Exercise 7.4 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.4$
Question 1.
Write the Maclaurin series expansion of the following functions:
(i) $\mathrm{e}^{\mathrm{x}}$
(ii) $\sin x$
(iii) $\cos x$
(iv) $\log (1-\mathrm{x}) ;-1 \leq \mathrm{x}<1$
(v) $\tan ^{-1}(\mathrm{x}) ;-1 \leq \mathrm{x} \leq 1$
(vi) $\cos ^2 \mathrm{x}$
Solution:
(i)
$
\begin{aligned}
& f(x)=e^x \quad ; \quad f(0)=e^0=1 \\
& f^{\prime}(x)=e^x \quad ; \quad f^{\prime}(0)=1 \\
& f^{\prime \prime}(x)=e^x \quad ; \quad f^{\prime \prime}(0)=1 \\
& f(x)=e^x=1+\frac{1 \cdot x}{1 !}+\frac{1}{2 !} x^2+\frac{1}{3 !} x^3 \ldots \\
& =1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !} \ldots \text { holds for all } x \\
&
\end{aligned}
$

(ii)
$
\begin{aligned}
& f(x)=\sin x ; \quad f(0)=0 \\
& f^{\prime}(x)=\cos x ; \quad f^{\prime}(0)=1 \\
& f^{\prime \prime}(x)=-\sin x ; \quad f^{\prime \prime}(0)=0 \\
& f^{\prime \prime \prime}(x)=-\cos x ; \quad f^{\prime \prime \prime}(0)=-1 \\
& f^4(x)=\sin x ; \quad f^4(0)=0 \\
& f^5(x)=\cos x ; \quad f^5(0)=1 \\
&
\end{aligned}
$
The Maclaurin expansion of $f(x)$ is
$
\begin{aligned}
f(x) & =f(0)+\frac{x^1}{\lfloor} f^{\prime}(0)+\frac{x^2}{\lfloor 2}+f^{\prime \prime}(0)+\ldots \\
f(x) & =\sin x=\frac{x^1}{\lfloor}(1)+\frac{x^3}{\lfloor}(-1)+\frac{x^5}{\lfloor}(1)+\ldots \\
& =x-\frac{x^3}{\lfloor 3}+\frac{x^5}{\lfloor 5}-\ldots
\end{aligned}
$

(iii)
$
\begin{aligned}
& f(x)=\cos x ; \quad f(0)=1 \\
& f^{\prime}(x)=-\sin x ; \quad f^{\prime}(0)=0 \\
& f^{\prime \prime}(x)=-\cos x ; \quad f^{\prime \prime}(0)=-1 \\
& f^3(x)=\sin x ; \quad f^3(0)=0 \\
& f^4(x)=\cos x ; \quad f^4(0)=1 \\
&
\end{aligned}
$
The Maclaurin's expansion $f(x)=f(0)+\frac{x^1}{\lfloor} f^{\prime}(0)+\frac{x^2}{\lfloor 2} f^{\prime \prime}(0)+\ldots$.
$
\begin{aligned}
f(x) & =\cos x=1+\frac{x^2}{\lfloor 2}(-1)+\frac{x^4}{\lfloor 4}(1)-\ldots \\
& =1-\frac{x^2}{\lfloor 2}+\frac{x^4}{\lfloor 4}-\ldots
\end{aligned}
$

(iv)
$
\begin{aligned}
f(x) & =\log (1-x) \\
f(0) & =0 \\
f^{\prime}(x) & =\frac{1}{1-x}(-1)=\frac{-1}{1-x} \\
f^{\prime}(0) & =-1 \\
f^{\prime \prime}(x) & =-\frac{-1}{(1-x)^2}(-1)=\frac{-1}{(1-x)^2} \\
f^{\prime \prime}(0) & =-1 \\
f^{\prime \prime}(x) & =-\left(\frac{-2}{(1-x)^3}(-1)\right)=\frac{-2}{(1-x)^3} \\
f^{\prime \prime}(0) & =-2
\end{aligned}
$
$f^{\prime \prime \prime}(0)=-2$ The Maclaurin's expansion of $f(x)$ is $f(x)=f(0)+\frac{x^1}{\lfloor 1} f^{\prime}(0)+\frac{x^2}{\lfloor 2} f^{\prime \prime}(0)+\ldots$.
Here $f(x)=\log (1-x)$
So $\log (1-x)=0+\frac{x}{\lfloor}(-1)+\frac{x^2}{\lfloor 2}(-1)+\frac{x^4}{\lfloor 4}(-2)$
$
=-x-\frac{x^2}{2}-\frac{x^3}{3} \cdots
$

(v)
$
\begin{aligned}
f(x) & =\tan ^{-1} x ; f(0)=0 \\
f^{\prime}(x) & =\frac{1}{1+x^2}=1-x^2+x^4-x^6 \ldots ; f^{\prime}(0)=1=1 ! \\
f^{\prime \prime}(x) & =-2 x+4 x^3-6 x^5 \ldots ; f^{\prime \prime}(0)=0 \\
f^{\prime \prime \prime}(x) & =-2+12 x^2-30 x^4 \ldots ; f^{\prime \prime \prime}(0)=-2=-(2 !) \\
f^{i v}(x) & =24 x-120 x^3 \ldots ; f^{i v}(0)=0 \\
f^v(x) & =24-360 x^2 \ldots ; f^v(0)=24=4 ! \\
\tan ^{-1} x & =0+\frac{1}{1 !} x+\frac{0}{2 !} x^2-\frac{2}{3 !} x^3+\frac{0}{4 !} x^4+\frac{4 !}{5 !} x^5+\ldots . \\
& =x-\frac{1}{3} x^3+\frac{1}{5} x^5-\ldots .
\end{aligned}
$

(vi) $f(x)=\cos ^2 x$
$f(0)=1$
$f^{\prime}(x)=2 \cos x(-\sin x)=-\sin 2 x$
$f^{\prime}(0)=0$
$f^{\prime}(x)=(-\cos 2 x)(2)$
$\mathrm{f}^{\prime \prime}(0)=-2$
$f^{\prime \prime}(x)=-2[-\sin 2 x](2)=4 \sin 2 x$
$\mathrm{f}^{\prime \prime}(0)=0$
$f^4(x)=4(\cos 2 x)(2)=8 \cos 2 x$
$f^4(0)=8$

Now the Maclaurin's expansion of $f(x)$ is $f(x)=f(0)+\frac{x^1}{\lfloor 1} f^{\prime}(0)+\frac{x^2}{\lfloor 2} f^{\prime \prime}(0)+\ldots .$.
(i.e.,)
$
\begin{aligned}
& \cos ^2 x=1+\frac{x}{\lfloor}(0)+\frac{x^2}{\lfloor 2}(-2)+\frac{x^3}{\lfloor 3}(0)+\frac{x^4}{\lfloor 4}(8) \ldots \\
& \cos ^2 x=1-\frac{2 x^2}{\lfloor 2}+\frac{2^3 x^4}{\lfloor 4} \ldots
\end{aligned}
$

Question 2.
Write down the Taylor series expansion, of the function $\log \mathrm{x}$ about $\mathrm{x}=1$ upto three non-zero terms for $\mathrm{x}$ $>0$.
Solution:
$
\begin{aligned}
& f(x)=f(a)+(x-a) f^{\prime}(a)+\frac{(x-a)^2}{\lfloor 2} f^{\prime \prime}(a) \ldots \\
& \text { Here } f(x)=\log x \text { and } a=1 \\
& \text { so } f(x)=\log x \Rightarrow f(1)=0 \\
& f^{\prime}(x)=\frac{1}{x} \Rightarrow f^{\prime}(1)=1 \\
& f^{\prime \prime}(x)=\frac{-1}{x^2} \Rightarrow f^{\prime \prime}(1)=-1 \\
& f^{\prime \prime \prime}(x)=-\left(\frac{-2}{x^3}\right)=\frac{2}{x^3} \Rightarrow f^{\prime \prime \prime}(1)=2 \\
& =(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\ldots \\
&
\end{aligned}
$
Question 3.
Expand $\sin \mathrm{x}$ in ascending powers $\mathrm{x}-\frac{\pi}{4}$ upto three non-zero terms.
Solution:
$
f(x)=\sin x
$

$\begin{aligned}
& x=\left(x-\frac{\pi}{4}+\frac{\pi}{4}\right) \\
& \therefore f\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\
& f^{\prime}(x)=\cos x \quad \Rightarrow \quad f^{\prime}\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\
& f^{\prime \prime}(x)=-\sin x \Rightarrow f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \\
& f^{\prime \prime \prime}(x)=-\cos x \Rightarrow f^{\prime \prime \prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \\
& f^4(x)=\sin x \Rightarrow f^4\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\
& \sin x=\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{\left\lfloor\frac{1}{\sqrt{2}}\right.}\left(\frac{1}{\sqrt{2}}\right)+\frac{\left(x-\frac{\pi}{4}\right)^2}{\sqrt{2}}\left(-\frac{1}{\sqrt{2}}\right)-\ldots . \\
& =\frac{1}{\sqrt{2}}\left\{1+\frac{\left(x-\frac{\pi}{4}\right)}{\lfloor}-\frac{\left(x-\frac{\pi}{4}\right)^2}{\lfloor 2}-\ldots .\right. \\
& =\frac{\sqrt{2}}{2}\left[1+\frac{\left(x-\frac{\pi}{4}\right)}{\lfloor 1}-\frac{\left(x-\frac{\pi}{4}\right)^2}{\lfloor 2}-\ldots\right] \\
&
\end{aligned}$

Question 4.
Expand the polynomial $f(x)=x^2-3 x+2$ in powers of $x-1$

Solution:
$
\begin{aligned}
& f(x)=x^2-3 x+2=(x-1)(x-2) \\
& f(1)=0 \\
& f^{\prime}(x)=2 x-3 ; f^{\prime}(1)=-1 \\
& \mathrm{f}^{\prime}(\mathrm{x})=2 ; \mathrm{f}^{\prime}(1)=2 \\
& f(x)=f(1)+\frac{x-1}{\lfloor} f^{\prime}(1)+\frac{(x-1)^2}{\lfloor 2} f^{\prime \prime}(1) \ldots \\
& =0+\frac{x-1}{\lfloor}(-1)+\frac{(x-1)^2}{\lfloor 2}(2) \ldots . \\
& =-(x-1)+(x-1)^2 \\
&
\end{aligned}
$