{aligned} &\frac{d v}{d x}=\frac{d}{d x}\left(\left(\frac{4}{3}\right) \pi x^{3}\right) \\\\ &\frac{d v}{d x}=\frac{4 \pi}{3} \frac{d}{d x}\left(x^{3}\right) \\\\ &\frac{d}{d x}\left(x^{3}\right)=n x^{n-1} \end{aligned}

\begin{aligned} &\frac{d v}{d x}=4 \pi x^{2} \text { when } \frac{d v}{d x}=4 \pi \times 100=400 \\\\ &(x=10) \\\\ &\Rightarrow \Delta y=\left(\frac{d y}{d x}\right) \Delta y \end{aligned}

\begin{aligned} &\frac{d v}{d x}=400 \pi \text { and } \Delta x=-0.2 \\\\ &\Delta x=(400 \pi)(-0.2)=-80 \pi \end{aligned}

Differentials, errors and approximations exercise 13.1 question 3

Answer:$2 k \pi c m^{2}$
Hint: Here, Area of a circular plate of radius $x$ is given by $A=\pi x^{2}$
Given: the radius of a circular plate initially is 10 cm and it increase by K%
Solution: Suppose $x$ be the radius of the plate, and $\Delta x$ is the change in the value of $x$
Thus we have $x=10 \text { and } \Delta x=\frac{K}{100} \times 10$
So, $\Delta x=0.1 K$
Differentiating A with respect to $x$
\begin{aligned} &\frac{d A}{d x}=\frac{d}{d x}\left(\pi x^{2}\right) \\\\ &\frac{d A}{d x}=\pi \frac{d}{d x}\left(x^{2}\right)=n x^{n-1} \\\\ &\Rightarrow \frac{d A}{d x}=2 \pi x \end{aligned}
\begin{aligned} &\Rightarrow \text { when } x=10 \text { and } \frac{d A}{d x}=2 \pi(10) \text { so, } \frac{d A}{d x}=20 \pi \\\\ &\Rightarrow \Delta y=\frac{d y}{d x} \Delta x \end{aligned}
\begin{aligned} &\text { Here, } \frac{d A}{d x}=20 \pi \text { and } \Delta x=0.1 K \\\\ &\Delta A=(20 \pi)(0.1 K) \\\\ &\Delta A=2 K \pi \end{aligned}

Differentials, errors and approximations exercise 13.1 question 4

Hint: area of a cubical box of radius $x$ is given by $s=6 x^{2}$
Given:$\Delta x$ is the error in the value of $x \; \; \Delta x=0.01 x$
Solution: suppose $x$ be the of cubical box
So,$\Delta x=0.01 x$
Differentiate $s$ with respect to $x$
\begin{aligned} &\frac{d s}{d x}=\frac{d}{d x}\left(6 x^{2}\right) \\\\ &\Rightarrow \frac{d s}{d x}=6 \times \frac{d}{d x}\left(x^{2}\right) \\\\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \Rightarrow \frac{d s}{d x}=6 x 2 x=12 x \end{aligned}
⇒ As we know that if $y=F(x) \Delta x$ is a small increment in $x$ , then the corresponding increment in $y$
$\Delta x=F(x+\Delta y)-F(x)$ , is approximately given as $\Delta y=\left(\frac{d y}{d x}\right) \Delta x$
\begin{aligned} &\Rightarrow \text { Here, } \frac{d s}{d x}=12 x \Delta x=0.01 x \\\\ &\Delta s=12 x \times 0.01 x \\\\ &\Delta s=0.12 x^{2} \end{aligned}
$\Rightarrow$ Percentage of error is
$\text { Error }=\frac{0.12 x^{2}}{6 x^{2}} \times 100 \%=0.02 \times 100 \%=2 \%$

Differentials, errors and approximations exercise 13.1 question 5

Hint: Surface volume of a sphere of radius $x$ is given by $v=\frac{4}{3} \pi x^{3}$
Given: let $x$ be the radius and $\Delta x$ be the error in the value $x$
Solution: Suppose $x$ be the radius of the sphere and $\Delta x$ be the error in the value $x$
⇒ Thus, we have $\Delta x=\left(\frac{0.1}{100}\right) \times(x)$
So, $\Delta x=0.001 x$
⇒ Volume of a sphere, $v=\frac{4}{3} \pi x^{3}$
So, Differentiate v with respect $x$
\begin{aligned} &\Rightarrow \frac{d v}{d x}=\frac{d}{d x}\left(\frac{4}{3} \pi x^{3}\right) \Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3} \frac{d}{d x}\left(x^{3}\right) \\\\ \\\\ &\Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3}\left(4 x^{2}\right)=4 \pi x^{2} \end{aligned}
$\Rightarrow$ As we know, $y=f(x)$ and $\Delta x$ is a smaller increment,
\begin{aligned} &\Delta y=\left(\frac{d v}{d x}\right) \Delta x=0.001 x \\\\ &\Delta x=0.004 \pi x^{3} \\\\ &\text { Here, } \frac{d v}{d x}=4 \pi x^{2} \text { and } \Delta x=0.001 x \end{aligned}

$\Rightarrow$ Percentage error $=\frac{0.004 \pi x^{3}}{\frac{4}{3} \pi x^{3}} \times 100=0.003 \times 100=0.3 \%$

Differentials, errors and approximations exercise 13.1 question 6

Hint: Here we use the basic concept of area and logarithm
Given:$pv^{1.4}$ =constant
Solution: Given as $pv^{1.4}$=constant and the decrease in $v$ is $\frac{1}{2}%$
$\Rightarrow$ Thus, we have $\Delta v=\left(\frac{-1}{2}\right) 100 x v$
So, $\Delta v=-0.005 v$
$\Rightarrow$ Now, $pv^{1.4}$ =constant
taking log on both sides,
$\log \left(p v^{1.4}\right)=\log (\text { constant })$
$\Rightarrow$ Differentiate both sides with respect to v
\begin{aligned} &\frac{d}{d p}(\log p) x \frac{d p}{d v}+\frac{d}{d v}(1.4 \log v)=0 \\\\ &\Rightarrow \frac{d}{d p}(\log p) x \frac{d p}{d v}+1.4 \frac{d}{d v}(\log v)=0 \end{aligned}
\begin{aligned} &\Rightarrow \frac{d}{d x}(\log x)=\frac{1}{x} \\\\ &\Rightarrow \frac{d}{d v}=\frac{-1.4}{v} P \end{aligned}
\begin{aligned} &\Delta p=\left(\frac{-1.4}{v} p\right)(-0.005 x) \\\\ &\Delta p=0.007 p \end{aligned}
$\Rightarrow$ Percentage of error is
$\text { Error }=\frac{0.007 p}{p} \times 100=0.7 \%$

Differentials, errors and approximations exercise 13.1 question 7

Hint: total surface area of the cone is given by $s=\pi r^{2}+\pi r l$
Given: $\Delta x=\left(\frac{k}{100}\right) \times(x)$
$\Delta x=0.01 k x$
Solution: Suppose $x$ be the height of the cone and $\Delta x$ be the change in the value of $x$
⇒ Thus we have $\Delta x=\left(\frac{k}{100}\right) \times x$
$\Delta x=0.01 k x$
⇒ suppose us assume that radius, the semi height and semi vertical angle of the come to be $r$, 1 and respectively as shown

from above figure, using trigonometry

$\tan d=\frac{O B}{O A}=\frac{r}{x}$
⇒ we have also
$\cos x=\frac{O A}{A B}=\frac{x}{l}$
$1=\frac{x}{\cos x}$
⇒ Now the total surface area of the cone is given by
$s=\pi x^{2}+\pi r l$
⇒ from the above, we have $r=x \tan x \text { and } l=x \sec x$
$\Rightarrow s=\pi x^{2}+\pi(x \tan x \cdot x \sec x)$
⇒ Differentiate s with respect to $x$
\begin{aligned} &\frac{d s}{d x}=\frac{d}{d x}\left[\pi x^{2} \tan x(\tan x+\sec x)\right] \\\\ &\Rightarrow \frac{d s}{d x}=\pi \tan x(\tan x+\sec x) \frac{1}{d x}\left(x^{2}\right) \end{aligned}
\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\\\ &\Rightarrow \frac{d s}{d x}=2 \pi x \tan x(\tan x+\sec x) \end{aligned}
\begin{aligned} &\Rightarrow \text { so, } \Delta x=\left(\frac{d y}{d x}\right) \Delta x \\\\ &\Rightarrow \text { Here } \frac{d s}{d x}=2 \pi x(\tan x+\sec x) \text { and } \Delta x=0.01 k x \\\\ &\Rightarrow \Delta s=(2 \pi x \tan (x)[\tan (x)+\sec (x)](0.01 k x) \end{aligned}
⇒ Percentage of increase in S
\begin{aligned} &\text { Increase }=\frac{\Delta s}{s} \times 100 y \\\\ &=100+\times 0 . .03 k \pi x^{2} \tan x[\tan x+\sec x] \\\\ &\pi x^{2} \tan x(\tan x+\sec x) \end{aligned}
\begin{aligned} &\text { Increase }=0.02 k \times 100 \\\\ &=2 k \% \end{aligned}

Differentials, errors and approximations exercise 13.1 question 8

Hint: volume of a sphere of radius $x$ is given by $v=\frac{4}{3} \pi x^{3}$
Given: we have $\Delta x=0.01 k x$
Solution: Suppose the error in measuring the radius of a sphere be $k$
⇒ Suppose $x$ be the radius of the sphere and $\Delta x$ be the error in the value of $x$
⇒ Thus, we have $\Delta x=\left(\frac{k}{100}\right) \times(x)$
So, $\Delta x=0.01 k x$
⇒ Differentiate v with respect to $x$
\begin{aligned} &\frac{d v}{d x}=\frac{d}{d x}\left(\frac{4}{3}\right) \pi x^{3} \\\\ &\Rightarrow \frac{d v}{d x}=\frac{4 \pi}{3} \frac{d}{d x}\left(x^{3}\right) \Rightarrow \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}
\begin{aligned} &\frac{d v}{d x}=\frac{4 \pi}{3}\left(3 x^{2}\right) \frac{d v}{d x}=4 \pi x^{2} \\\\ &\Rightarrow \Delta y=\frac{d y}{d x} \Delta x, \text { Here } \frac{d v}{d x}=4 \pi x^{2} \Delta x=0.01 k x \end{aligned}
$\therefore \Delta x=0.04 k \pi x^{3}$
⇒ percentage of error is,
\begin{aligned} &\text { Error }=\frac{0.04 k \pi x^{3}}{\frac{4}{3} \pi x^{3}} \times 100 \% \\\\ \end{aligned}
$=\frac{0.04 k \pi x^{3}}{4 \pi x^{3}} \times 100 \times 3=3 k \%$

Differentials, errors and approximations exercise 13.1 question 9 (i)

Hint: Here we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\sqrt{25.3}$
Solution: let $y=\sqrt{x}$
\begin{aligned} &x=25 \& \Delta x=0.3 \\\\ &\Rightarrow \text { Since } y=\sqrt{x} \\\\ &\frac{d y}{d x}=\frac{d \sqrt{x}}{d x}=\frac{1}{2 \sqrt{x}} \end{aligned}
⇒ Now,
\begin{aligned} &\Delta y=\frac{d y}{d x} \Delta x \\\\ &=\frac{1}{2 \sqrt{x}} \times 0.3 \\\\ &=\frac{1}{2 \sqrt{x}} \times 0.3=0.03 \end{aligned}
\begin{aligned} &\Rightarrow \text { Also, } \Delta y= \frac{f(x+\Delta x)-f(x)}\ ={\sqrt{25+0.3}-\sqrt{25}} \\\\ &\sqrt{25.3}=5.03 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (ii)

Hint: here we use the formula
Given:$(0.009)^{\frac{1}{3}}$
Solution: consider $y=x^{\frac{1}{3}} \text { Let } x=0.009 \text { and } \Delta x=0.001$$\Delta x=(x+\Delta x)^{\frac{1}{3}}=(0.009)^{\frac{1}{3}}-(0.008)^{\frac{1}{3}}(0.009)^{\frac{1}{3}}=0.2+\Delta y$

Now, $dy$ is approximately equal to $\Delta y$ and is given by,

\begin{aligned} &d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \\\\ &=\frac{3}{3 \times 0.04}(0.001)=\frac{0.001}{0.12} \\\\ &=0.008 \end{aligned}
Hence, the approximate value of $(0.004)^{\frac{1}{3}} i \text { is } 0.2+0.08=0.208$

Differentials, errors and approximations exercise 13.1 question 9 (iii)

Hint: Here we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given:$(0.007)^{\frac{1}{3}}$
Solution: consider the function $f(x)=3 \sqrt{x}$
Let:
\begin{aligned} &x=0.008 \\\\ &x+\Delta x=0.007 \end{aligned}
Then,
\begin{aligned} &\Delta x=-0.001 \\\\ &\text { For } x=0.008, \end{aligned}
\begin{aligned} &y=\sqrt{0.008}=0.2 \\\\ &\text { Let, } d x=\Delta x=0.001 \\\\ &\text { Now } y=3 \sqrt{x} \end{aligned}
\begin{aligned} &\frac{d y}{d x}=\frac{1}{3(x)^{\frac{2}{3}}} \\\\ &\left(\frac{d y}{d x}\right) x=0.008=\frac{1}{0.12} \times 0.001=\frac{1}{120} \end{aligned}\begin{aligned} &\Rightarrow \Delta y=\frac{1}{120}=0.008333 \\\\ &(0.007)^{\frac{1}{3}}=y+\Delta y=0.191667 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (iv)

Hint: Here, we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\sqrt{401}$
Solution:
Let $y=x^{\frac{1}{2}}$
Where, $x=400$
\begin{aligned} &x+\Delta x=401 \\\\ &\Delta x=1 \end{aligned}
Now, $\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}$
Using, $\Delta y=\frac{d y}{d x} \Delta x=\frac{1}{2 \sqrt{x}} \times 1$
Putting the value of $x$
\begin{aligned} \sqrt{401} &=y+\Delta y \\\\ &=20+0.025 \\\\ &=20.025 \end{aligned}
Hint : Here, we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $(15)^{\frac{1}{4}}$
Solution:
Let $y=x^{\frac{1}{4}}=(16)^{\frac{1}{4}}=2$
Where , $x=16$
\begin{aligned} &x+\Delta x=15 \\\\ &\Delta x=-1 \end{aligned}
Now, $y=x^{\frac{1}{4}}$
Differentiating w.r.t $x$
$\frac{d y}{d x}=\frac{1}{4} x^{\frac{-3}{4}}$
Using, $\Delta y=\frac{d y}{d x} \Delta x=\frac{1}{4 x^{\frac{3}{4}}} \times-1$
Putting the value of $x$
$\Delta y=\frac{-1}{4(16)^{\frac{3}{4}}}=\frac{-1}{4 \times 2^{3}}=-0.03125$
\begin{aligned} (15)^{\frac{1}{4}} &=y+\Delta y \\\\ &=2+(-0.03125) \\\\ &=1.96875 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (vi)

Hint: Here, we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $(255)^{\frac{1}{4}}$
Solution:
Let $y=x^{\frac{1}{4}}=(256)^{\frac{1}{4}}=4$
where $x=16$
\begin{aligned} &x+\Delta x=255 \\\\ &\Delta x=-1 \end{aligned}
Now, $y=x^{\frac{1}{4}}$
Differentiating w.r.t $x$
$\frac{d y}{d x}=\frac{1}{4} x^{\frac{-3}{4}}$
Using,
$\Delta y=\frac{d y}{d x} \Delta x=\frac{1}{4 x^{\frac{3}{4}}} \times-1$
Putting the value of $x$
\begin{aligned} &\Delta y=\frac{-1}{4(256)^{\frac{3}{4}}}=\frac{-1}{4 \times 4^{4 \times \frac{3}{4}}}=\frac{-1}{256}=-0.0039 \\\\ &(255)^{\frac{1}{4}}=y+\Delta y \end{aligned}
\begin{aligned} &=4+(-0.0039) \\\\ &=3.9961 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (vii)

Hint: Here, we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\frac{1}{(2.002)^{2}}$
Solution:
consider $y=\frac{1}{x^{2}}$
Here $x=2 \text { and } \Delta x=0.002$
On differentiating wrt $x$,
$\frac{d y}{d x}=\frac{-2}{x^{3}}$
So we get,
$\Delta y=\frac{d y}{d x} \cdot \Delta x$
On substituting the value we get,
$\Delta \mathrm{y}=\frac{-2}{8}(0.002)$
On further calculating we get,
$\Delta \mathrm{y}=\frac{-0.5}{1000}=-0.0005$
On substitution we get,
$-0.005=\frac{1}{(2.002)^{2}}-\frac{1}{4}$
We get
$\frac{1}{(2.002)^{2}}=0.2495$

Differentials, errors and approximations exercise 13.1 question 9 (viii)

Hint: Here, we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\log _{10} 4=0.6021$
$\log _{10} \mathrm{e}=0.4343$
Solution:
consider $\mathrm{y}=\log _{10} x$
Or $\mathrm{y}=\frac{\log _{10} \mathrm{x}}{\log _{10} \mathrm{e}}$
We get
$y=0.4343 \log _{e} x$
Here
$\mathrm{x}=4 \text { and } \Delta \mathrm{x}=.04$
On differentiating wrt $x$
\begin{aligned} &\frac{d y}{d x}=\frac{0.4343}{x} \Delta \mathrm{x} \\\\ &\text { At } \mathrm{x}=4 \\\\ &\frac{d y}{d x}=\frac{1}{4} \end{aligned}
\begin{aligned} &\text { So }, \Delta y=\frac{d y}{d x} d x=\frac{1}{4} 0.04=0.01 \\\\ &\text { i.e } \Delta y=0.01 \\\\ &\text { Therefore } \log _{\mathrm{e}} 4.04=\mathrm{y}+\Delta \mathrm{y}=1.396368 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (ix)

Answer: $2.3046$
Hint: Here we use
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\log e 10.2=2.3026$
Solution:
Consider $y=\operatorname{loge} x$
Here $x=10 \text { and } \Delta x=0.02$
By Differentiating w.r.t x,
So we get
$\Delta y=\frac{d y}{d x} \Delta x$
$\Rightarrow \Delta y=\frac{d y}{d x} \Delta x=\frac{1}{10} \times 0.02$
on further calculation
\begin{aligned} &\Delta y=\frac{0.02}{10}=0.002 \\\\ &\Rightarrow \Delta y=f(x+\Delta x)-f(x) \\\\ &0.002=\operatorname{loge}(10+0.02)-\operatorname{loge} 10 \end{aligned}
It can be written as
$0.002=\operatorname{loge} 10.02=2.3026$
We get,
$\log 10.02=2.3026$

Differentials, errors and approximations exercise 13.1 question 9 (x)

Answer: $1.004343$
Hint: Here we use
$\Delta y=f(x+\Delta x)-f(x)$
Given:
Solution:
let $x=10$
$x+\Delta x=10.1$
Then
\begin{aligned} &\Delta x=0.1 \\\\ &\text { For } x=y \log _{10}=1 \\\\ &\text { Let: } d x=\Delta x=0.1 \end{aligned}\begin{aligned} &\Delta x=0.1 \\\\ &\text { For } x=y \log _{10}=1 \\\\ &\text { Let: } d x=\Delta x=0.1 \end{aligned}\begin{aligned} &\Delta x=0.1 \\\\ &\text { For } x=y \log _{10}=1 \\ &\text { Let: } d x=\Delta x=0.1 \end{aligned}
Now, $y=\log _{10} x=\frac{\log e x}{\log e 10}=\frac{1}{2.3025 x}$
\begin{aligned} &\left(\frac{d y}{d x}\right)_{x=10}=0.04343 \\\\ &\Delta y=d y=\frac{d y}{d x} x d x=0.04343 \times 0.1 \\\\ &\Delta y=0.004343 \end{aligned}

$\log _{10} 10.1=y+\Delta y=1.004343$

Differentials, errors and approximations exercise 13.1 question 9 (xi)

Answer: $0.4849$

Hint: $f(x+\Delta x)=f(x)+\Delta f(x)$
Given: $f(x)=\cos x=\cos 60^{\circ}$
$\Delta x=1=0.01745 \text { radius }$
Solution:
$\cos 61^{\circ}=f\left(61^{\circ}\right)=f\left(60^{\circ}+1^{\circ}\right)=f\left(60^{\circ}\right)+\Delta x f^{1}\left(60^{\circ}\right)=0.5+[0.01745]$
$\left[-\sin 60^{\circ}\right]$
\begin{aligned} &{\left[\therefore f^{1}(x)=-\sin x\right]} \\\\ &\Rightarrow \cos 61^{\circ}-0.5+(0.01745)(-0.86603) \\\\ &\therefore \cos 61^{\circ}=0.4848 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xiv)

Answer: $0.57560442$
Hint: Here, we use $f(x+\Delta y)-f(x)$
Given: $\cos \left(\frac{11 \pi}{36}\right)$
Solution: Let $x=\frac{12 \pi}{36}=\frac{\pi}{3}$
\begin{aligned} &\text { So, } x+\Delta x=\frac{\pi}{36} \\\\ &\Rightarrow \Delta x=\frac{-\pi}{36}=\frac{-22}{7}=-0.0873 \end{aligned}

Differentiating $f(x)$
$\frac{d f}{d x}=\frac{d}{d x}(\cos x)$
we know
\begin{aligned} &\frac{d}{d x}(\cos x)=-\sin x \\\\ &\frac{d f}{d x}=-\sin x \end{aligned}
when, $x=\frac{\pi}{3}$
we have
$\frac{d f}{d x}=-\sin \left(\frac{\pi}{3}\right)$
\begin{aligned} &\Rightarrow\left(\frac{d f}{d x}\right)_{x=\frac{\pi}{3}}=-0.86603 \\\\ &\Rightarrow f(x+\Delta x)-f(x) \\\\ &\Delta y=\left(\frac{d y}{d x}\right) \Delta x \end{aligned}
\begin{aligned} &\text { Here, } \frac{d f}{d x}=-0.86603 \\\\ &\Delta x=0.0873 \\\\ &\Rightarrow \Delta f=0.07560442 \end{aligned}
now, we have
\begin{aligned} &f\left(\frac{11 \pi}{36}\right)=f\left(\frac{\pi}{3}\right)+\Delta f \\\\ &f\left(\frac{11 \pi}{36}\right)=\cos \left(\frac{\pi}{3}\right)+0.07560442 \end{aligned}
\begin{aligned} &f\left(\frac{11 \pi}{36}\right)=0.5+0.07560442 \\\\ &f\left(\frac{11 \pi}{36}\right)=0.57560442 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xvi)

Answer: $3.074$
Hint: Here, we use
$\Delta y=f(x)+\Delta x)-f(x)$
Given: $29^{\frac{1}{2}}$
Solution:$y=f(x)=(x)^{\frac{1}{3}}$
\begin{aligned} &\Rightarrow \text { let } \\\\ &x=27 \\\\ &x+\Delta x=29 \end{aligned}
Thus,
\begin{aligned} &\Delta x=2 \\\\ &\text { For } x=27 \\\\ &y=(27)^{\frac{1}{3}}=3 \end{aligned}
Let $d x=\Delta x=2$
Now, $y=(x)^{\frac{1}{3}}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{3 x^{\frac{1}{3}}}$
\begin{aligned} &\left(\frac{d y}{d x}\right)_{x=27}=\frac{1}{27} \\\\ &\Delta y=d y=\frac{d y}{d x} d x=\frac{1}{27} \times 2=0.074 \end{aligned}
\begin{aligned} &\Delta y=0.074 \\\\ &\text { Hence, } 29^{\frac{1}{3}}=3+0.074=3.074 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xvii)

Answer: $4.042$
Hint: here we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $66^{\frac{1}{3}}$
Solution: Consider the function $y=f(x)=x^{\frac{1}{3}}$
Let,
\begin{aligned} &x=64 \\\\ &x+\Delta x=66 \end{aligned}
Then,
\begin{aligned} &\Delta x=2 \\\\ &\Rightarrow \text { for } x=64 \\\\ &y=(64)^{\frac{1}{3}}=4 \end{aligned}
Let,
\begin{aligned} &d x=\Delta x=2 \\\\ &\text { Now, } y=\left(x^{\frac{1}{3}}\right) \\\\ &\frac{d y}{d x}=\frac{1}{3(x)^{\frac{2}{3}}} \\\\ &\left(\frac{d y}{d x}\right)=\frac{1}{48} \end{aligned}
\begin{aligned} &\Delta y=d y=\frac{d y}{d x} d x=\frac{1}{48} \times 2=0.042 \\\\ &\Delta y=0.042 \\\\ &(66)^{\frac{1}{3}}=y+\Delta y=4.042 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9(xviii)

Hint:Here, we use the formula
$\Delta y=f(x+\Delta x)-f(x)$
Given:$\sqrt{26}$
Solution: Consider the function $y=f(x)=\sqrt{x}$
Let
\begin{aligned} &x=25 \\\\ &x+\Delta x=26 \end{aligned}
Then,
\begin{aligned} &\Delta x=2 \\\\ &\text { For } x=25 \\\\ &y=\sqrt{25}=5 \end{aligned}
Let,
$d x=\Delta x=1$
Now
$y=(x)^{\frac{1}{2}}$
\begin{aligned} &\frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \\\\ &\left(\frac{d y}{d x}\right)_{x=25}=\frac{1}{10} \\\\ &\Delta y=d y=\frac{d y}{d x} \times d x=\frac{1}{10} \times 1=0.1 \end{aligned}
\begin{aligned} &\Delta y=0.1 \\\\ &\sqrt{26}=y+\Delta y=5.1 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xix)

Hint: $\Delta y=f(x+\Delta x)-f(x)$
Let's use this formula
Given: $\sqrt{37}$
Solution: $\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}$
So, we get
$\Delta y=\frac{d y}{d x} \Delta x$
So , we get
$\Delta y=\frac{1}{2 \sqrt{x}} \times 1$
on further calculation
$\Delta y=\frac{1}{12}=0.08$
we know that
$\Delta y=f(x+\Delta x)-f(x)$
By substituting the values,
\begin{aligned} &0.08=\sqrt{36+1}-\sqrt{36} \\\\ &0.08=\sqrt{37}-6 \\\\ &\sqrt{37}=6.08 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xx)

Hint: $\Delta y=f(x+\Delta x)-f(x)$
we use this formula
Given:$\sqrt{0.48}$
Solution: consider the function $y=f(x)=\sqrt{x}$
Let
\begin{aligned} &x=0.49 \\\\ &x+\Delta x=0.48 \end{aligned}
Then,
\begin{aligned} &\Delta x=-0.01 \\\\ &\Rightarrow \text { for } x=0.49 \\\\ &y=\sqrt{0.19}=0.7 \end{aligned}
Let,
\begin{aligned} &d x=\Delta x=0.01 \\\\ &\Rightarrow \text { now } y=(x)^{\frac{1}{2}} \\\\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{2 x}} \\\\ &\left(\frac{d y}{d x}\right)_{x=0.49}=\frac{1}{1.4} \end{aligned}
\begin{aligned} &\Delta y=d y=\frac{d y}{d x} \times d x=\frac{1}{1.4} \times(-0.01) \\\\ &\Delta y=-0.007143 \\\\ &\sqrt{0.48}=y+\Delta y=0.693 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxi)

Answer: $3.00926$

Hint: Here we use below formula,
$\Delta y=f(x+\Delta x)-f(x)$
Given:$(82)^{\frac{1}{4}}$
Solution: $x=81$
$\Delta x=1$
on differentiating $f(x)$ with respect to $x$
$\frac{d f}{d x}=\frac{d}{d x}\left(x \frac{1}{4}\right)$
we know $\frac{d}{d x}\left(x^{3}\right)=n x^{n-1}$
\begin{aligned} &\frac{d f}{d x}=\frac{1}{4} x^{\frac{1}{4}-1} \\\\ &\frac{d f}{d x}=\frac{1}{4} x^{\frac{3}{4}}=\frac{1}{4 x^{\frac{3}{4}}} \end{aligned}
\begin{aligned} &\Rightarrow \text { when } x=81 \text { we have } \frac{d f}{d x}=\frac{1}{4(81)^{\frac{3}{4}}} \\\\ &\Rightarrow \frac{d f}{d x}=\frac{1}{4(34)^{\frac{3}{4}}}=\frac{1}{4\left(3^{3}\right)}=\frac{1}{4 \times 27}=\frac{1}{108}=0.00926 \end{aligned}
\begin{aligned} &\Rightarrow \Delta y=\left(\frac{d y}{d x}\right) \Delta x \\\\ &\Delta f=0.00926 \\\\ &\Rightarrow \text { now, } f(82)=f(81)+\Delta f \end{aligned}
\begin{aligned} &f(82)=3+0.00926 \\\\ &f(82)=3.00926 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxii)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\left(\frac{17}{81}\right)^{\frac{1}{4}}$
Solution: Let $y=x \frac{1}{4}$
where $x=16 \text { and } \Delta x=1$
Since $y=x^ \frac{1}{4}$
$\frac{d y}{d x}=\frac{d\left(x ^\frac{1}{4}\right)}{d x}=\frac{1}{4} x^{\frac{1}{4}-1}=\frac{1}{4} x^{\frac{-3}{4}}=\frac{1}{4 x^{\frac{3}{4}}}$
\begin{aligned} &\Rightarrow \text { Now, } \Delta y=\frac{d y}{d x} \Delta x=\frac{1}{4} \times \frac{1}{16^{\frac{3}{4}}} \times 1 \\\\ &=\frac{1}{4} \times \frac{1}{24^{\frac{3}{4}}}=\frac{1}{4} \times \frac{1}{2^{3}}=\frac{1}{32} \end{aligned}
Now,
$(17)^{\frac{1}{4}}=y+\Delta y$
Putting values,
\begin{aligned} &(17)^{\frac{1}{4}}=(16)^{\frac{1}{4}}+\Delta y \\\\ &(17)^{\frac{1}{4}}=(24)^{\frac{1}{4}}+\Delta y \end{aligned}
\begin{aligned} &(17)^{\frac{1}{4}}=2+\frac{1}{32} \\ &(17)^{\frac{1}{4}}=2.03125 \end{aligned}
Now,
$\left(\frac{17}{81}\right)^{\frac{1}{4}}=\frac{2.03125}{3}=0.677$

Differentials, errors and approximations exercise 13.1 question 9 (xxiii)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $(33)^{\frac{1}{5}}$
Solution: $(33)^{\frac{1}{5}}=(32+1)^{\frac{1}{5}}$
let $y=f(x)=x^{\frac{1}{5}}$
\begin{aligned} &\Rightarrow y+\Delta y=(x+\Delta x)^{\frac{1}{5}} \\\\ &\Rightarrow \Delta y=(x+\Delta x)^{\frac{1}{5}}-x^{\frac{1}{5}} \end{aligned}
Also,
\begin{aligned} &\Delta y=f^{1}(x) \Delta x \\\\ &(x+\Delta x)^{\frac{1}{5}}-x^{\frac{1}{5}}=\frac{1}{5} x^{\frac{-4}{5}} \times \Delta x \end{aligned}
Put $x=32, \Delta x=1$
\begin{aligned} &(33)^{\frac{1}{5}}-(32)^{\frac{1}{5}}=\frac{1}{5}(2)^{4}(1) \\\\ &\Rightarrow(33)^{\frac{1}{5}}=2+0.0125=2.0125 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxiv)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\sqrt{36.6}$
Solution: $y=\sqrt{x} \text { let } x=36 \text { and } A x=0.6$then,
\begin{aligned} &\Delta y=\sqrt{x+\Delta x}-\sqrt{x} \\\\ &=\sqrt{36.6}-\sqrt{36} \\\\ &=\sqrt{36.6}-6 \\\\ &\Delta y=\sqrt{36.6}-6 \end{aligned}
now, $dy$ is approximate equal to $Ay$ and is given by.
\begin{aligned} &d y=\left(\frac{d y}{d x}\right) d x=\frac{1}{2 \sqrt{x}}(0.6) \\\\ &\frac{1}{2 \sqrt{36}} \times 0.6=0.05 \end{aligned}
The approximate value of $\sqrt{36.6} \text { is } 6+0.05=6.05$

Differentials, errors and approximations exercise 13.1 question 9 (xxv)

Answer: $25^{\frac{1}{3}}=2.926$
Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $25^{\frac{1}{3}}$
Solution: Let $y=x^{\frac{1}{3}}$
where $x=27 \text { and } \Delta x=-2$
Since $y=x^{\frac{1}{3}}$
\begin{aligned} &\frac{d y}{d x}=\frac{d\left(x \frac{1}{3}\right)}{d x} \\\\ &=\frac{1}{3} x^{\frac{1}{3}-1} \\\\ &=\frac{1}{3} x^{\frac{-2}{3}}=\frac{1}{3 x^{\frac{2}{3}}} \end{aligned}
Now,
$\Delta y=\frac{d y}{d x} \Delta x$
Putting values
\begin{aligned} &\Delta y=\frac{1}{3(27)^{\frac{2}{3}}} \times(-2) \\\\ &\Delta y=\frac{-2}{3 \times\left(3^{3}\right)^{\frac{2}{3}}}=\frac{2}{3 \times 3^{2}}=\frac{2}{27}=-0.074 \end{aligned}
\begin{aligned} &(25)^{\frac{1}{3}}=3-0.074 \\\\ &(25)^{\frac{1}{3}}=2.926 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxvi)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\sqrt{49.5}$
Solution: $\text { let } f(x)=\sqrt{x} \quad \text { where } x=49$
$\text { Let } \Delta x=0.5$
$f(x+\Delta x)=\sqrt{x+\Delta x}=\sqrt{49.5}$
Now by definition approximately we can write
\begin{aligned} &f{}'(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x} \\\\ &f(x)=\sqrt{x}=\sqrt{49}=7 \\\\ &\Delta x=0.5 \\\\ &\Rightarrow f(x)=\frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{49}}=\frac{1}{14} \end{aligned}
putting values in (i) we get,
\begin{aligned} &\frac{1}{14}=\frac{\sqrt{495}-7}{0.5} \\\\ &\sqrt{49.5}=\frac{0.5}{14}+7=\frac{05+98}{14}=\frac{485}{14} \\\\ &=7.036 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxvii)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $(3.968)^{\frac{3}{2}}$
Solution: $\text { let } y=f(x)=x^{\frac{3}{2}}, x=4$
\begin{aligned} &x+\Delta x=3.968 \\\\ &\Delta x=-0.032 \\\\ &\Delta y=\left[\frac{d y}{d x}\right]_{x=4} \times \Delta x \end{aligned}
\begin{aligned} &\Delta y=\left[\frac{3}{2} x^{\frac{1}{2}}\right]_{x=4} \times \Delta x \\\\ &\Delta y=\frac{3}{2} \times 2(-0.032)(0.096) \\\\ &(3.968)^{\frac{3}{2}}=f(x+\Delta x) \end{aligned}
\begin{aligned} &=f(x)+\Delta y \\\\ &=8-0.096 \\\\ &=7.904 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxviii)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $(1.999)^{5}$
Solution: $\operatorname{let} x=2 \text { and } \Delta x=-0.001$
$\text { Let } y=x^{5}$
on differentiating both side wrt we get
\begin{aligned} &\frac{d y}{d x}=5 x^{4} \\\\ &\text { Now, } \Delta y=\frac{d y}{d x} \Delta x=5 x^{4} \times \Delta x \\\\ &=5 \times 2^{4} \times[-0.001]=-80 \times 0.001 \end{aligned}
\begin{aligned} &=-0.080 \\\\ &(1.999)^{5}=y+\Delta y=2^{5}+(-0.080) \\\\ &=32-0.080 \\\\ &=31.420 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 9 (xxix)

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\sqrt{0.082}$
Solution: $\text { let } f(x)=\sqrt{x}$
Using $f(x+\Delta x)=f(x)+\Delta x \times f(x)$
Taking $x=09 \text { and } \Delta x=-0.008$
We get $f(0.09-0.008)=f(0.09)+(-0.008)(0.09)$
\begin{aligned} &\Rightarrow \sqrt{0.082}=\sqrt{0.09}-0.0008 \times \frac{1}{2 \sqrt{0.09}} \\\\ &=0.3-\frac{0.008}{0.6} \\\\ &=0.3-0.01333=0.2867 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 10

Hint: Here we use this below formula
$f(x+\Delta x)-f(x)=\Delta y$
Given:$f(2.01), f(x)=4 x^{2}+5 x+2$
Solution: $f 2.01=f(x+\Delta x)$
$=4(x+\Delta x)^{2}+5(x+\Delta x)+2$
\begin{aligned} &\Delta y=f(x+\Delta x)-f(x) \\\\ &\Rightarrow f(x)+f(x) x \Delta x \\\\ &f(2.01)=\left(4 x^{2}+5 x+2\right)+(8 x+5) \Delta x \end{aligned}
\begin{aligned} &=\left(4(2)^{2}+5((2)+2)[8(2)+5](0.01)\right. \\\\ &=(16+10+2)+(16+5)(0.01) \\\\ &=28+0.21 \\\\ &=28.21 \end{aligned}
Hence, the approximate value of $f(2.01) \text { is } 28.21$

Differentials, errors and approximations exercise 13.1 question 11

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $x^{3}-7 x^{2}+15$
Solution: $x=5 \Delta x=0.001$
⇒ then we have,
\begin{aligned} &f(5.001)=f(x+\Delta x)=(x+\Delta x)^{2}-7(x+\Delta x)^{2} \\ &=f 15 \end{aligned}
Now, $\Delta y=f(x+\Delta x)-f(x)$
\begin{aligned} &\therefore f(x+\Delta x)=f(x)+\Delta y \\\\ &=f(x)+f(x) \Delta x \\\\ &\Rightarrow f(5.001)=\left(x^{3}-7 x^{2}+15\right)+\left(3 x^{2}-14 x\right) \Delta x \\\\ &=\left[(5)^{3}-7(5)^{2}+15\right]+\left[3(5)^{2}-14(5)\right](0.001) \end{aligned}
\begin{aligned} &=-35+(s)(0.001) \\\\ &=-35+0.005 \\\\ &=-34.995 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 12

Hint: Here we use this below formula
$\Delta y=f(x+\Delta x)-f(x)$
Given: $\log _{10} e=0.4343$
Solution: Let: $y=f(x)=\log _{10} x$
⇒ Here,
\begin{aligned} &x=1000 \\\\ &x+\Delta x=1005 \\\\ &\Delta x=5 \\\\ &d x=\Delta x=5 \end{aligned}
\begin{aligned} &\text { For } x=1000 \\\\ &y=\log _{10} 1000=\log _{10}(10)^{3}=3 \\\\ &\Rightarrow \text { Now } y=\log _{10} x=\frac{\log e x}{\log e 10} \\\\ &\frac{d y}{d x}=\frac{0.4343}{x} \end{aligned}
\begin{aligned} &\left(\frac{d y}{d x}\right)_{x=1000}=\frac{0.4343}{1000}=0.00043 \\\\ &\Delta y=d y=\frac{d y}{d x} \times d x=0.0004343 \times 5=3.0021715 \end{aligned}

Differentials, errors and approximations exercise 13.1 question 13

Answer: $2.16 \pi m^{2}$
Hint: Here, we use the surface area of the sphere (s)
$s=4 \pi r^{2}$
Given: $r=9 \mathrm{~m} \text { and } \Delta r=0.03 \mathrm{~m}$
Solution: $s=4 \pi r^{2}$
\begin{aligned} &\frac{d s}{d x}=8 \pi r \\\\ &d s=\left(\frac{d s}{d x}\right) \Delta r \\\\ &=(8 \pi r) \Delta r \end{aligned}
\begin{aligned} &=8 \pi(9)(0.03) m^{2} \\\\ &=2.16 \pi m^{2} \end{aligned}

Hence, the approximate error in circulating the surface area is $2.16 \pi m^{2}$

Differentials, errors and approximations exercise 13.1 question 14

Hint: Here, we use the formula of surface area of cube
$s=6 x^{2}$
Given: $\Delta x=1 \% \text { decrease }$
Solution:
\begin{aligned} &\Delta x=(-0.01) x \\\\ &d s=\frac{d s}{d x} \times \Delta x=6 \times 2 \Delta x \end{aligned}
\begin{aligned} &=6 \times 2 x \times(-0.01 x) \\\\ &=-0.12 x^{2} \end{aligned}
Therefore $-0.12x^{2}$ is the approximate change in surface area of cube

Differentials, errors and approximations exercise 13.1 question 15

Answer: $3.92 \pi \mathrm{m}^{3}$
Hint: Here volume $V$ the so here is given by $V=\frac{4}{3} \pi r^{3}$
Given: $r=7 \mathrm{~m} \text { and } \Delta r=0.02 \mathrm{~m}$
Solution: $\frac{d v}{d x}=4 \pi r^{2}$
\begin{aligned} &\Delta v=d v\left(\frac{d v}{d r}\right) \Delta r=4 \pi r^{2} \Delta r \\\\ &=4 \pi(7)^{2}(0.02)=3.92 \pi m^{3} \end{aligned}

Differentials, errors and approximations exercise 13.1 question 16

Answer: $0.03 x^{3} m^{3}$
Hint: Here we use the formula of a cube of side $x$ is given by $v=x^{3}$
Given: 1% increasing
Solution:$\left(\frac{d v}{d r}\right) \Delta x$
\begin{aligned} &=3 x^{2} \Delta x \\\\ &=3 x^{2} \times 0.01 x \\\\ &=0.03 x^{3} \end{aligned}
⇒ hence the approximate change in the volume of the cube is $0.03 x^{3} m^{3}$

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