MHT CET 2024 10th May Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}𝐼& ={\int }_{\frac{𝜋}{2}}^0|\sin 𝑥-\cos 𝑥|𝑑𝑥\\ & ={\int }_{\frac{𝜋}{4}}^0-(\sin 𝑥-\cos 𝑥)𝑑𝑥+{\int }_{\frac{𝜋}{2}}^{\frac{𝜋}{4}}(\sin 𝑥-\cos 𝑥)𝑑𝑥\end{array}$

$...\left\{\begin{array}{cc}𝑓\left(𝑥\right)=𝑓(-𝑥)& 0\le 𝑥\le \frac{𝜋}{4}\\ =𝑓\left(𝑥\right)& \frac{𝜋}{4}\le 𝑥\le \frac{𝜋}{2}\end{array}\right\}$

$\begin{array}{cc}& ={\int }_{\frac{𝜋}{4}}^0(\cos 𝑥-\sin 𝑥)𝑑𝑥+{\int }_{\frac{𝜋}{2}}^{\frac{𝜋}{4}}(\sin 𝑥-\cos 𝑥)𝑑𝑥\\ & =(\sin 𝑥+\cos 𝑥{)}_{\frac{𝜋}{4}}^0+(-\cos 𝑥-\sin 𝑥{)}_{\frac{𝜋}{2}}^{\frac{𝜋}{4}}\end{array}$

$\begin{array}{cc}& =[\sin \frac{𝜋}{4}+\cos \frac{𝜋}{4}-\sin 0-\cos 0]\\ & -[\cos \frac{𝜋}{2}+\sin \frac{𝜋}{2}-\cos \frac{𝜋}{4}-\sin \frac{𝜋}{4}]\end{array}$

$\begin{array}{cc}& =[\frac{1}{\sqrt{\surd 2}}+\frac{1}{\sqrt{\surd 2}}-0-1]-[0+1-\frac{1}{\sqrt{\surd 2}}-\frac{1}{\sqrt{\surd 2}}]\\ & =[\frac{2}{\sqrt{\surd 2}}-1]-[1-\frac{2}{\sqrt{\surd 2}}]\\ & =\frac{4}{\sqrt{\surd 2}}-2\\ & =\frac{4\sqrt{\surd 2}}{2}-2\\ & =2\sqrt{\surd 2}-2\\ & =2(\sqrt{\surd 2}-1)\end{array}$

Let p : Payment will be made

q : Work is finished in time.

Given statement is

$(𝑝\to 𝑞)\wedge (𝑞\to 𝑝)$

Negation of above statement is

$(𝑝\wedge \sim 𝑞)\vee (𝑞\wedge \sim 𝑝)$

$\therefore$ Option (D) is correct.

$𝑦=\frac{\sin 𝑥}{1+\frac{\cos 𝑥}{1+\frac{\sin 𝑥}{1+\frac{\cos 𝑥}{...}}}}$

$𝑦=\frac{\sin 𝑥}{1+\frac{\cos 𝑥}{1+𝑦}}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑦=\frac{(1+𝑦)\sin 𝑥}{(1+𝑦)+\cos 𝑥}\\ & 𝑦\left(\right(1+𝑦)+\cos 𝑥)=(1+𝑦)\sin 𝑥\\ & 𝑦+{𝑦}^2+𝑦\cos 𝑥=\sin 𝑥+\sin 𝑥\cdot 𝑦\end{array}\\ & \text{ Differentiating w.r.to }𝑥\text{, we get }\end{array}$

$\begin{array}{cc}\frac{d𝑦}{d𝑥}+2𝑦\frac{d𝑦}{d𝑥}+𝑦(-\sin 𝑥)+\cos 𝑥& \frac{d𝑦}{d𝑥}\\ & =\frac{d𝑦}{d𝑥}\sin 𝑥+(1+𝑦)\cos 𝑥\end{array}$

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}+2𝑦\frac{d𝑦}{d𝑥}-𝑦\sin 𝑥+\cos 𝑥\frac{d𝑦}{d𝑥}\\ & =\sin 𝑥\frac{d𝑦}{d𝑥}+(1+𝑦)\cos 𝑥\\ & \frac{d𝑦}{d𝑥}(1+2𝑦+\cos 𝑥-\sin 𝑥)=\cos 𝑥+𝑦\cos 𝑥+𝑦\sin 𝑥\\ & \frac{d𝑦}{d𝑥}=\frac{𝑦\sin 𝑥+(1+𝑦)\cos 𝑥}{1+2𝑦+\cos 𝑥-\sin 𝑥}\end{array}$

$\begin{array}{cc}& \left[\bar{a}\bar{b}\bar{c}\right]=\mid \begin{array}{ccc}1& 0& -1\\ 𝑥& 1& 1-𝑥\\ 𝑦& 𝑥& 1+𝑥-𝑦\end{array}\mid \\ & \text{ Applying }{𝐶}_3\Rightarrow {𝐶}_3+{𝐶}_1\text{, we get }\\ & \left[\bar{a}\bar{b}\bar{c}\right]=\mid \begin{array}{ccc}1& 0& 0\\ 𝑥& 1& 1\\ 𝑦& 𝑥& 1+𝑥\end{array}\mid =1(1+𝑥-𝑥)=1\end{array}$

$\begin{array}{cc}& \text{ Area of }△𝐴𝑂𝐵=12\text{ sq. units }\\ & \begin{array}{cc}& \frac{1}{2}𝑎𝑏=12\\ \therefore & 𝑎𝑏=24\\ \therefore & 𝑏=\frac{24}{𝑎}\text{.... (i)}\end{array}\end{array}$

Equation of line in point slope form is

$\begin{array}{cc}& 𝑦-𝑎=\frac{-𝑎}{𝑏}(𝑥-0)\\ & 𝑦-𝑎=\frac{-𝑎𝑥}{\frac{24}{𝑎}}\text{...[from (i)]}\end{array}$

$\begin{array}{cc}& 𝑦-a=\frac{-a^2𝑥}{24}\text{.... (ii)}\\ & \text{ Since line passes through }(2,3)\end{array}$

$\begin{array}{cc}\therefore & 3-𝑎=\frac{-{𝑎}^2\left(2\right)}{24}\\ & \Rightarrow 3-𝑎=\frac{-{𝑎}^2}{12}\\ & \Rightarrow {𝑎}^2-12𝑎+36=0\\ & \Rightarrow (𝑎-6{)}^2=0\\ & \Rightarrow 𝑎=6\\ \therefore & 𝑏=\frac{24}{6}=4\text{.... [from (i)]}\end{array}$

Required equation of line is

$\begin{array}{cc}& 𝑦-6=\frac{-36𝑥}{24}\text{.... [from (ii)]}\\ & 2𝑦-12=-3𝑥\\ & \Rightarrow 3𝑥+2𝑦-12=0\\ & \Rightarrow 3𝑥+2𝑦=12\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & {\sin }^2𝑥-\cos 2𝑥=2-\sin 2𝑥\\ & \Rightarrow 1-{\cos }^2𝑥-(2{\cos }^2𝑥-1)-2+\sin 2𝑥=0\\ & \Rightarrow 1-3{\cos }^2𝑥+1-2+2\sin 𝑥\cdot \cos 𝑥\\ & \Rightarrow \cos 𝑥(2\sin 𝑥-3\cos 𝑥)=0\\ & \Rightarrow \cos 𝑥=0\text{ or }2\sin 𝑥-3\cos 𝑥=0\\ & \text{ But }2\sin 𝑥\ne 3\cos 𝑥\\ & \Rightarrow \cos 𝑥=0\\ & \Rightarrow 𝑥=(2𝑛+1)\frac{𝜋}{2},𝑛\in 𝑍\end{array}$

In a game, 3 coins are tossed.

$P($ getting all heads or all tails $)=\frac{2}{8}=\frac{1}{4}$

$P($ getting one head or two heads $)=\frac{6}{8}=\frac{3}{4}$

Let $𝑥$ : number of rupees the person gets.

$\begin{array}{cc}& 𝑃(𝑋=100)=\frac{1}{4}\\ & 𝑃(𝑋=-40)=\frac{3}{4}\end{array}$

$\therefore$ The amount he can expect to $\mathrm{win}=$ mean

$\begin{array}{cc}& =\sum n_i{p}_i\\ & =100\times \frac{1}{4}+(-40)\times \frac{3}{4}\\ & =\frac{100}{4}-\frac{120}{4}\\ & =-5\\ & =5\text{ loss }\end{array}$

$\begin{array}{cc}& \text{ Given equation of curve is }\\ & {𝑥}^4-2𝑥{𝑦}^2+{𝑦}^2+3𝑥-3𝑦=0\text{.... (i)}\end{array}$

Slope of tangent $=𝑚=\frac{𝑑𝑦}{𝑑𝑥}$ at $(0,0)$

Differentiating (i) w.r.to $𝑥$, we get

$\begin{array}{cc}& 4{𝑥}^3-2𝑥\cdot 2𝑦\frac{d𝑦}{d𝑥}-{𝑦}^2\left(2\right)+2𝑦\frac{d𝑦}{d𝑥}+3-3\frac{d𝑦}{d𝑥}=0\\ \therefore & \frac{d𝑦}{d𝑥}=\frac{4{𝑥}^3-2{𝑦}^2+3}{4𝑥𝑦-2𝑦+3}\\ \therefore & m={\frac{d𝑦}{d𝑥}\mid }_{(0,0)}=\frac{4\left(0\right)-2\left(0\right)+3}{4\left(0\right)\cdot \left(0\right)-2\left(0\right)+3}=\frac{3}{3}=1\\ \therefore & m=1\\ \therefore & \tan 𝜃=1\\ \therefore & 𝜃=\frac{𝜋}{4}\end{array}$

Let $A\equiv ({𝑥}_1,{𝑦}_1)$ and $B\equiv ({𝑥}_2,{𝑦}_2)$.

According to the given condition,

$\begin{array}{cc}& {𝑥}_1+{𝑥}_2=-2a,{𝑥}_1{𝑥}_2=-b^2\\ & {𝑦}_1+{𝑦}_2=-2p,{𝑦}_1{𝑦}_2=-q^2\end{array}$

The equation of the circle with $A({𝑥}_1,{𝑦}_1)$ and $B({𝑥}_2,{𝑦}_2)$ as the end points of diameter is

$\begin{array}{cc}& (𝑥-{𝑥}_1)(𝑥-{𝑥}_2)+(𝑦-{𝑦}_1)(𝑦-{𝑦}_2)=0\\ & \Rightarrow {𝑥}^2+{𝑦}^2-𝑥({𝑥}_1+{𝑥}_2)-𝑦({𝑦}_1+{𝑦}_2)+{𝑥}_1{𝑥}_2+{𝑦}_1{𝑦}_2=0\\ & \Rightarrow {𝑥}^2+{𝑦}^2+2a𝑥+2p𝑦-b^2-q^2=0\end{array}$

Radius of hemispherical bowl $\left(r\right)=180\mathrm{cm}$

Rate of flow $\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right)=108{\mathrm{dm}}^3/\min$

$\begin{array}{cc}& =108000{\mathrm{cm}}^3/\min \\ & =\frac{108000}{60}{\mathrm{cm}}^3/\sec \\ & =1800{\mathrm{cm}}^3/\sec \end{array}$

Let depth of water in bowl be $𝑥$.

$\therefore$ Volume of water in hemispherical

$\begin{array}{cc}& \text{ bowl }\left(V\right)=\frac{𝜋}{3}{𝑥}^2(3r-𝑥)\\ & V=\frac{𝜋}{3}{𝑥}^2(3\times 180-𝑥)\\ & \therefore V=180𝜋{𝑥}^2-\frac{𝜋}{3}{𝑥}^3\end{array}$

$\begin{array}{cc}& \text{ Differentiating w.r.t. }𝑥\text{, we get }\\ & \begin{array}{cc}& \frac{𝑑𝑉}{𝑑𝑡}=360𝜋𝑥\frac{𝑑𝑥}{𝑑𝑡}-𝜋{𝑥}^2\frac{𝑑𝑥}{𝑑𝑡}\\ & {\frac{𝑑𝑉}{𝑑𝑡}\mid }_{𝑥=120}=\frac{𝑑𝑥}{𝑑𝑡}(360𝜋\times 120-{120}^2𝜋)\\ \therefore & 1800=\frac{𝑑𝑥}{𝑑𝑡}(360𝜋\times 120-{120}^2𝜋)\\ \therefore & 15=\frac{𝑑𝑥}{𝑑𝑡}(360𝜋-120𝜋)\\ & \frac{𝑑𝑥}{𝑑𝑡}=\frac{15}{240𝜋}=\frac{1}{16𝜋}\mathrm{cm}/\sec \end{array}\end{array}$

$\begin{array}{cc}& {𝑆}_1=\sum 𝑟=\frac{𝑟(𝑟+1)}{2}\\ & {𝑆}_2=\sum {𝑟}^2=\frac{𝑟(𝑟+1)(2𝑟+1)}{6}\\ & \text{ and }{𝑆}_3=\sum {𝑟}^3={\left(\frac{𝑟(𝑟+1)}{2}\right)}^2\\ \therefore & \underset{𝑟\to \infty }{lim}\frac{{𝑆}_1(1+\frac{{𝑆}_3}{4})}{{𝑆}_2^2}\end{array}$

$\begin{array}{cc}& =\underset{𝑟\to \infty }{lim}\frac{\frac{𝑟(𝑟+1)}{2}(1+\frac{{𝑟}^2(𝑟+1{)}^2}{16})}{\frac{{𝑟}^2(𝑟+1{)}^2(2𝑟+1{)}^2}{36}}\\ & =\underset{𝑟\to \infty }{lim}\frac{𝑟(𝑟+1)}{2}(1+\frac{{𝑟}^2(𝑟+1{)}^2}{16})÷\frac{{𝑟}^2(𝑟+1{)}^2(2𝑟+1{)}^2}{36}\\ & =\underset{𝑟\to \infty }{lim}\frac{𝑟(𝑟+1)}{2}(1+\frac{{𝑟}^2(𝑟+1{)}^2}{16})\times \frac{36}{{𝑟}^2(𝑟+1{)}^2(2𝑟+1{)}^2}\\ & =\underset{𝑟\to \infty }{lim}1(1+\frac{{𝑟}^2(𝑟+1{)}^2}{16})\times \frac{18}{𝑟(𝑟+1)(2𝑟+1{)}^2}\\ & =18\underset{𝑟\to \infty }{lim}\left(\frac{16+{𝑟}^2(𝑟+1{)}^2}{16}\right)\cdot \frac{1}{𝑟(𝑟+1)(2𝑟+1{)}^2}\\ & =\frac{18}{16}\underset{𝑟\to \infty }{lim}(\frac{16}{{𝑟}^4}+{(1+\frac{1}{𝑟})}^2)\cdot \frac{1}{\left(1\right)(1+\frac{1}{𝑟}){(2+\frac{1}{𝑟})}^2}\end{array}$

$\begin{array}{cc}& =\frac{18}{16}{(0+(1+0{)}^2)}^2\left(\frac{1}{1(1+0)(2+0{)}^2}\right)\\ & =\frac{18}{16}\left(1\right)\times \left(\frac{1}{4}\right)\\ & =\frac{18}{64}=\frac{9}{32}\end{array}$

$\begin{array}{cc}& \text{ Let }2{\tan }^{-1}\left(\frac{1}{5}\right)=𝑥\\ \therefore & {\tan }^{-1}\left(\frac{1}{5}\right)=\frac{𝑥}{2}\\ \therefore & \tan \frac{𝑥}{2}=\frac{1}{5}\\ & \text{Using}\\ & \tan 2𝜃=\frac{2\tan 𝜃}{1-{\tan }^2𝜃}\\ \therefore & \tan 𝑥=\frac{2{\tan }^{\frac{𝑥}{2}}}{1-{\tan }^2\frac{𝑥}{2}}\end{array}$

$\begin{array}{cc}& \Rightarrow \tan 𝑥=\frac{2\times \frac{1}{5}}{1-\frac{1}{25}}\\ & \Rightarrow \tan 𝑥=\frac{5}{12}\text{.... (i)}\end{array}$

$\begin{array}{cc}& \text{ Consider }\tan (2{\tan }^{-1}\left(\frac{1}{5}\right)+\frac{𝜋}{4})\\ & =\tan (𝑥+\frac{𝜋}{4})\\ & =\frac{\tan 𝑥+\tan \frac{𝜋}{4}}{1-\tan 𝑥\cdot \tan \frac{𝜋}{4}}\\ & =\frac{\frac{5}{12}+1}{1-\frac{5}{12}}\\ & =\frac{\frac{17}{12}}{\frac{7}{12}}=\frac{17}{7}\end{array}$

$\begin{array}{cc}& \text{ Given, }\left|\overrightarrow{𝑎}\right|=1,\left|\overrightarrow{𝑏}\right|=1\text{ and }\mid \overrightarrow{𝑐}=2\\ & \begin{array}{cc}& \text{ Also, }\bar{a}\times (\bar{a}\times \bar{c})+\bar{b}=\bar{0}\\ & \Rightarrow (\bar{a}\cdot \bar{c})\bar{a}-(\bar{a}\cdot \bar{a})\bar{c}+\bar{b}=\bar{0}\\ & \Rightarrow (\bar{a}\cdot \bar{c})\bar{a}-\bar{c}+\bar{b}=\bar{0}\dots [\because \bar{a}\cdot \bar{a}=|\bar{a}{|}^2=1]\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow (\bar{a}\cdot \bar{c})\bar{a}-\bar{c}=-\bar{b}\\ & \Rightarrow \left|\right(\bar{a}\cdot \bar{c})\bar{a}-\bar{c}|=|-\bar{b}|\\ & \Rightarrow \left|\right(\bar{a}\cdot \bar{c})\bar{a}-\bar{c}{|}^2=|\bar{b}{|}^2\end{array}$

$\begin{array}{cc}& \Rightarrow \left|\right(\bar{a}\cdot \bar{c})\bar{a}{|}^2+|\bar{c}{|}^2-2\left\{\right(\bar{a}\cdot \bar{c})\bar{a}\cdot \bar{c}\}=|\bar{b}{|}^2\\ & \Rightarrow (\bar{a}\cdot \bar{c}{)}^2|\bar{a}{|}^2+|\bar{c}{|}^2-2(\bar{a}\cdot \bar{c}\left)\right(\bar{a}\cdot \bar{c})=|\bar{b}{|}^2\\ & \Rightarrow (\bar{a}\cdot \bar{c}{)}^2\left\{\right|\bar{a}{|}^2-2\}+|\bar{c}{|}^2=|\bar{b}{|}^2\end{array}$

$\Rightarrow -(\bar{a}\cdot \bar{c}{)}^2+4=1\dots [\because |\bar{b}{|}^2=1,|\bar{c}{|}^2=4]$

$\begin{array}{cc}& \Rightarrow (\bar{a}\cdot \bar{c}{)}^2=3\\ & \Rightarrow \bar{a}\cdot \bar{c}=\pm \sqrt{\surd 3}\\ & \Rightarrow \left|\bar{a}\right|\bar{c}\mid \cos 𝜃=\sqrt{\surd 3},\end{array}$

$\begin{array}{cc}& \text{ where }𝜃\text{ is an acute angle between }\bar{a}\text{ and }\bar{c}\\ & \Rightarrow \cos 𝜃=\frac{\sqrt{\surd 3}}{2}\Rightarrow 𝜃=\frac{𝜋}{6}\end{array}$

$\begin{array}{cc}& {𝑥}^2+{𝑦}^2=4\\ & \Rightarrow {𝑦}^2=4-{𝑥}^2\end{array}$

$\begin{array}{cc}& \text{ Required area }\\ & \begin{array}{cc}& =\underset{0}{\overset{2}{\int }}\sqrt{\surd 4-{𝑥}^2}𝑑𝑥\\ & ={[\frac{𝑥}{2}\sqrt{\surd 4-{𝑥}^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{𝑥}{2}\right)]}_2^0\\ & =[\frac{2}{2}\sqrt{\surd 4-(2{)}^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{2}{2}\right)-\frac{0}{2}\sqrt{\surd 4-0}-\frac{4}{2}{\sin }^{-1}\left(\frac{0}{2}\right)]\\ & =2\times \frac{𝜋}{2}\\ & =𝜋\text{ sq. units }\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }𝐴+𝐵=\left[\begin{array}{cc}1& \tan \frac{𝜃}{2}\\ -\tan \frac{𝜃}{2}& 1\end{array}\right]\text{.... (i)}\\ & {𝐴}^{𝑇}+{𝐵}^{𝑇}=\left[\begin{array}{cc}1& -\tan \frac{𝜃}{2}\\ \tan \frac{𝜃}{2}& 1\end{array}\right]\end{array}$

Since $𝐴$ is symmetric and $𝐵$ is skew symmetric

$\begin{array}{cc}& {𝐴}^{𝑇}=𝐴,{𝐵}^{𝑇}=-𝐵\\ \therefore & 𝐴-𝐵=\left[\begin{array}{cc}1& -\tan \frac{𝜃}{2}\\ \tan \frac{𝜃}{2}& 1\end{array}\right]\text{.... (ii)}\end{array}$

Adding (i) and (ii), we get

$\begin{array}{cc}2𝐴& =\left[\begin{array}{cc}1& \tan \frac{𝜃}{2}\\ -\tan \frac{𝜃}{2}& 1\end{array}\right]+\left[\begin{array}{cc}1& -\tan \frac{𝜃}{2}\\ \tan \frac{𝜃}{2}& 1\end{array}\right]\\ 2𝐴& =\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\\ \therefore 𝐴& =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

$\begin{array}{cc}& \text{ Now, subtracting (ii) from (i), we get }\\ & \begin{array}{cc}& 2𝐵=\left[\begin{array}{cc}1& \tan \frac{𝜃}{2}\\ -\tan \frac{𝜃}{2}& 1\end{array}\right]-\left[\begin{array}{cc}1& -\tan \frac{𝜃}{2}\\ \tan \frac{𝜃}{2}& 1\end{array}\right]\\ & 2𝐵=\left[\begin{array}{cc}0& 2\tan \frac{𝜃}{2}\\ -2\tan \frac{𝜃}{2}& 0\end{array}\right]\\ \therefore & 𝐵=\left[\begin{array}{cc}0& \tan \frac{𝜃}{2}\\ -\tan \frac{𝜃}{2}& 0\end{array}\right]\\ \therefore & {𝐵}^{-1}=\frac{1}{{\tan }^2\frac{𝜃}{2}}\left[\begin{array}{cc}0& -\tan \frac{𝜃}{2}\\ \tan \frac{𝜃}{2}& 0\end{array}\right]\end{array}\end{array}$

$\begin{array}{cc}& \therefore {𝐵}^{-1}=⎡⎣⎢\begin{array}{cc}0& \frac{-1}{\tan \frac{𝜃}{2}}\\ \frac{1}{\tan \frac{𝜃}{2}}& 0\end{array}⎤⎦⎥\\ & \therefore \text{ Now, }(A^{-1}B+{\mathrm{AB}}^{-1})\end{array}$

$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}0& \tan \frac{𝜃}{2}\\ -\tan \frac{𝜃}{2}& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]⎡⎣⎢\begin{array}{cc}0& \frac{-1}{\tan \frac{𝜃}{2}}\\ \frac{1}{\tan \frac{𝜃}{2}}& 0\end{array}⎤⎦⎥$

$=\left[\begin{array}{cc}0& \tan \frac{𝜃}{2}\\ -\tan \frac{𝜃}{2}& 0\end{array}\right]+⎡⎣⎢\begin{array}{cc}0& \frac{-1}{\tan \frac{𝜃}{2}}\\ \frac{1}{\tan \frac{𝜃}{2}}& 0\end{array}⎤⎦⎥$

$=⎡⎣⎢\begin{array}{cc}0& \frac{{\tan }^2\frac{𝜃}{2}-1}{\tan \frac{𝜃}{2}}\\ \frac{1-{\tan }^2\frac{𝜃}{2}}{\tan \frac{𝜃}{2}}& 0\end{array}⎤⎦⎥$

$=2⎡⎣⎢\begin{array}{cc}0& (-1)\frac{1-{\tan }^2\frac{𝜃}{2}}{2\tan \frac{𝜃}{2}}\\ \frac{1-{\tan }^2\frac{𝜃}{2}}{2\tan \frac{𝜃}{2}}& 0\end{array}⎤⎦⎥$

$\begin{array}{cc}& =2\left[\begin{array}{cc}0& \frac{-1}{\tan 𝜃}\\ \frac{1}{\tan 𝜃}& 0\end{array}\right]\\ \therefore & (A^{-1}B+{\mathrm{AB}}^{-1})\text{ at }𝜃=\frac{𝜋}{6}\\ & =2⎡⎣⎢\begin{array}{cc}0& \frac{-1}{\tan \frac{𝜋}{6}}\\ \frac{1}{\tan \frac{𝜋}{6}}& 0\end{array}⎤⎦⎥\\ & =2\left[\begin{array}{cc}0& -\sqrt{\surd 3}\\ \sqrt{\surd 3}& 0\end{array}\right]\\ & =\left[\begin{array}{cc}0& -2\sqrt{\surd 3}\\ 2\sqrt{\surd 3}& 0\end{array}\right]\end{array}$

$\begin{array}{cc}\text{ Let }𝐼& =\int \frac{𝑑𝜃}{{\cos }^2𝜃(\tan 2𝜃+\sec 2𝜃)}\\ & =\int \frac{{\sec }^2𝜃𝑑𝜃}{\left(\frac{2\tan 𝜃}{1-{\tan }^2𝜃}\right)+\left(\frac{1+{\tan }^2𝜃}{1-{\tan }^2𝜃}\right)}\\ & =\int \frac{{\sec }^2𝜃(1-{\tan }^2𝜃)𝑑𝜃}{(1+\tan 𝜃{)}^2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Put }\tan 𝜃=t\Rightarrow {\sec }^2𝜃𝑑𝜃=\mathrm{dt}\\ & \therefore \begin{array}{cc}I& =\int \frac{(1-t^2)\mathrm{dt}}{(1+t{)}^2}=\int \frac{1-t}{1+t}\mathrm{dt}\\ & =\int \frac{2-(1+t)}{1+t}\mathrm{dt}\\ & =2\log |1+t|-t+C\\ & =2\log |1+\tan 𝜃|-\tan 𝜃+𝐶\end{array}\end{array}\\ & \text{ Comparing with }𝜆\tan 𝜃+2\log \left|𝑓\right(𝜃\left)\right|+𝐶\text{, we get }\\ & \therefore 𝜆=-1,f\left(𝜃\right)=1+\tan 𝜃\end{array}$

$\begin{array}{cc}& 2^{𝑥}+2^{𝑦}=2\\ & \Rightarrow 2^{𝑦}=2-2^{𝑥}\\ & \Rightarrow 𝑦={\log }_2(2-2^{𝑥})\text{ is defined, if }2-2^{𝑥}>0\\ & \Rightarrow 2^{𝑥}<2\\ & \Rightarrow 2^{𝑥-1}<1\\ & \Rightarrow 𝑥-1<0\\ & \Rightarrow -\infty <𝑥<1\end{array}$