AP EAPCET 2023 16th May Evening Shift Question Paper with Solutions

The function $𝑓$ is a bijection, which mean $𝑓$ is one-one and onto.

$\because 𝑓\left(𝑥\right)={𝑥}^2-3𝑥+2={(𝑥-\frac{3}{2})}^2-\frac{1}{4}$

So, domain at which the function can be one-one are $[\frac{3}{2},\infty )$ and $(-\infty ,\frac{3}{2}]$ and the range of $𝑓$ is $[-\frac{1}{4},\infty )$.

Now, it can be conclude that $𝐴=[\frac{3}{2},\infty )$ or $𝐴=(-\infty ,\frac{3}{2}]$ and $𝐵=[-\frac{1}{4},\infty )$

Given, $𝑓(𝑥+𝑦)=𝑓\left(𝑥\right)+𝑓\left(𝑦\right)\forall 𝑥,𝑦\in 𝑅$ and $𝑓\left(1\right)=7$

Now, $𝑓(𝑥+1)=𝑓\left(𝑥\right)+𝑓\left(1\right)$

$\begin{array}{cc}\Rightarrow & 𝑓(𝑥+1)=𝑓\left(𝑥\right)+7\\ \Rightarrow & 𝑓(1+𝑟)=7+𝑓\left(𝑟\right)\end{array}$

$\begin{array}{cc}& \therefore \sum _{𝑟=1}^{𝑛}𝑓\left(𝑟\right)=𝑓\left(1\right)+𝑓\left(2\right)+\dots +𝑓\left(𝑛\right)\\ & =7+(7+7)+(7+7+7)+\dots +\left(\underset{𝑛\text{-times }}{\underset{}{7+7+7+\dots +7}}\right)\\ & =7+7\left(2\right)+7\left(3\right)+\dots +7\left(𝑛\right)\\ & =7[1+2+3+\dots +𝑛]\\ & =7\left[\frac{𝑛(𝑛+1)}{2}\right]=\frac{7𝑛(𝑛+1)}{2}\end{array}$

Given, $𝐴=\left[\begin{array}{cc}-\cot 𝜃& \mathrm{cosec}𝜃\\ \mathrm{cosec}𝜃& -\cot 𝜃\end{array}\right]$

$\Rightarrow \left|𝐴\right|={\cot }^2𝜃-{\mathrm{cosec}}^2𝜃=-1$

Now, ${𝐴}^{-1}=\frac{1}{\left|𝐴\right|}(\mathrm{adj}𝐴)$

$=\frac{1}{-1}\left[\begin{array}{cc}-\cot 𝜃& -\mathrm{cosec}𝜃\\ -\mathrm{cosec}𝜃& -\cot 𝜃\end{array}\right]$

Given, ${𝐴}^{-1}=𝐴$ at $𝜃={𝜃}_1$

$\begin{array}{cc}& \\ & \Rightarrow \left[\begin{array}{cc}\cot {𝜃}_1& \mathrm{cosec}{𝜃}_1\\ \mathrm{cosec}{𝜃}_1& \cot {𝜃}_1\end{array}\right]=\left[\begin{array}{c}-\cot {𝜃}_1\mathrm{cosec}{𝜃}_1\\ \mathrm{cosec}{𝜃}_1-\cot {𝜃}_1\end{array}\right]\\ & \Rightarrow \cot {𝜃}_1=-\cot {𝜃}_1\Rightarrow 2\cot {𝜃}_1=0\\ & \Rightarrow \cot {𝜃}_1=0\\ & \Rightarrow {𝜃}_1=\frac{𝜋}{2}\end{array}$

Given, ${𝐴}^{-1}+𝐴=0$ at $𝜃={𝜃}_2$

$\begin{array}{cc}& \left(\begin{array}{cc}\cot {𝜃}_2& \mathrm{cosec}{𝜃}_2\\ \mathrm{cosec}{𝜃}_2& \cot {𝜃}_2\end{array}\right)+\left(\begin{array}{cc}-\cot {𝜃}_2& \mathrm{cosec}{𝜃}_2\\ \mathrm{cosec}{𝜃}_2& -\cot {𝜃}_2\end{array}\right)\\ & =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ & \Rightarrow \left[\begin{array}{cc}0& 2\mathrm{cosec}{𝜃}_2\\ 2\mathrm{cosec}{𝜃}_2& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

Which is an incorrect statement because $0\ne 1$.

So, such ${𝜃}_2$ does not exist.

Given $𝐴={\left[{𝑎}_{𝑖𝑗}\right]}_{3\times 3}$ such that

${𝑎}_{𝑖𝑗}=\frac{1}{3}|𝑖-5𝑗|$

So, ${𝑎}_{11}=\frac{1}{3}|1-5(1\left)\right|=\frac{1}{3}|1-5|=\frac{1}{3}|-4|$

$=\frac{1}{3}\left(4\right)=\frac{4}{3}$

$\begin{array}{cc}{𝑎}_{12}& =\frac{1}{13}|1-5(2\left)\right|=\frac{1}{3}|1-10|=\frac{1}{3}|-9|\\ & =\frac{1}{3}\left(9\right)=3\\ {𝑎}_{13}& =\frac{1}{3}|1-5(3\left)\right|=\frac{1}{3}|1-15|=\frac{1}{3}|-14|\\ & =\frac{1}{3}\left(14\right)=\frac{14}{3}\\ {𝑎}_{21}& =\frac{1}{3}|2-5(1\left)\right|=\frac{1}{3}|2-5|=\frac{1}{3}|-3|\\ & =\frac{1}{3}\left(3\right)=1\end{array}$

$\begin{array}{cc}{𝑎}_{22}& =\frac{1}{3}|2-5(2\left)\right|=\frac{1}{3}|2-10|=\frac{1}{3}|-8|\\ & =\frac{1}{3}\left(8\right)=\frac{8}{3}\\ {𝑎}_{23}& =\frac{1}{3}|2-5(3\left)\right|=\frac{1}{3}|2-15|=\frac{1}{3}|-13|\\ & =\frac{1}{3}\left(13\right)=\frac{13}{3}\\ {𝑎}_{31}& =\frac{1}{3}|3-5(1\left)\right|=\frac{1}{3}|3-5|=\frac{1}{3}|-2|\\ & =\frac{1}{3}\left(2\right)=\frac{2}{3}\\ {𝑎}_{32}& =\frac{1}{3}|3-5(2\left)\right|=\frac{1}{3}|3-10|=\frac{1}{3}|-7|\\ & =\frac{1}{3}\left(7\right)=\frac{7}{3}\\ {𝑎}_{33}& =\frac{1}{3}|3-5(3\left)\right|=\frac{1}{3}|3-15|=\frac{1}{3}|-12|\\ & =\frac{1}{3}\left(12\right)=4\end{array}$

$\text{ Thus, }𝐴=⎡⎣⎢\begin{array}{ccc}4/3& 3& 14/3\\ 1& 8/3& 13/3\\ 2/3& 7/3& 4\end{array}⎤⎦⎥$

Given,

$\begin{array}{cc}& ⎡⎣⎢\begin{array}{ccc}3{𝑥}^2+𝑥+3& 2{𝑥}^2-𝑥+4& 7{𝑥}^2+8𝑥+5\\ 5{𝑥}^2+3𝑥+2& 4{𝑥}^2-2𝑥-1& 7{𝑥}^2+5𝑥+8\\ 3{𝑥}^2+2𝑥+5& 4{𝑥}^2-𝑥-2& 3{𝑥}^2+8𝑥+7\end{array}⎤⎦⎥\\ & =𝑃{𝑥}^2+𝑄𝑥+𝑅\\ & \text{ SO, }𝑅=⎡⎣⎢\begin{array}{ccc}3& 4& 5\\ 2& -1& 8\\ 5& -2& 7\end{array}⎤⎦⎥\end{array}$

Now, $\det 𝑅=3(-7+16)-4(14-40)+5(-4+5)$

$\begin{array}{cc}& =3\left(9\right)-4(-26)+5\left(1\right)\\ & =27+104+5\\ & =136\end{array}$

Given, $𝐴(-2+𝑖)$ and $𝐵(3-4𝑖)$ and $\frac{𝐴𝐶}{𝐶𝐵}=\frac{1}{2}$, where $𝐶$ is a point lie. On the straight line joining the points $𝐴$ and $𝐵$.

$\begin{array}{cc}\text{ So, }𝐶& =\frac{1\left(3\right)+2(-2)}{1+2}+\frac{1(-4)+2\left(1\right)}{1+2}𝑖\\ & =-\frac{1}{3}-\frac{2}{3}𝑖\end{array}$

The point $𝐶$ lie on quadrant III.

$\begin{array}{cc}\text{ So, Arg }𝐶& ={\tan }^{-1}\left(\frac{-2/3}{-1/3}\right)-𝜋\\ & ={\tan }^{-1}\left(2\right)-𝜋\end{array}$

Given, ${𝑧}_1=4+3𝑖$ and $\left|{𝑧}_1\right|=𝛼$

Now, ${\overline{𝑧}}_1=4-3𝑖$ and

$𝛼=\left|{𝑧}_1\right|=\sqrt{\surd 4^2+3^2}=5$

The distance between $𝑧$ and ${\overline{𝑧}}_1$ is 6 units. The region $|𝑧-{\overline{𝑧}}_1|\le 𝛼$ represents the collection of all those points which lies either on the circle $𝐶$ or inside of the circle $𝐶$, where $𝐶$ is a circle centred at ${\overline{𝑧}}_1$ and having the radius 5 .

Clearly, ${𝑧}_1$ does not lie in the region

$|𝑧-\overline{𝑧}|\le 𝛼$

Given, ${𝑧}_1,{𝑧}_2,{𝑧}_3$ are the vertices of an equilateral triangle and $𝑧$ is its circumcenter.

So, $|𝑧-{𝑧}_1|=|𝑧-{𝑧}_2|=|𝑧-{𝑧}_3|=$ Radius of circum, circle

$\Rightarrow \frac{|𝑧-{𝑧}_1|}{|𝑧-{𝑧}_2|}=\frac{|𝑧-{𝑧}_3|}{|{𝑧}_1-{𝑧}_1|}$

Given, $𝑎$ and $𝑏$ are the roots of the equation ${𝑦}^2+𝑦+1=0$

So, $𝑎+𝑏=-\frac{1}{1}=-1$ and $𝑎𝑏=\frac{1}{1}=1$

Now, $(𝑎+𝑏{)}^2={𝑎}^2+{𝑏}^2+2𝑎𝑏$

$\begin{array}{cc}& \Rightarrow (-1{)}^2={𝑎}^2+{𝑏}^2+2(1)\\ & \Rightarrow {𝑎}^2+{𝑏}^2=1-2=-1\end{array}$

Also, $(𝑎+𝑏{)}^4={𝑎}^4+4{𝑎}^3𝑏+6{𝑎}^2{𝑏}^2+4𝑎{𝑏}^3+{𝑏}^4$

$\begin{array}{cc}& \Rightarrow (𝑎+𝑏{)}^4={𝑎}^4+{𝑏}^4+4𝑎𝑏(𝑎+𝑏)+6(𝑎𝑏{)}^2\\ & \Rightarrow {𝑎}^4+{𝑏}^4=(𝑎+𝑏{)}^4-4𝑎𝑏(𝑎+𝑏)-6(𝑎𝑏{)}^2\\ & \Rightarrow {𝑎}^4+{𝑏}^4+{𝑎}^{-1}{𝑏}^{-1}=(𝑎+𝑏{)}^4-4𝑎𝑏(𝑎+𝑏)-6(𝑎𝑏{)}^2+(𝑎𝑏{)}^{-1}\\ & =(-1{)}^4-4(1\left)\right(-1)-6(1{)}^2+(1{)}^{-1}\\ & =1+4-6+1=6-6=0\end{array}$

Given, ${𝑥}^2+𝑥-6$ is a factor of

$2{𝑥}^3+{𝑥}^2+𝑎𝑥+𝑏$

So, the solutions of the equation ${𝑥}^2+𝑥-6=0$ will be the solutions of the equation

$2{𝑥}^3+{𝑥}^2+𝑎𝑥+𝑏=0...\left(𝑖\right)$

Now, ${𝑥}^2+𝑥-6=0$

$\begin{array}{cc}\Rightarrow & (𝑥+3)(𝑥-2)=0\\ \Rightarrow & 𝑥=2-3\end{array}$

On putting $𝑥=2$ and $𝑥=-3$ into the

Eq. (i), we get

$\begin{array}{cc}& 2𝑎+𝑏+20=0...\left(𝑖𝑖\right)\\ & -3𝑎+𝑏-45=0...\left(𝑖𝑖𝑖\right)\end{array}$

On solving Eqs. (ii) and (iii), we get

$𝑎=-13\text{ and }𝑏=6$

Now, $6𝑎+13𝑏=6(-13)+13\left(6\right)$

$=-78+78=0$

Given, $𝑐$ and $𝑑$ are the roots of

${𝑥}^2+𝑎𝑥+𝑏=0.$

So, $𝑐+𝑑=-\frac{𝑎}{1}=-𝑎$ and $𝑐𝑑=\frac{𝑏}{1}=𝑏$

On putting $𝑎=-(𝑐+𝑑)$ and $𝑏=𝑐𝑑$ into the equation

${𝑥}^2+(4𝑐+𝑎)𝑥+(𝑏+2𝑎𝑐+4{𝑐}^2)=0,$

we get

$\begin{array}{c}{𝑥}^2+(4𝑐-𝑐-𝑑)𝑥+(𝑐𝑑+2\\ (-𝑐-𝑑)𝑐+4{𝑐}^2)=0\\ \Rightarrow {𝑥}^2+(3𝑐-𝑑)𝑥+(𝑐𝑑-2{𝑐}^2-2𝑐𝑑+4{𝑐}^2)=0\\ \Rightarrow {𝑥}^2+(3𝑐-𝑑)𝑥+(2{𝑐}^2-𝑐𝑑)=0\dots \text{ (i) }\end{array}$

Now, it can be concluded that $𝑑-2𝑐$ and - $𝑐$ are the root of Eq. (i), because,

$\begin{array}{cc}& (𝑑-2𝑐)+(-𝑐)=-\frac{(3𝑐-𝑑)}{1}\\ & \text{ and }(𝑑-2𝑐)(-𝑐)=\frac{2{𝑐}^2-𝑐𝑑}{1}\end{array}$

Hence, option (d) is correct.

The equation,

${𝑥}^4+7{𝑥}^3+18{𝑥}^2+20𝑥+8=0\text{ can be }$

factored as

$\begin{array}{cc}& {𝑥}^4+7{𝑥}^3+18{𝑥}^2+20𝑥+8=0\\ & \Rightarrow (𝑥+2{)}^3(𝑥+1)=0\\ & \Rightarrow 𝑥=-2,-2,-2,-1\end{array}$

Given equation is

$2{𝑥}^3+5{𝑥}^2-4𝑥-12=0$

⇒ Let $𝑎,𝑎,𝑏$ are root of

$2{𝑥}^3+5{𝑥}^2-4𝑥-12=0$

$\text{Now,}2𝑎+𝑏=-\frac{5}{2}...\left(𝑖\right)$

$\begin{array}{cc}& {𝑎}^2+2𝑎𝑏=-\frac{4}{2}=-2...\left(𝑖\right)\\ {𝑎}^2𝑏=& -\frac{(-12)}{2}=6...\left(𝑖𝑖𝑖\right)\end{array}$

On solving Eqs. (i), (ii) and (iii), we get $𝑎=-2,𝑏=3/2$

Now, the quadratic equation whose roots are $-2,\frac{3}{2}$ be $(𝑥-(-2\left)\right)(𝑥-\frac{3}{2})=0$

On simplifying the equation $(𝑥-(-2\left)\right)(𝑥-\frac{3}{2})=0$, we get the equation $2{𝑥}^2+𝑥-6=0$.

So, the required constant term be -6 .

Given, $𝑟$ is the greatest integer such that ${30}^{𝑟}$ divides 30 !

Factorisation of 30 is $2\times 3\times 5$.

Now, 30! $=30\times 29\times 28\times \dots \times 3\times 2\times 1$

$=(2\times 3\times 5)\times \left(29\right)\times (2^2\times 7)\times \left(3^3\right)$

$\begin{array}{c}\times (2\times 13)\times (5\times 5)\times (2^3\times 3)\times \left(23\right)\times (2\times 11)\\ \times (3\times 7)\times (2^2\times 5)\times \left(19\right)\times (2\times 3^2)\times \left(17\right)\end{array}$

$\left(2^4\right)\times (3\times 5)\times (2\times 7)\times \left(13\right)\times (2^2\times 3)\times \left(11\right)$

$\times (2\times 5)\times \left(3^2\right)\times \left(2^3\right)\times \left(7\right)\times (2\times 3)$

$\left(5\right)\times \left(2^2\right)\times \left(3\right)\times \left(2\right)\times \left(1\right)$

$=2^{26}\times 3^{14}\times 5^7\times {11}^2\times {13}^2$

$\times 17\times 19\times 23\times 29$

$=(2\times 3\times 5{)}^7\times 2^{19}\times 3^7\times {11}^2\times {13}^2\times 17$

$\times 19\times 23\times 29$

$=(30{)}^7\times 2^{19}\times 3^7\times {11}^2\times {13}^2\times 17\times 19$

$\times 23\times 29$

Thus, value of $𝑟$ be 7 .

$\begin{array}{cc}& \text{ The coefficient of }{𝑥}^3\text{ in the }\\ & \begin{array}{cc}& \text{ expansion of }(1-2𝑥{)}^{\frac{1}{2}}(1+3𝑥{)}^{-\frac{1}{3}}\\ & =\frac{1\times (-\frac{1}{3})(-\frac{1}{3}-1)(-\frac{1}{3}-2)}{3!}(3{)}^3\end{array}\end{array}$

$\begin{array}{cc}& +\frac{1}{2}(-2)\times \frac{(-\frac{1}{3})(-\frac{1}{3}-1)}{2!}\\ & (3{)}^2+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-2{)}^2\times (-\frac{1}{3})\left(3\right)\\ & +\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-2{)}^3\times 1\\ =-& \frac{14}{3}-2+\frac{1}{2}-\frac{1}{2}\\ =-& \frac{20}{3}\end{array}$

In the word ARRANGEMENT there are 2A, 2E, 2R, 2N, 1G, 1T and 1M.

The total number of arrangement is $\frac{11!}{2!2!2!2!}$

The total number of arrangement in which 2E comes together is $\frac{10!}{2!2!2!}$

Now, total required arrangement be

$\begin{array}{cc}& \frac{11!}{2!2!2!2!}-\frac{10!}{2!2!2!}\\ =& \frac{11\times 10!}{16}-\frac{10!}{8}=(\frac{11}{16}-\frac{1}{8})10!=\frac{9}{16}(10!)\end{array}$

The number of letters in the word VOWEL is 5 in which O and E are vowels rest are constants.

Consider O and E as a single letter, which can be written as OE and EO.

So, required numbers of arrangement $=4!\times 2=24\times 2=48$

Given, ${}^{20-2𝑥}{𝐶}_{𝑥-3}\in 𝑁$ and $𝑥\in 𝑁$

$\begin{array}{cc}& {\because }^{20-2𝑥}{𝐶}_{𝑥-3}\in 𝑁\\ & \therefore 20-2𝑥\in 𝑁\text{ and }0\le 𝑥-3\le 20-2𝑥\\ & \Rightarrow 𝑥\in \{1,2,3,\dots ,9\}\text{ and }3\le 𝑥\le \frac{23}{3}[\because 𝑥\in 𝑁]\\ & \Rightarrow 𝑥\in \{3,4,5,6,7\}\end{array}$

Thus, the number of elements in the given set will be 5 .

$\begin{array}{cc}& \text{Given, }\frac{\cot 𝐴}{1-\tan 𝐴}+\frac{\tan 𝐴}{1-\cot 𝐴}\\ & \begin{array}{cc}& =\frac{{\cos }^2𝐴}{\sin 𝐴(\cos 𝐴-\sin 𝐴)}\\ & +\frac{{\sin }^2𝐴}{\cos 𝐴(\sin 𝐴-\cos 𝐴)}\\ & =\frac{{\cos }^3𝐴-{\sin }^3𝐴}{(\sin 𝐴\cos 𝐴)(\cos 𝐴-\sin 𝐴)}\\ & =\frac{(\cos 𝐴-\sin 𝐴)({\cos }^2𝐴+{\sin }^2𝐴+\cos 𝐴\sin 𝐴)}{(\sin 𝐴\cos 𝐴)(\cos 𝐴-\sin 𝐴)}\\ & =\frac{1+\cos 𝐴\sin 𝐴}{\sin 𝐴\cos 𝐴}\\ & =\sec 𝐴\mathrm{cosec}𝐴+1\end{array}\end{array}$

We know that,

$\begin{array}{cc}& \tan 3𝜃=\frac{3\tan 𝜃-{\tan }^3𝜃}{1-3{\tan }^2𝜃}\\ & \Rightarrow (\tan 3𝜃)(1-3{\tan }^2𝜃)=(3\tan 𝜃-{\tan }^3𝜃)\end{array}$

On taking square on both sides, we get

$\begin{array}{cc}& \begin{array}{cc}& {\tan }^{2}3𝜃[1+9{\tan }^4𝜃-6{\tan }^2𝜃]\\ =& 9{\tan }^2𝜃+{\tan }^6𝜃-6{\tan }^4𝜃\\ \Rightarrow & 3+27{\tan }^4𝜃-18{\tan }^2𝜃=9{\tan }^2𝜃+{\tan }^6𝜃-6{\tan }^4𝜃\end{array}\\ & \begin{array}{cc}& [\because 𝜃=\frac{𝜋}{9}\Rightarrow 3𝜃=\frac{𝜋}{3}\Rightarrow \tan 3𝜃=\sqrt{\surd 3}\\ & \Rightarrow {\tan }^{2}3𝜃=3]\end{array}\\ & \Rightarrow 1+27{\tan }^2𝜃-33{\tan }^4𝜃+{\tan }^6𝜃=4\end{array}$

We know that,

$\begin{array}{cc}& \cot {18}^{\circ }=\frac{\sqrt{\surd 10+2\sqrt{\surd 5}}}{\sqrt{\surd 5}-1}\text{ and }\\ & \cot {36}^{\circ }=\frac{\sqrt{\surd 5}+1}{\sqrt{\surd 10-2\sqrt{\surd 5}}}\end{array}$

$\begin{array}{cc}𝑆𝑜,\cot {18}^{\circ }\cdot \cot {36}^{\circ }+1=& \frac{\sqrt{\surd 10+2\sqrt{\surd 5}}}{\sqrt{\surd 5}-1}\times \frac{\sqrt{\surd 5}+1}{\sqrt{\surd 10-2\sqrt{\surd 5}}}+1\end{array}$

$\begin{array}{cc}& =\sqrt{\surd \frac{10+2\sqrt{\surd 5}}{10-2\sqrt{\surd 5}}\times \frac{10+2\sqrt{\surd 5}}{10+2\sqrt{\surd 5}}}\times \frac{\sqrt{\surd 5}+1}{\sqrt{\surd 5}-1}\times \frac{\sqrt{\surd 5}+1}{\sqrt{\surd 5}+1}+1\\ & =\frac{10+2\sqrt{\surd 5}}{\sqrt{\surd 100-20}}\times \frac{(\sqrt{\surd 5}+1{)}^2}{5-1}+1\\ & =\frac{10+2\sqrt{\surd 5}}{\sqrt{\surd 80}}\times \frac{6+2\sqrt{\surd 5}}{4}+1\\ & =\frac{60+20+32\sqrt{\surd 5}}{16\sqrt{\surd 5}}+1\\ & =\frac{80+32\sqrt{\surd 5}}{16\sqrt{\surd 5}}+1=\frac{5}{\sqrt{\surd 5}}+2+1\\ & =3+\sqrt{\surd 5}\end{array}$

Given, ${\cos }^3𝑥\sin 4𝑥=\sum _{𝑥}^{𝑟=0}{𝑐}_{𝑟}\sin 𝑟𝑥$

$\begin{array}{cc}& \Rightarrow {\cos }^3𝑥\sinh 𝑥=\left(\frac{\cos 3𝑥+3\cos 𝑥}{4}\right)\sin 4𝑥\\ & =\frac{1}{8}[2\sin 4𝑥\cos 3𝑥+3\cdot 2(\sin 4𝑥\cos 𝑥\left)\right]\\ & =\frac{1}{8}[\sin 7𝑥+\sin 𝑥+3\sin 5𝑥+3\sin 3𝑥]\\ & =\frac{1}{8}[\sin 𝑥+3\sin 3𝑥+3\sin 5𝑥+\sin 7𝑥]\end{array}$

clearly, ${𝑎}_1=\frac{1}{8},{𝑎}_3=\frac{3}{8},{𝑎}_5=\frac{3}{8},{𝑎}_7=\frac{1}{8}$