TG EAPCET 2025 2nd May Morning Shift Question Paper with Solutions

Given, $𝑓\left(𝑥\right)=\tan \left(\frac{𝜋}{\sqrt{\surd 𝑥+1}+4}\right)$

For $\sqrt{\surd 𝑥+1}$ to be defined,

$\begin{array}{cc}\therefore & \sqrt{\surd 𝑥+1}+4\ge 0+4=4\\ \Rightarrow & 0<\frac{1}{\sqrt{\surd 𝑥+1}+4}\le \frac{1}{4}\\ \Rightarrow & 0<\frac{𝜋}{\sqrt{\surd 𝑥+1}+4}\le \frac{𝜋}{4}\\ \therefore & 𝑓\left(𝑥\right)=\tan (\frac{𝜋}{\sqrt{\surd 𝑥+1}}+4),\text{ where }\\ & 0<\frac{𝜋}{\sqrt{\surd 𝑥+1}+4}\le \frac{𝜋}{4}\end{array}$

So, the function $𝑓\left(𝑥\right)$ is strictly increasing in the interval $(0,\frac{𝜋}{4}]$

And the range of the given function is $(\tan (0),\tan \left(\frac{𝜋}{4}\right)]$, which is $(0,1]$

$\therefore$ The range of $𝑓\left(𝑥\right)$ is $(0,1]$

Given, $𝑓\left(𝑥\right)={\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right)$

The domain of ${\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right)$ is $[-1,1]$

$\therefore -1\le \sqrt{\surd {𝑥}^2+𝑥+1}\le 1$

Since $\sqrt{\surd {𝑥}^2+𝑥+1}$ is always,

Let $𝑔\left(𝑥\right)={𝑥}^2+𝑥+1={(𝑥+\frac{1}{2})}^2+\frac{3}{4}$

The minimum value of $𝑔\left(𝑥\right)$ is $3/4$, which occurs at $𝑥=-\frac{1}{2}$

$\begin{array}{cc}\therefore & {𝑥}^2+𝑥+1\ge \frac{3}{4}\\ \Rightarrow & \sqrt{\surd {𝑥}^2+𝑥+1}\ge \sqrt{\surd \frac{3}{4}}=\frac{\sqrt{\surd 3}}{2}\\ \Rightarrow & \frac{\sqrt{\surd 3}}{2}\le \sqrt{\surd {𝑥}^2+𝑥+1}\le 1\\ \Rightarrow & {\sin }^{-1}\left(\frac{\sqrt{\surd 3}}{2}\right)\le {\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right)\\ \Rightarrow & \frac{𝜋}{3}\le 𝑓\left(𝑥\right)\le \frac{𝜋}{2}\le {\sin }^{-1}\end{array}$

Thus, the range of $𝑓\left(𝑥\right)={\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right)$ is $[\frac{𝜋}{3},\frac{𝜋}{2}]$

Given series

$1+(1+3)+(1+3+5)+(1+3+5+7)+\dots$

to 10 terms.

The $𝑛$th term of the series is the sum of first $𝑛$ odd numbers.

$\begin{array}{cc}𝑆𝑜,{𝑇}_{𝑛}& =1+3+5+\dots +(2𝑛-1)={𝑛}^2\\ \therefore {𝑇}_1& =1^2=1\\ {𝑇}_2& =2^2=4\\ {𝑇}_3& =3^2=9\\ ⋮& \\ {𝑇}_{10}& ={10}^2=100\end{array}$

Thus, ${𝑆}_{10}=\sum _{10}^{𝑛=1}{𝑛}^2$

We know that the sum of the first $𝑘$ squares is

$\sum _{𝑘}^{𝑛=1}{𝑛}^2=\frac{𝑘(𝑘+1)(2𝑘+1)}{6}$

For $𝑘=10$

$\begin{array}{cc}{𝑆}_{10}& =\frac{10(10+1)(2\times 10+1)}{6}\\ & =\frac{10\times 11\times 21}{6}=\frac{2310}{6}=385\\ \therefore {𝑆}_{10}& =385\end{array}$

Given equations are $𝑥+𝑦-𝑧=1$,

$2𝑥+4𝑦-𝑧=0\text{ and }3𝑥+4𝑦+5𝑧=18$

The augmented matrix for the system is

$𝐴=⎡⎣⎢\begin{array}{cccc}1& 1& -1& 1\\ 2& 4& -1& 0\\ 3& 4& 5& 18\end{array}⎤⎦⎥$

We need to transform this matrix into

$𝐵=⎡⎣⎢\begin{array}{cccc}1& 𝑎& 0& -1\\ 0& 2& 1& 𝑏\\ 0& 0& 𝑐& 32\end{array}⎤⎦⎥$

Now, perform the row operation for matrix $𝐴$, we get

$\begin{array}{c}{𝑅}_2\to {𝑅}_2-2{𝑅}_1\text{ and }{𝑅}_3\to {𝑅}_3-3{𝑅}_1\\ 𝐴=⎡⎣⎢\begin{array}{cccc}1& 1& -1& 1\\ 0& 2& 1& -2\\ 0& 1& 8& 15\end{array}⎤⎦⎥\end{array}$

Now, ${𝑅}_3\to {𝑅}_3-\frac{1}{2}{𝑅}_2$

$𝐴=⎡⎣⎢\begin{array}{cccc}1& 1& -1& 1\\ 0& 2& 1& -2\\ 0& 0& 7.5& 16\end{array}⎤⎦⎥$

Apply ${𝑅}_3\to 2{𝑅}_3$, we get

$𝐴=⎡⎣⎢\begin{array}{cccc}1& 1& -1& 1\\ 0& 2& 1& -2\\ 0& 0& 15& 32\end{array}⎤⎦⎥$

Again apply ${𝑅}_1\to {𝑅}_1+{𝑅}_2$, we get

$𝐴=⎡⎣⎢\begin{array}{cccc}1& 3& 0& -1\\ 0& 2& 1& -2\\ 0& 0& 15& 32\end{array}⎤⎦⎥$

$\begin{array}{cc}& \text{ Thus, }𝑎=3,𝑏=-2\text{ and }𝑐=15\\ & \begin{array}{cc}\Rightarrow & 𝑐=15\\ \therefore & \sqrt{\surd 𝑎+𝑏+𝑐}=\sqrt{\surd 3+(-2)+15}\\ & =\sqrt{\surd 1+15}=\sqrt{\surd 16}=4\end{array}\end{array}$

Given, $𝐾=\mid \begin{array}{ccc}9& 25& 16\\ 16& 36& 25\\ 25& 49& 36\end{array}\mid$

$\begin{array}{cc}& =9\mid \begin{array}{cc}36& 25\\ 49& 36\end{array}\mid -25\mid \begin{array}{cc}16& 25\\ 25& 36\end{array}\mid +16\mid \begin{array}{cc}16& 36\\ 25& 49\end{array}\mid \\ & =9(1296-1225)-25(576-625)+16(784-900)\\ & =9\left(71\right)-25(-49)+16(-116)\\ & =639+1225-1856=8\\ & \therefore 𝐾=8\end{array}$

So, the roots are $𝐾$ and $𝐾+1$, i.e., 8 and 9

$\begin{array}{cc}\text{ Sum of roots }& =8+9=17\\ & =\frac{-𝑏}{𝑎}=𝛼+𝛽\end{array}$

And product of roots $=8\times 9=72$

$=\frac{𝑐}{𝑎}=𝛼𝛽$

So, the general form of a quadratic equation is ${𝑥}^2-(𝛼+𝛽)𝑥+𝛼𝛽=0$

$\Rightarrow {𝑥}^2-17𝑥+72=0$

Given, $𝐴=⎡⎣⎢\begin{array}{ccc}1& -3& -5\\ -2& 4& -6\\ 7& -11& 13\end{array}⎤⎦⎥$

$\begin{array}{cc}& \left|𝐴\right|=\mid \begin{array}{ccc}1& -3& -5\\ -2& 4& -6\\ 7& -11& 13\end{array}\mid \\ & =1(4\times 13-(-6)\times (-11\left)\right)-(-3)(-2\times 13-(-6)\times 7)-5(-2\times (-11)-7\times 4)\\ & =1(52-66)+3(-26+42)-5(22-28)\\ & =-14+48+30=64\end{array}$

Since, $|\mathrm{adj}𝐴|=|𝐴{|}^{𝑛-1}$, for $𝑛\times 𝑛$ matrix $𝐴$

So, for $3\times 3$ matrix,

$\therefore \begin{array}{cc}|\mathrm{adj}𝐴|& =|𝐴{|}^{3-1}=|𝐴{|}^2\\ \therefore |\mathrm{adj}𝐴|& =|𝐴{|}^2=(64{)}^2=4096\\ \sqrt{\surd |\mathrm{adj}𝐴|}& =\sqrt{\surd 4096}=64\end{array}$

$\begin{array}{cc}\Delta 𝑟& =\mid \begin{array}{cc}\frac{1}{3𝑟-2}& \frac{2}{3𝑟-5}\\ 0& \frac{3}{3𝑟+1}\end{array}\mid \\ & =\frac{1}{(3𝑟-2)}\times \frac{3}{(3𝑟+1)}-\frac{2}{3𝑟-5}\times 0\\ & =\frac{3}{(3𝑟-2)(3𝑟+1)}\end{array}$

Using partial fractions

$\begin{array}{cc}\Delta 𝑟& =\frac{𝐴}{3𝑟-2}+\frac{𝐵}{3𝑟+1}\\ \Rightarrow 3& =𝐴(3𝑟+1)+𝐵(3𝑟-2)\end{array}$

If $𝑟=\frac{2}{3}$, then

$\begin{array}{cc}& \Rightarrow 3=𝐴(3\times \frac{2}{3}+1)+𝐵(3\times \frac{2}{3}-2)\\ & \Rightarrow 3𝐴=3\\ & \Rightarrow 𝐴=1\end{array}$

If $𝑟=-\frac{1}{3}$, then

$\begin{array}{cc}& 3=𝐵(3\times (-\frac{1}{3})-2)=𝐵(-1-2)\\ & \Rightarrow 𝐵=-1\\ & \text{ So, }\Delta 𝑟=\frac{1}{3𝑟-2}-\frac{1}{3𝑟+1}\\ & \therefore \sum _{𝑟=1}^{33}\Delta 𝑟=\sum _{𝑟=1}^{33}(\frac{1}{3𝑟-2}-\frac{1}{3𝑟+1})\\ & \Rightarrow (\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+\\ & \dots +(\frac{1}{97}-\frac{1}{100})\\ & \Rightarrow 1-\frac{1}{100}=\frac{99}{100}=0.99\end{array}$

Given, $\frac{2+3𝑖}{𝑖-2}-\frac{4𝑖-3}{3+4𝑖}=𝑥+𝑖𝑦$

Now, $\frac{2+3𝑖}{𝑖-2}=\frac{2+3𝑖}{-2+𝑖}\times \frac{(-2-𝑖)}{(-2-𝑖)}$

$\Rightarrow \frac{-4-2𝑖-6𝑖-3{𝑖}^2}{(-2{)}^2-{𝑖}^2}=\frac{-1-8𝑖}{5}$

And, $\frac{4𝑖-3}{3+4𝑖}\times \frac{3-4𝑖}{3-4𝑖}$

$\Rightarrow \frac{-9+12𝑖+12𝑖-16{𝑖}^2}{3^2-(4𝑖{)}^2}=\frac{7+24𝑖}{25}$

So, $\frac{2+3𝑖}{𝑖-2}-\frac{4𝑖-3}{3+4𝑖}$

$\begin{array}{cc}& =\frac{-1}{5}-\frac{8𝑖}{5}-\frac{7}{25}-\frac{24}{25}𝑖\\ & =\frac{-12}{25}-\frac{64}{25}𝑖\end{array}$

On comparing, we get

$𝑥=\frac{-12}{25},𝑦=\frac{-64}{25}$

So, $3𝑥+𝑦=3\left(\frac{-12}{25}\right)-\frac{64}{25}$

$=-\frac{36}{25}-\frac{64}{25}=\frac{-100}{25}=-4$

Given, arg $\left(\frac{𝑧-3𝑖}{𝑧+2𝑖}\right)=\frac{𝜋}{4}$

Let $𝑧=𝑥+𝑖𝑦$

So, $\frac{𝑧-3𝑖}{𝑧+2𝑖}=\frac{𝑥+𝑖𝑦-3𝑖}{𝑥+𝑖𝑦+2𝑖}$

$\Rightarrow \frac{𝑥+𝑖(𝑦-3)}{𝑥+𝑖(𝑦+2)}\times \frac{𝑥-𝑖(𝑦+2)}{𝑥-𝑖(𝑦+2)}$

$\begin{array}{cc}& \Rightarrow \frac{{𝑥}^2-𝑖𝑥(𝑦+2)+𝑖𝑥(𝑦-3)+(𝑦-3)(𝑦+2)}{{𝑥}^2+(𝑦+2{)}^2}\\ & \Rightarrow \frac{{𝑥}^2+{𝑦}^2-𝑦-6-𝑖5𝑥}{{𝑥}^2+(𝑦+2{)}^2}\end{array}$

Let the complex numbers be $𝐴+𝑖𝐵$, where

$𝐴=\frac{{𝑥}^2+{𝑦}^2-𝑦-6}{{𝑥}^2+(𝑦+2{)}^2}\text{ and }𝐵=\frac{-5𝑥}{{𝑥}^2+(𝑦+2{)}^2}$

So, $\arg \left(\frac{𝑧-3𝑖}{𝑧+2𝑖}\right)=\left(\frac{𝜋}{4}\right)$

$\begin{array}{cc}& \Rightarrow \tan \left(\frac{𝐵}{𝐴}\right)=\frac{𝜋}{4}\\ & \Rightarrow \frac{-5𝑥}{{𝑥}^2+{𝑦}^2-𝑦-6}={\tan }^{-1}\left(\frac{𝜋}{4}\right)=1\\ & \Rightarrow {𝑥}^2+{𝑦}^2-𝑦-6=-5𝑥\\ & \Rightarrow {𝑥}^2+5𝑥+{𝑦}^2-𝑦-6=0,\text{ where }\\ & 𝑥\ne 0\text{ and }𝑦\ne -2\end{array}$

Given, $𝑥={𝜔}^2-𝜔+2$ where $𝑤$ is a complex cube root of unity.

We know that $1+𝜔+{𝜔}^2=0$

$\Rightarrow {𝜔}^2=-1-𝜔$

So, $𝑥=(-1-𝜔)-𝜔+2=1-2𝜔$

$\Rightarrow 𝜔=\frac{1-𝑥}{2}$

So, $1+𝜔+{𝜔}^2=0$

$\begin{array}{cc}& \Rightarrow 1+\left(\frac{1-𝑥}{2}\right)+{\left(\frac{1-𝑥}{2}\right)}^2=0\\ & \Rightarrow 1+\frac{(1-𝑥)}{2}+\frac{(1-𝑥{)}^2}{4}=0\\ & \Rightarrow 4+2(1-𝑥)+1-2𝑥+{𝑥}^2=0\\ & \Rightarrow 4+2-2𝑥+1-2𝑥+{𝑥}^2=0\\ & \Rightarrow {𝑥}^2-4𝑥+7=0\end{array}$

Let $𝑧=\sqrt{\surd 3}-𝑖$

$\left|𝑧\right|=\sqrt{\surd (\sqrt{\surd 3}{)}^2+(-1{)}^2}=\sqrt{\surd 3+1}=2$

And $𝜃={\tan }^{-1}\left(\frac{-1}{\sqrt{\surd 3}}\right)=-\frac{𝜋}{6}$ or $\frac{11𝜋}{6}$

So, $𝑧=2(\cos (-\frac{𝜋}{6})+𝑖\sin (-\frac{𝜋}{6}))$

Now, ${𝑧}^{\frac{3}{7}}=(\sqrt{\surd 3}-𝑖{)}^{\frac{3}{7}}$

First, ${𝑧}^3=(\sqrt{\surd 3}-𝑖{)}^3$

$\begin{array}{cc}& =2^3(\cos \left(\frac{-3𝜋}{6}\right)+𝑖\sin \left(\frac{-3𝜋}{6}\right))\\ & =8(\cos \frac{𝜋}{2}-𝑖\sin \frac{𝜋}{2})...\left(𝑖\right)\end{array}$

Now, ${𝑧}^{\frac{3}{7}}=(\sqrt{\surd 3}-𝑖{)}^{\frac{3}{7}}$

$=(8{)}^{\frac{1}{7}}(\cos \left(\frac{-\frac{𝜋}{2}+2𝑘𝜋}{7}\right)+𝑖\sin \left(\frac{-\frac{𝜋}{2}+2𝑘𝜋}{7}\right))...(𝑖𝑖)$

for $𝑘=0,1,\dots ,6$

So, ${𝑧}^3=8(0-𝑖)=-8𝑖$ [Using Eq. (i)]

Now, we know that the product of $𝑛𝑡$ h root of a complex number $𝐴$ is given by $(-1{)}^{𝑛-1}𝐴$, if $𝐴$ is a real number.

Since, $𝑛=7$ is odd, the product of the 7 roots of $𝐴$ is $𝐴$ itself.

$\therefore$ The product of the 7th roots of $𝐴$ is $𝐴$ itself.

$\therefore (\sqrt{\surd 3}-𝑖{)}^{\frac{3}{7}}={𝑧}^3$

So, product $=-8𝑖$

Given, ${\sin }^2𝑥+𝑏\sin 𝑥+𝑐=0$ and

$𝛼+𝛽=\frac{𝜋}{2}$

Let $𝑦=\sin 𝑥$, thus we get

${𝑦}^2+𝑏𝑦+𝑐=0$

The roots of this equation are $\sin 𝛼$ and $\sin 𝛽$.

So, sum of the roots $=\sin 𝛼+\sin 𝛽=-\frac{𝑏}{1}$

Product of the roots $=\sin 𝛼\cdot \sin 𝛽=𝑐$

Since, $𝛼+𝛽=\frac{𝜋}{2}$

$\Rightarrow 𝛽=\frac{𝜋}{2}-𝛼$

So, $\sin 𝛼+\sin (\frac{𝜋}{2}-𝛼)=-𝑏$

$\begin{array}{cc}\Rightarrow \sin 𝛼+\cos 𝛼=-𝑏& \\ & [\because \sin (\frac{𝜋}{2}-𝜃)=\cos 𝜃]\end{array}$

$\Rightarrow (\sin 𝛼+\cos 𝛼{)}^2=(-𝑏{)}^2$

$\Rightarrow {\sin }^2𝛼+{\cos }^2𝛼+2\sin 𝛼\cos 𝛼=(-𝑏{)}^2$

$\Rightarrow 1+2𝑐={𝑏}^2$$[\because {\sin }^2𝜃+{\cos }^2𝜃=1]$

$\Rightarrow {𝑏}^2-1=2𝑐$

Given, quadratic equation is

$a x^2+a x+5=0$

For a quadratic equation

$\begin{gathered} A x^2+B x+C=0 \text { to have no real roots, } \\ \Delta=B^2-4 A C<0 \end{gathered}$

So, $\quad \Delta=a^2-4(a)(5)<0$

$$\begin{aligned} & \Rightarrow \quad a^2-20 a<0 \\ & \Rightarrow \quad a(a-20)<0 \end{aligned}$$

So, $\quad 0<a<20$

Thus, $a$ must be an integer.

So, the integral values of $a$ in this range are $1,2,3, \ldots, 19$.

The number of such value is

$19-1+1=19$

Given equation is

$32{𝑥}^3-48{𝑥}^2+22𝑥-3=0$

Since, the roots of this equation are in AP .

So, sum of roots $=𝑎-𝑑+𝑎+𝑎+𝑑=3𝑎$

$=\frac{-(-48)}{32}=\frac{3}{2}\Rightarrow 𝑎=\frac{1}{2}$

Now, sum of the products of roots $=\frac{22}{32}$

$\Rightarrow (𝑎-𝑑)𝑎+𝑎(𝑎+𝑑)+(𝑎-𝑑)(𝑎+𝑑)=\frac{11}{16}$

$\begin{array}{cc}& \Rightarrow {𝑎}^2-𝑎𝑑+{𝑎}^2+𝑎𝑑+{𝑎}^2-{𝑑}^2=\frac{11}{16}\\ & \Rightarrow 3{𝑎}^2-{𝑑}^2=\frac{11}{16}\Rightarrow 3{\left(\frac{1}{2}\right)}^2-{𝑑}^2=\frac{11}{16}\\ & \Rightarrow {𝑑}^2=\frac{3}{4}-\frac{11}{16}=\frac{12-11}{16}=\frac{1}{16}\\ & \Rightarrow {𝑑}^2=\frac{1}{16}\end{array}$

So, the square of the common difference of the root is $\frac{1}{16}$.

Given, equation is

${𝑥}^4-2{𝑥}^3+{𝑥}^2+4𝑥-6=0$

Let the roots of this equation be $𝛼,𝛽,𝛾$ and $𝛿$.

Sum of the roots

$=𝛼+𝛽+𝛾+𝛿=\frac{-(-2)}{1}=2...\left(𝑖\right)$

Sum of the products of roots

$\begin{array}{cc}& =𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿\\ & =\frac{1}{1}=1\end{array}$

Sum of products of roots taken three at a time

$\begin{array}{cc}& =𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿\\ & =\frac{-4}{1}=-4\end{array}$

Product of roots $=𝛼𝛽𝛾𝛿=\frac{-6}{1}=-6$

Since, the sum of two roots be zero, i.e.,

$\begin{array}{cccc}& & 𝛼+𝛽& =0\\ \Rightarrow & & 𝛽& =-𝛼\end{array}$

So, $𝛾+𝛿=2$        [from Eq. (i)]

And,

$\begin{array}{cc}& 𝛼(-𝛼)+𝛼𝛾+𝛼𝛿+(-𝛼𝛾)+(-𝛼)𝛿+𝛾𝛿=1\\ & \Rightarrow -{𝛼}^2+𝛼𝛾+𝛼𝛿-𝛼𝛾-𝛼𝛿+𝛾𝛿=1\\ & \Rightarrow -{𝛼}^2+𝛾𝛿=1...\left(𝑖𝑖\right)\end{array}$

And $𝛼(-𝛼)𝛾𝛿=-6$

$\Rightarrow -{𝛼}^2𝛾𝛿=-6...\left(𝑖𝑖𝑖\right)$

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

$\begin{array}{cc}& 𝛾𝛿=1+{𝛼}^2\\ \Rightarrow & {𝛼}^2(1+{𝛼}^2)=6\\ \Rightarrow & {𝛼}^4+{𝛼}^2=6\end{array}$

Let $𝑦={𝛼}^2$, then we get

$\begin{array}{cc}& {𝑦}^2+𝑦-6=0\\ \Rightarrow & (𝑦+3)(𝑦-2)=0\\ \Rightarrow & 𝑦=-3\text{ and }𝑦=2\end{array}$

Since, ${𝛼}^2$ must be non-negative,

$𝑦={𝛼}^2=2$

Then, $𝛾𝛿=1+{𝛼}^2=1+2=3$

Now, $(𝛾+𝛿{)}^2={𝛾}^2+{𝛿}^2+2𝛾𝛿$

$\begin{array}{cc}\Rightarrow & {𝛾}^2+{𝛿}^2=(𝛾+𝛿{)}^2-2𝛾𝛿\\ & =2^2-2\left(3\right)=4-6=-2\end{array}$

The total number of ways to choose any subset of the 5 options $=2^5=32$

These 32 subsets include choosing 2, 3, 4 or 5 options.

But choosing options 0 and 1 are not allowed.

Now, subsets with 0 options

$={}^5{𝐶}_0=\frac{5!}{0!5!}=1$

Subsets with 1 options

$={}^5{𝐶}_1=\frac{5!}{1!4!}=5$

So, total subsets with options 0 and 1

$=1+5=6$

Thus, the number of ways in which the student can answer that questions

$=32-6=26$

Let the triangle be $𝐴𝐵𝐶$ with point $(𝑥,𝑦)$,

where $0\le 𝑥\le 4$ and $0\le 𝑦\le 4$

So, the points $(𝑥,𝑦)$ are $(0,0),(0,1)$, $(0,2),(0,3),(0,4),(1,0),(1,1),(1,2)$, $(1,3),(1,4),(2,0),(2,1),(2,2),(2,3)$, $(2,4),(3,0),(3,1),(3,2),(3,3),(3,4)$, $(4,0),(4,1),(4,2),(4,3),(4,4)$.

Total coordinates are 25 , but we need to select only 3 .

Now, the number of triangles with sets of 3 non-collinear points. Total number of ways to choose 3 points $={}^{25}{𝐶}_3$

But, we must subtract the number of collinear triplets.

Case I Horizontal lines

Number of ways to choose 3 collinear points

$\Rightarrow {}^5{𝐶}_3=10$

So, in 5 rows $=5\times 10=50$

Case II Vertical lines

Same as above $\to 5$ columns

$=5\times 10=50$

Now, diagonal from bottom-left to top-right

$=$ Length $3:2$ such diagonals + Length

4 : 2 diagonals + Length 5:1 diagonal

$\begin{array}{cc}=& 2\left({}^3{𝐶}_3\right)+2\left({}^4{𝐶}_3\right)+{}^5{𝐶}_3\\ & 2\left(1\right)+2\left(4\right)+10\\ =& 2+8+10=20\end{array}$

And, diagonals from top-left to bottom-right $=20$

Total collinear triplets

$=50+50+20+20=140$

So, total number of triangles $={}^{25}{𝐶}_3-140$

$\Rightarrow 2300-140=2160$

Total Alphabets are A, D, E, H, L, N Letters before H are A, D, E.

So, the number of letters before $H=3\times 5$ !

$=3\times 120=360$

Letters before $𝐸$ are A, D.

So, the number of letters before $𝐸=2\times 4$ !

$=2\times 24=48$

Letters before L are $A,D$.

So, the number of letters before $𝐿$

$=2\times 3!=12$

Letters before N is D

So, the number of letters with N only

$\Rightarrow 1\times 1!=1$

So, total words $=360+48+12+1=421$

Now, only 1 letter remaining, that is D .

So, Rank of HELAND $=421+1=422$

Given, coefficient of 3rd term from beginning in the expansion ${(𝑎{𝑥}^2-\frac{8}{𝑏𝑥})}^9=$ Coefficient of 3rd term from end in the expansion ${(𝑎𝑥-\frac{2}{𝑏{𝑥}^2})}^9$

For ${(𝑎{𝑥}^2-\frac{8}{𝑏𝑥})}^9$

$\begin{array}{cc}{𝑇}_3& ={}^9{𝐶}_2{\left(𝑎{𝑥}^2\right)}^7{\left(\frac{-8}{𝑏𝑥}\right)}^2\\ & =\frac{9!}{2!7!}{𝑎}^7{𝑥}^{14}\cdot \frac{64}{{𝑏}^2{𝑥}^2}\\ & =\frac{36\cdot {𝑎}^7}{{𝑏}^2}\cdot {𝑥}^{12}\cdot 64\end{array}$

So, its coefficient is $\frac{36\cdot 64{𝑎}^7}{{𝑏}^2}$

For ${(𝑎𝑥-\frac{2}{𝑏{𝑥}^2})}^9$

$\begin{array}{cc}{𝑇}_8& ={}^9{𝐶}_7(𝑎𝑥{)}^2{\left(\frac{-2}{𝑏{𝑥}^2}\right)}^7\\ & =\frac{9!}{2!7!}{𝑎}^2{𝑥}^2\cdot \frac{-128}{{𝑏}^7{𝑥}^{14}}\end{array}$

So, if coefficient is

$=-36\times 128\frac{{𝑎}^2}{{𝑏}^7}$

Now, $36\cdot \frac{64{𝑎}^7}{{𝑏}^2}=-36\cdot 128\frac{{𝑎}^2}{{𝑏}^7}$

$\begin{array}{cccc}\Rightarrow & & 64\cdot {𝑎}^5{𝑏}^5& =-128\\ \Rightarrow & & {𝑎}^5{𝑏}^5& =-2\end{array}$

Given, $5^{2𝑛}-48𝑛+𝑘$ is divisible by 24

Let $𝐸\left(𝑛\right)=5^{2𝑛}-48𝑛+𝑘$

and $𝐸\left(𝑛\right)\equiv 0(mod24),\forall 𝑛\in 𝑁$

$\Rightarrow 5^{2𝑛}-48𝑛+𝑘\equiv 0(mod24)$

$\Rightarrow 𝑘=48𝑛-5^{2𝑛}(mod24)$