TG EAPCET 2023 13th May Evening Shift Question Paper with Solutions

Consider the function $𝑓\left(𝑥\right)=\sqrt{\surd 16-{𝑥}^2}$

Clearly, $𝑓:[0,4]\to [0,4]$ defined by $𝑓\left(𝑥\right)=\sqrt{\surd 16-{𝑥}^2}$ is a bijection.

Given, $𝑓\left(𝑥\right)=\frac{\sqrt{\surd \left|𝑥\right|-𝑥}}{\sqrt{\surd 𝑥-\left[𝑥\right]}}$

Case I $\left|𝑥\right|-𝑥\ge 0$

$\therefore 𝑥\in 𝑅...\left(𝑖\right)$

Case II $𝑥-\left[𝑥\right]>0$

$\therefore \left\{𝑥\right\}>0...\left(𝑖𝑖\right)$

From Eqs. (i) and (ii), we get

Domain $=𝑅-𝑍$

For interval $𝑥\in (-\infty ,-1)$

Range of ( $2𝑥-3$ )

$\Rightarrow (-\infty ,-5)...\left(𝑖\right)$

For interval $𝑥\in [-1,1]$

Range of $1-{𝑥}^2$

$\Rightarrow [0,1]...\left(𝑖𝑖\right)$

For interval $𝑥\in (1,\infty )$

Range of $3{𝑥}^2+2$

$\begin{array}{cc}& \Rightarrow (5,\infty )...\left(𝑖𝑖𝑖\right)\\ & (-\infty ,-5)\cup [0,1]\cup (5,\infty )\end{array}$

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}𝑏& 𝑎& 0\\ 𝑐& 0& 𝑏\\ 𝑎& 𝑎& 𝑏\end{array}⎤⎦⎥,𝐵=⎡⎣⎢\begin{array}{ccc}0& 𝑎& 𝑏\\ 𝑏& 0& 𝑐\\ 𝑏& 𝑎& 𝑎\end{array}⎤⎦⎥\\ 𝐴𝐵& =⎡⎣⎢\begin{array}{ccc}𝑏& 𝑎& 0\\ 𝑐& 0& 𝑏\\ 𝑎& 𝑎& 𝑏\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}0& 𝑎& 𝑏\\ 𝑏& 0& 𝑐\\ 𝑏& 𝑎& 𝑎\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}0+𝑎𝑏+0& 𝑎𝑏+0+0& {𝑏}^2+𝑐𝑎+0\\ 0+0+{𝑏}^2& 𝑎𝑐+0+𝑎𝑏& 𝑏𝑐+0+𝑎𝑏\\ 0+𝑎𝑏+{𝑏}^2& {𝑎}^2+0+𝑎𝑏& 𝑎𝑏+𝑎𝑐+𝑎𝑏\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}𝑎𝑏& 𝑎𝑏& {𝑏}^2+𝑎𝑐\\ 𝑎𝑏+{𝑏}^2& 𝑎𝑐+𝑎𝑏& 𝑏𝑐+𝑎𝑏\\ 𝑎& 𝑎𝑏+𝑎𝑐+𝑎𝑏\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}2& 2& 7\\ 1& 8& 5\\ 3& 6& 10\end{array}⎤⎦⎥\end{array}$

${𝑏}^2=1,𝑎𝑏=2$

$\begin{array}{cc}\therefore & {𝑎}^2+𝑎𝑏=6\\ \Rightarrow & {𝑎}^2+2=6\Rightarrow {𝑎}^2=4\end{array}$

$\begin{array}{cccc}\Rightarrow & & 𝑎& =\pm 2\\ & \therefore & {𝑏}^2+𝑎𝑐& =7\\ \Rightarrow & & 1+𝑎𝑐& =7\Rightarrow 𝑎𝑐=6\end{array}$

$\begin{array}{cc}𝑐& =\frac{6}{\pm 2}=\pm 3\Rightarrow {𝑐}^2=9\\ \therefore {𝑎}^2+{𝑏}^2+{𝑐}^2& =9+1+4=14\end{array}$

$𝐴=⎡⎣⎢\begin{array}{ccc}1& 𝑎& 3\\ 𝑏& 2& 𝑐\\ 3& 𝑑& 4\end{array}⎤⎦⎥$

Since, ${𝐴}^{𝑇}=𝐴$

$\begin{array}{c}⎡⎣⎢\begin{array}{ccc}1& 𝑎& 3\\ 𝑏& 2& 𝑐\\ 3& 𝑑& 4\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 𝑏& 3\\ 𝑎& 2& 𝑑\\ 3& 𝑐& 4\end{array}⎤⎦⎥\\ 𝑎=𝑏\text{ and }𝑐=𝑑...\left(𝑖\right)\\ 𝐵=⎡⎣⎢\begin{array}{ccc}0& 5& 𝑏\\ -5& 0& -7\\ 6& 𝑐& 0\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Since, }{𝐵}^{𝑇}=-𝐵\\ & \Rightarrow ⎡⎣⎢\begin{array}{ccc}0& -5& 6\\ 5& 0& 𝑐\\ 𝑏& -7& 0\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}0& -5& -𝑏\\ 5& 0& 7\\ -6& -𝑐& 0\end{array}⎤⎦⎥\end{array}\\ & \text{ So, }𝑏=-6,𝑎=-6,𝑐=7\text{ and }𝑑=7\\ & \begin{array}{cc}𝐴𝐵& =⎡⎣⎢\begin{array}{ccc}1& -6& 3\\ -6& 2& 7\\ 3& 7& 4\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}0& 5& -6\\ -5& 0& -7\\ 6& 7& 0\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}0+30+18& 5+0+21& -6+42\\ 0-10+42& -30+49& 22\\ -11& 43& -67\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}48& 26& 36\\ 32& 19& 22\\ -11& 43& -67\end{array}⎤⎦⎥\end{array}\end{array}$

$𝐴=⎡⎣⎢\begin{array}{ccc}-1& -3& -2\\ 0& 1& 2\\ 3& 4& 5\end{array}⎤⎦⎥$

Co-factor matrix of $𝐴$

$\begin{array}{cc}& =⎡⎣⎢\begin{array}{ccc}-3& +6& -3\\ +7& 1& -5\\ -4& 2& -1\end{array}⎤⎦⎥\\ \mathrm{adj}\left(𝐴\right)& =⎡⎣⎢\begin{array}{ccc}-3& 7& -4\\ 6& 1& 2\\ -3& -5& -1\end{array}⎤⎦⎥\\ \left|𝐴\right|& =-1(5-8)+3(0-6)-2(0-3)\end{array}$

$\begin{array}{c}=3-18+6=-9\\ \therefore {𝐴}^{-1}=\frac{\mathrm{Adj}𝐴}{\left|𝐴\right|}=\mid \begin{array}{ccc}1/3& -7/9& 4/9\\ -2/3& -1/9& -2/9\\ 1/3& 5/9& 1/9\end{array}\mid \\ \therefore {𝑎}_1+{𝑐}_2+{𝑏}_3=\frac{1}{3}+\frac{5}{9}+(-\frac{2}{9})\\ =\frac{3+5-2}{9}=\frac{6}{9}=\frac{2}{3}\end{array}$

Given, the system of linear equations

$\begin{array}{cc}& 2𝑥-3𝑦+5𝑧=12\\ & 5𝑥+2𝑦+3𝑧=11\end{array}$

and $𝑥+2𝑦-3𝑧=-3$

has the unique solution $(𝛼,𝛽,𝛾)$.

Now, by using the Cramer's rule, we get

$\begin{array}{cc}𝛼& =\frac{\mid \begin{array}{ccc}12& -3& 5\\ 11& 2& 3\\ -3& 2& -3\end{array}\mid }{\mid \begin{array}{ccc}2& -3& 5\\ 5& 2& 3\\ 1& 2& -3\end{array}\mid }=\frac{-76}{-38}=2\\ 𝛽& =\frac{\mid \begin{array}{ccc}2& 12& 5\\ 5& 11& 3\\ 1& -3& -3\end{array}\mid }{\mid \begin{array}{ccc}2& -3& 5\\ 5& 2& 3\\ 1& 2& -3\end{array}\mid }=\frac{38}{-38}=-1\\ \text{ and }𝛾& =\frac{\mid \begin{array}{ccc}2& -3& 12\\ 5& 2& 11\\ 1& 2& -3\end{array}\mid }{\mid \begin{array}{ccc}2& -3& 5\\ 5& 2& 3\\ 1& 2& -3\end{array}\mid }=\frac{-38}{-38}=1\end{array}$

$\begin{array}{c}\therefore 2𝛼+5𝛽+3𝛾=2\left(2\right)+5(-1)+3\left(1\right)\\ =4-5+3=7-5=2\end{array}$

$\begin{array}{cc}& \text{ Here, }\\ & \begin{array}{cc}{2}^{2022}(\frac{1}{2}& {+\frac{\sqrt{\surd 3}}{2}𝑖)}^{2022}-2^{2022}{(\frac{\sqrt{\surd 3}}{2}-\frac{𝑖}{2})}^{2022}\\ & =2^{2022}{𝑒}^{𝑖\left(\frac{𝜋}{3}\right)\left(2022\right)}-2^{2022}{𝑒}^{𝑖(-\frac{𝜋}{6})\left(2022\right)}\\ & =2^{2022}{𝑒}^{𝑖\left(674\right)𝜋}-2^{2022}{𝑒}^{𝑖(-337)𝜋}\\ & =2^{2022}(1+0𝑖)-2^{2022}(-1+0𝑖)\\ & =2^{2022}+2^{2022}=2^{2023}\end{array}\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ Given, }{\left(\frac{\sqrt{\surd 3}+𝑖}{\sqrt{\surd 3}-𝑖}\right)}^4+{\left(\frac{\sqrt{\surd 3}-𝑖}{\sqrt{\surd 3}+𝑖}\right)}^4=𝑟\text{ cis }𝜃\\ & \Rightarrow \frac{(\sqrt{\surd 3}+𝑖{)}^8+(\sqrt{\surd 3}-𝑖{)}^8}{\left[\right(\sqrt{\surd 3}-𝑖\left)\right(\sqrt{\surd 3}+𝑖){]}^4}=𝑟\text{ cis }𝜃\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{{(2\mathrm{cis}\frac{𝜋}{6})}^8+{(2\mathrm{cis}(-\frac{𝜋}{6}))}^8}{2^8}=𝑟\mathrm{cis}𝜃\\ & \begin{array}{cc}& \Rightarrow \mathrm{cis}\frac{8𝜋}{6}+\mathrm{cis}(-\frac{8𝜋}{6})=𝑟\mathrm{cis}𝜃\\ & \Rightarrow \cos \frac{8𝜋}{6}+𝑖\sin \frac{8𝜋}{6}+\cos (-\frac{8𝜋}{6})+𝑖\sin (-\frac{8𝜋}{6})=𝑟\mathrm{cis}𝜃\end{array}\\ & \Rightarrow 2\cos \left(\frac{8𝜋}{6}\right)=𝑟\mathrm{cis}𝜃\Rightarrow 2(-\frac{1}{2})=𝑟\mathrm{cis}𝜃\\ & \Rightarrow -1=𝑟\mathrm{cis}𝜃\Rightarrow \pm \sqrt{\surd -1}=\sqrt{\surd 𝑟\mathrm{cis}𝜃}\\ & \Rightarrow \sqrt{\surd 𝑟\mathrm{cis}𝜃}=\pm 𝑖\\ & \Rightarrow \sqrt{\surd 𝑟\mathrm{cis}𝜃}=\mathrm{cis}\left(\frac{𝜋}{2}\right),\mathrm{cis}\left(\frac{3𝜋}{2}\right)\end{array}$

$\begin{array}{cc}& \text{ Given equation, }\\ & \begin{array}{cc}& |𝑧-2|+|𝑧-2𝑖|=4\\ & \Rightarrow \left|\right(𝑥+𝑖𝑦)-2|+\left|\right(𝑥+𝑖𝑦)-2𝑖|=4\\ & \Rightarrow \sqrt{\surd (𝑥-2{)}^2+{𝑦}^2}+\sqrt{\surd {𝑥}^2+(𝑦-2{)}^2}=4...\left(𝑖\right)\\ & \text{ Let, }𝑙=(𝑥-2{)}^2+{𝑦}^2={𝑥}^2+{𝑦}^2-4𝑥+4\\ & \text{ and }𝑚={𝑥}^2+(𝑦-2{)}^2={𝑥}^2+{𝑦}^2-4𝑦+4\end{array}\end{array}$

Eq. (i), can be written as

$\sqrt{\surd 𝑙}+\sqrt{\surd 𝑚}=4...\left(𝑖𝑖\right)$

Now, $𝑙-𝑚=4𝑦-4𝑥$

$\begin{array}{cccc}\Rightarrow & & (\sqrt{\surd 𝑙}+\sqrt{\surd 𝑚})(\sqrt{\surd 𝑙}-\sqrt{\surd 𝑚})& =4(𝑦-𝑥)\\ \Rightarrow & & 4(\sqrt{\surd 𝑙}-\sqrt{\surd 𝑚})& =4(𝑦-𝑥)\\ & \Rightarrow & (\sqrt{\surd 𝑙}-\sqrt{\surd 𝑚})& =𝑦-𝑥...\left(𝑖𝑖𝑖\right)\end{array}$

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

$2\sqrt{\surd 1}=𝑦-𝑥+4$

On squaring both sides, we get

$\begin{array}{c}4𝑙=(𝑦-𝑥+4{)}^2\\ \Rightarrow 4({𝑥}^2+{𝑦}^2-4𝑥+4)={𝑦}^2+{𝑥}^2+4^2-2𝑥𝑦-8𝑥+8𝑦\\ \Rightarrow 3{𝑥}^2+2𝑥𝑦+3{𝑦}^2-8𝑥-8𝑦=0\end{array}$

$\begin{array}{cc}& \text{ Given, }(\sqrt{\surd 3}-𝑖{)}^{2/5}={\left[2(\frac{\sqrt{\surd 3}}{2}-\frac{𝑖}{2})\right]}^{2/5}\\ & =2^{2/5}{[\cos (-\frac{𝜋}{6})+𝑖\sin (-\frac{𝜋}{6})]}^{2/5}\\ & =2^{2/5}{[\cos (-\frac{𝜋}{6}+6𝜋)+𝑖\sin (-\frac{𝜋}{6}+6𝜋)]}^{2/5}\\ & =2^{2/5}{[\cos \left(\frac{35𝜋}{6}\right)+𝑖\sin \left(\frac{35𝜋}{6}\right)]}^{2/5}\\ & =2^{2/5}[(\cos \frac{35𝜋}{6}\times \frac{2}{5})+𝑖\sin (\frac{35𝜋}{6}\times \frac{2}{5})]\\ & =2^{2/5}[\cos \left(\frac{7𝜋}{3}\right)+𝑖\sin \left(\frac{7𝜋}{3}\right)]\\ & =2^{2/5}(\cos \frac{𝜋}{3}+𝑖\sin \frac{𝜋}{3})=2^{2/5}[\frac{1}{2}+𝑖\left(\frac{\sqrt{\surd 3}}{2}\right)]\\ & =\frac{2^{2/5}}{2}(1+\sqrt{\surd 3}𝑖)=2^{-3/5}(1+\sqrt{\surd 3𝑖})\end{array}$

Given, $𝛼,𝛽,𝛾$ and $𝛿$ are the roots of the equation ${𝑥}^4+{𝑥}^2+1=0$

Also, given that, $𝛼+𝛽=-1,𝛾+𝛿=1$, ${𝛼}^2=𝛽$ and ${𝛾}^2=-𝛿$

We know that, roots of the equation ${𝑥}^4+{𝑥}^2+1=0$ are $\pm {𝜔}^{1/2}$ and $\pm 𝜔$

Now, it can be conclude that

$\begin{array}{cc}& 𝛼=-{𝜔}^{1/2},𝛽=𝜔,𝛾={𝜔}^{1/2}\text{ and }𝛿=-𝜔\\ & \therefore {𝛼}^{2023}+{𝛽}^{2023}+{𝛾}^{2022}+{𝛿}^{2022}\\ & ={(-{𝜔}^{1/2})}^{2023}+(𝜔{)}^{2023}+{\left({𝜔}^{1/2}\right)}^{2022}+(-𝜔{)}^{2022}\\ & =-{𝜔}^{1011}\cdot {𝜔}^{1/2}+𝜔+1+1\\ & \begin{array}{c}=-\left(1\right)\cdot {𝜔}^{1/2}+2+𝜔=-{𝜔}^{1/2}+2+𝜔\\ ={𝜔}^2+𝜔+2[\because -{𝜔}^{1/2}={𝜔}^2]\\ =1[\because 1+𝜔+{𝜔}^2=0]\\ \end{array}\end{array}$

Given, $𝑎{𝑥}^2-7𝑥+𝑐=0$ and $𝑎{𝑥}^2+5𝑥-𝑐=0$ have a common roots.

$\begin{array}{cc}& 𝑐=6𝑥\\ & \\ 𝑆𝑜,& 𝑥=\frac{𝑐}{6}\\ & \begin{array}{cc}& \therefore 𝑎\left(\frac{{𝑐}^2}{36}\right)-7\left(\frac{𝑐}{6}\right)+𝑐=0\\ & \Rightarrow 𝑎{𝑐}^2-42𝑐+36𝑐=0\Rightarrow 𝑎{𝑐}^2-6𝑐=0\\ & \Rightarrow 𝑐=0,(𝑎𝑐-6)=0\\ & \Rightarrow 𝑎=\frac{6}{𝑐},𝑎𝑐=6\end{array}\end{array}$

Now, according to the question, 3 is a root of $𝑎{𝑥}^2-7𝑥+𝑐=0$

$\begin{array}{cc}\Rightarrow & 9𝑎-21+𝑐=0\\ \Rightarrow 9\left(\frac{6}{𝑐}\right)-21+𝑐& =0\Rightarrow 54-21𝑐+{𝑐}^2=0\\ \Rightarrow {𝑐}^2-21𝑐+54& =0\\ 𝑐& =18,3...\left(𝑖\right)\\ 𝑥& =\frac{𝑐}{6}\Rightarrow 𝑥=3,\frac{1}{2}\end{array}$

So, common root $=\frac{1}{2}$

Given, inequality ${𝑥}^2-7𝑥+10\ge 0$ and $2𝑥+3-{𝑥}^2>0$

Case $𝐈(𝑥-5)(𝑥-2)\ge 0$

$\begin{array}{cc}& 𝑥\in (-\infty ,2]\cup [5,\infty )...\left(𝑖\right)\\ & \text{ Case }\text{ II }-{𝑥}^2+2𝑥+3>0\\ & -({𝑥}^2-2𝑥-3)>0\\ \Rightarrow & {𝑥}^2-2𝑥-3<0\\ \Rightarrow & {𝑥}^2-3𝑥+𝑥-3<0\\ \Rightarrow & 𝑥(𝑥-3)+1(𝑥-3)<0\\ & (𝑥-3)(𝑥+1)<0\end{array}$

$\begin{array}{cc}& 𝑥\in (-1,3)...\left(𝑖𝑖\right)\\ & \text{ From Eqs. (i) and (ii), we get }\\ & 𝑥\in (-1,2]\end{array}$

$\begin{array}{cc}& (𝑥-2)(𝑥+3)(2𝑥-1)=0\\ & 𝛼=2,𝛽=-3\text{ and }𝛾=\frac{1}{2}\\ & \therefore {𝛼}^3+{𝛽}^3+{𝛾}^3=8-27+\frac{1}{8}\\ & =\frac{64-216+1}{8}=-\frac{151}{8}\end{array}$

Given, $𝛼,𝛽$ and $𝛾$ are the real roots of the equation $18{𝑥}^3-15{𝑥}^2-4𝑥+4=0$ such that $𝛼=𝛽$ and $𝛼>𝛾$

So, $𝛼+𝛽+𝛾=-\frac{(-19)}{18}\Rightarrow 2𝛼+𝛾=\frac{5}{6}$

$\begin{array}{cc}\text{ and }& 𝛼𝛽+𝛽𝛾+𝛾𝛼\\ \Rightarrow & {𝛼}^2+2𝛼𝛾=-\frac{4}{18}\\ \Rightarrow & {𝛼}^2+2𝛼(\frac{5}{6}-2𝛼)=-\frac{2}{9}\\ \Rightarrow & 27{𝛼}^2-15𝛼-2=0\end{array}$

$\Rightarrow 𝛼=-\frac{1}{9},\frac{2}{3}$

If $𝛼=-\frac{1}{9}$, then $𝛾=\frac{5}{6}-2(-\frac{1}{9})=\frac{19}{18}$

But $𝛼>𝛾$.

So, $𝛼\ne -\frac{1}{9}$.

If $𝛼=\frac{2}{3}$, then $𝛾=\frac{5}{6}-2\left(\frac{2}{3}\right)=-\frac{1}{2}$

$\begin{array}{cc}\therefore 𝛼+{𝛽}^2& +{𝛾}^3\\ & =\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+{(-\frac{1}{2})}^3\\ & =\frac{2}{3}+\frac{4}{9}-\frac{1}{8}=\frac{71}{72}\end{array}$

Given, $𝛼$ is a multiple root of the equation

${𝑥}^5-6{𝑥}^4+11{𝑥}^3-2{𝑥}^2-12𝑥+8=0$

The possible rational roots of this equation be $\pm 1,\pm 2,\pm 4$ and $\pm 8$

Let $𝑓\left(𝑥\right)={𝑥}^5-6{𝑥}^4+11{𝑥}^3-2{𝑥}^2-12𝑥+8$

$\begin{array}{cc}& \text{ Now, }\\ & \begin{array}{cccc}𝒙& 𝒇\left(𝒙\right)& 𝒙& 𝒇\left(𝒙\right)\\ -1& 0& -4& -3240\\ +1& 0& 4& 120\\ -2& -192& -8& -63000\\ +2& 0& 8& 13608\end{array}\end{array}$

Thus, $-1,1$ and 2 be three roots of

${𝑥}^5-6{𝑥}^4+11{𝑥}^3-2{𝑥}^2-12𝑥+8=0$

So, $(𝑥+1)(𝑥-1)(𝑥-2)$ or ${𝑥}^3-2{𝑥}^2-𝑥+2$

be a factor of

$\begin{array}{cc}& {𝑥}^5-6{𝑥}^4+11{𝑥}^3-2{𝑥}^2-12𝑥+8\\ & \text{ Now, }{𝑥}^5-6{𝑥}^4+11{𝑥}^3-2{𝑥}^2-12𝑥+8\\ & ÷{𝑥}^3-2{𝑥}^2-𝑥+2={𝑥}^2-4𝑥+4\end{array}$

The roots of equation ${𝑥}^2-4𝑥+4=0$ are 2, 2.

Thus, the roots of the equation

${𝑥}^5-6{𝑥}^4+11{𝑥}^3-2{𝑥}^2-12𝑥+8=0$

be, $-1,1,2,2$ and 2 .

So, $𝛼=2$

$\begin{array}{cc}\therefore 3{𝛼}^2-2𝛼+1& =3(2{)}^2-2(2)+1\\ & =12-4+1=9\end{array}$

The number of words start with

$D=\frac{5!}{2!}=60$

The number of words start with

$𝐸=\frac{5!}{2!}=60$

The number of words start with

$𝐼=\frac{5!}{2!2!}=30$

The number of words start with

$\mathrm{ND}=\frac{4!}{2!}=12$

The number of words start with

$\mathrm{NE}=\frac{4!}{2!}=12$

Now, next word be 'NIDDEE'.

Thus, the rank of NIDDEE be

$60+60+30+12+12+1=175$

The sum of digits $1,2,3,5,6$ and 8 is 25 .

Since, required numbers are divisible by 3 but not by 6 . So, the sum of digits must be divisible by 3 and the numbers must be odd.

The sum of digits will be divisible by 3 only, when digits be $2,3,5,6$ and 8 .

Now,

$\begin{array}{cc}& \text{ Thus, the number of required number be }\\ & 4\times 3\times 2\times 1\times 2=48\end{array}$

$\begin{array}{cc}& \text{ Required number of ways }\\ & \begin{array}{c}={}^7{𝐶}_2{}^6{𝐶}_1{}^5{𝐶}_1{}^4{𝐶}_1+{}^7{𝐶}_1{}^6{𝐶}_2{}^5{𝐶}_1{}^4{𝐶}_1+\\ {}^7{𝐶}_1{}^6{𝐶}_1{}^5{𝐶}_2{}^4{𝐶}_1+{}^7{𝐶}_1{}^6{𝐶}_1{}^5{𝐶}_1{}^4{𝐶}_2\end{array}\end{array}$