MHT CET 2023 9th May Evening Shift (Mathematics) Question Paper with Solutions

Given curve is

$$\begin{array}{ll} & x y+a x+b y=0 \\ \therefore \quad & \text { Slope }=2=\frac{d y}{d x} \\ \therefore \quad & x y+a x+b y=0 \end{array}$$

Differentiating w.r.t. $x$, we get

$$\begin{array}{ll} & x \frac{\mathrm{d} y}{\mathrm{~d} x}+y+\mathrm{a}+\mathrm{b} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ \therefore \quad & (x+\mathrm{b}) \frac{\mathrm{d} y}{\mathrm{~d} x}=-(y+\mathrm{a}) \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-(y+\mathrm{a})}{x+\mathrm{b}} \\ \therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(1,1)}=2 \\ \therefore \quad & 2=\frac{-(1+\mathrm{a})}{1+\mathrm{b}} \\ \therefore \quad & \mathrm{a}+2 \mathrm{~b}=-3 \quad \text{..... (i)} \end{array}$$

$$\begin{aligned} & \text { Since }(1,1) \text { lies on } x y+a x+b y=0 \text {, we get } \\ & a+b=-1 \quad \text{.... (ii)} \end{aligned}$$

Solving (i), (ii), we get

$$\begin{aligned} & a=1, b=-2 \\ \therefore \quad & a-b=1-(-2)=3 \end{aligned}$$

$$\begin{aligned} f(x) & =x^3+6 x^2-36 x+7 \\ f^{\prime}(x) & =3 x^2+12 x-36 \\ & =3\left(x^2+4 x-12\right) \end{aligned}$$

For $\mathrm{f}(x)$ to be increasing,

$$\begin{aligned} & \mathrm{f}^{\prime}(x)>0 \\ & \Rightarrow 3\left(x^2+4 x-12\right)>0 \\ & \Rightarrow x^2+4 x-12>0 \\ & \Rightarrow(x+6)(x-2)>0 \\ & \Rightarrow x \in(-\infty,-6) \cup(2, \infty) \end{aligned}$$

$$\begin{aligned} & (\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})] \\ & =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\bar{a} \times \bar{a}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}-\bar{b} \times \bar{a}+\bar{b} \times \bar{b}+\bar{b} \times \bar{c}) \\ & =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\overline{0}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}+\bar{a} \times \bar{b}+\overline{0}+\bar{b} \times \bar{c}) \\ & =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\bar{c} \times \bar{a}+\bar{b} \times \bar{c}) \\ & =\bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \bar{b} \cdot(\bar{b} \times \bar{c}) \\ & +\bar{c} \cdot(\bar{c} \times \bar{a})+\bar{c} \cdot(\bar{b} \times \bar{c}) \end{aligned}$$

$$\begin{aligned} & =\overline{0}+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \times 0-0-0 \\ & =\left[\begin{array}{lll} \bar{a} & \bar{b} & \bar{c} \end{array}\right]+2\left[\begin{array}{lll} \bar{b} & \bar{c} & \bar{a} \end{array}\right] \\ & =\left[\begin{array}{lll} \bar{a} & \bar{b} & \bar{c} \end{array}\right]+2\left[\begin{array}{lll} \bar{a} & \bar{b} & \bar{c} \end{array}\right] \\ & =3\left[\begin{array}{lll} \bar{a} & \bar{b} & \bar{c} \end{array}\right] \end{aligned}$$

$$\left|\begin{array}{ccc} \cos (A+B) & -\sin (A+B) & \cos (2 B) \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \end{array}\right|=0$$

$\therefore \quad \cos (\mathrm{A}+\mathrm{B})[(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B})] +\sin (A+B)[\sin A \cos B+\sin B \cos A] +\cos 2 B\left[\sin ^2 A+\cos ^2 A\right]=0$

$$\begin{aligned} \therefore \quad & \cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}+\mathrm{B}) \\ & +\sin (\mathrm{A}+\mathrm{B}) \cdot \sin (\mathrm{A}+\mathrm{B})+\cos 2 \mathrm{~B}=0 \\ & \cos ^2(\mathrm{~A}+\mathrm{B})+\sin ^2(\mathrm{~A}+\mathrm{B})+\cos 2 \mathrm{~B}=0 \\ & 1+\cos 2 \mathrm{~B}=0 \end{aligned}$$

$$\begin{aligned} & 2 \cos ^2 \mathrm{~B}=0 \\ & \therefore \quad \cos \mathrm{B}=0 \end{aligned}$$

$\therefore \quad B=(2 n+1) \frac{\pi}{2} \text { for }(n \in Z)$

\begin{aligned} & y=x|x| \quad \ldots[\text { Given }]\\ & \text { Required area } \\ & =\int_\limits{-1}^1 x|x| \mathrm{d} x \\ & =2 \int_\limits0^1 x^2 \mathrm{~d} x \quad \text{...}[\because \text{Area is always positive]}\\ & =2 \times\left[\frac{x^3}{3}\right]_0^1 \\ & =2 \times\left(\frac{1}{3}-0\right)=\frac{2}{3} \text { sq.units} \end{aligned}

$$\begin{aligned} & \text { In } \triangle A B C \\ & \frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k \quad \text{... [Given]}\\ & \therefore \quad a+b=7 k \quad \text{.... (i)}\\ & b+c=8 k \quad \text{.... (ii)}\\ & c+a=9 k \quad \text{.... (iii)} \end{aligned}$$

Adding above equations,

$$\begin{aligned} & 2 \mathrm{a}+2 \mathrm{~b}+2 \mathrm{c}=24 \mathrm{k} \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}=12 \mathrm{k} \quad \text{.... (iv)} \end{aligned}$$

Solving equations (i), (ii), (iii), (iv) We get,

$$\begin{aligned} & \quad \mathrm{c}=5 \mathrm{k}, \mathrm{a}=4 \mathrm{k}, \mathrm{b}=3 \mathrm{k} \\ & \therefore \quad \mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2 \end{aligned}$$

$\therefore \quad \triangle \mathrm{ABC}$ is right angled triangle

$\therefore \quad \angle \mathrm{C}=90^{\circ}$

$$\begin{aligned} \begin{aligned} \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \mathrm{ab} \sin \mathrm{C} \\ =\frac{1}{2} \mathrm{ab} \sin 90 \\ =\frac{1}{2} \times 4 \mathrm{k} \times 3 \mathrm{k} =6 \mathrm{k}^2 \end{aligned} \\ \therefore \quad \text { Now, } \frac{[\mathrm{A}(\triangle \mathrm{ABC})]^2}{\mathrm{k}^4}=\frac{\left(6 \mathrm{k}^2\right)^2}{\mathrm{k}^4}=\frac{36 \mathrm{k}^4}{\mathrm{k}^4}=36 \end{aligned}$$

$$\begin{aligned} & \mathrm{f}(x) \text { is continuous at } x=0 \\ & \therefore \quad \text { L.H.L. }=\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{k x}{2}}{x^2} \\ & =2 \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{\mathrm{k} x}{2}}{\frac{\mathrm{k}^2 x^2}{4} \times 4} \cdot \mathrm{k}^2 \\ & =\frac{1}{2} \mathrm{k}^2 \lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{k} x}{2}}{\frac{\mathrm{kx}}{2}}\right)^2 \\ & \therefore \quad \text { L.H.L. }=\frac{1}{2} \mathrm{k}^2 \\ \end{aligned}$$

$$\begin{aligned} & \therefore \quad \text { R.H.L. }=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \\ & =\lim _{x \rightarrow 0} \frac{(\sqrt{x})(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} \\ & =\sqrt{16+\sqrt{0}}+4 \\ & \therefore \quad \text { R.H.L. }=8 \\ & \mathrm{f}(x) \text { is continuous at } x=0 \quad \text { ....[Given] } \\ & \therefore \quad \text { L.H.L. }=\text { R.H.L. } \\ & \therefore \quad \frac{1}{2} \mathrm{k}^2=8 \\ & \therefore \quad \mathrm{k}^2=16 \\ & \therefore \quad \mathrm{k}= \pm 4 \\ \end{aligned}$$

Given differential equation is

$$\begin{aligned} & \log \left(\frac{d y}{d x}\right)=a x+b y \\ \therefore \quad & \frac{d y}{d x}=e^{a x+b y} \\ \therefore \quad \frac{d y}{d x} & =e^{a x} \cdot e^{b y} \end{aligned}$$

$$\begin{array}{ll} \therefore \quad & \frac{\mathrm{d} y}{\mathrm{e}^{b y}}=\mathrm{e}^{a x} \cdot \mathrm{d} x \\ & \mathrm{e}^{-b y} \mathrm{~d} y-\mathrm{e}^{a x} \mathrm{~d} x=0 \end{array}$$

Integrating both sides, we get

$$\begin{aligned} & \int e^{-b y} d y-\int e^{a x} d x=0 \\ & \frac{e^{-b y}}{-b}-\frac{e^{a x}}{a}+c=0 \\ & \text { i.e., } \frac{e^{-b y}}{b}+\frac{e^{a x}}{a}=c \\ & a e^{-b y}+b e^{a x}=a b c \\ & a e^{-b y}+b e^{a x}=c_1, \text { where } c_1=a b c \end{aligned}$$

$$\begin{aligned} & \mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\} \\ & x^2+30 \leq 11 x \\ & x^2-11 x+30 \leq 0 \\ & (x-5)(x-6) \leq 0 \\ & x \in[5,6] \\ & \text { Now, } \mathrm{f}(x)=3 x^3-18 x^2+27 x-40 \\ & \mathrm{f}^{\prime}(x)=9 x^2-36 x+27 \\ & \mathrm{f}^{\prime}(x)=9\left(x^2-4 x+3\right) \\ & \quad=9\left[\left(x^2-4 x+4\right)-1\right] \\ & \quad=9(x-2)^2-9 \\ & \therefore \quad \mathrm{f}^{\prime}(x)>0 \forall x \in[5,6] \end{aligned}$$

$\therefore \quad \mathrm{f}(x)$ is strictly increasing in the interval $[5,6]$

$\therefore \quad$ Maximum value of $\mathrm{f}(x)$ when $x \in[5,6]$ is $f(6)=122$

$$\begin{aligned} & \mathrm{Z}_1=2+\mathrm{i} \\ & \mathrm{Z}_2=3-4 \mathrm{i} \\ & \overline{Z_1}=2-\mathrm{i} \\ & \overline{\mathrm{Z}_2}=3+4 \mathrm{i} \\ & \frac{\overline{Z_1}}{\bar{Z}_2}=\frac{2-\mathrm{i}}{3+4 \mathrm{i}} \\ & =\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)} \\ & =\frac{6-8 i-3 i+4 i^2}{(3)^2-(4 i)^2} \\ & =\frac{6-11 \mathrm{i}-4}{9+16} \quad \ldots\left[i^2=-1\right] \\ & \mathrm{a}+\mathrm{bi}=\frac{2-1 \mathrm{li}}{25} \\ & \therefore \quad a=\frac{2}{25}, b=\frac{-11}{25} \\ \end{aligned}$$

$$\begin{aligned} & \text { Now }-7 a+b \\ & =-7\left(\frac{2}{25}\right)-\frac{11}{25} \\ & =\frac{-14-11}{25} \\ & =\frac{-25}{25}=-1 \end{aligned}$$

$$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} \mathrm{d} x \\ & =\int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x}-\frac{\sin \alpha}{\cos \alpha}} d x \\ & =\int \frac{\sin x \cos \alpha+\sin \alpha \cos x}{\sin x \cos \alpha-\sin \alpha \cos x} \mathrm{~d} x \\ & =\int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} d x \\ & \text { Let } x-\alpha=\mathrm{t} \\ & \therefore \quad I=\frac{\sin (t+2 \alpha)}{\sin t} \\ & =\int \frac{\sin (t) \cos 2 \alpha+\cos (t) \sin 2 \alpha}{\sin (t)} d x \\ & =\cos 2 \alpha \int 1 d t+\sin 2 \alpha \int \cot (\mathrm{t}) \mathrm{dt} \\ & =\cos 2 \alpha \cdot \mathrm{t}+\sin 2 \alpha \cdot \log |\sin (\mathrm{t})|+\mathrm{c} \\ & \therefore \quad \mathrm{I}=(x-\alpha) \cos 2 \alpha+\log |\sin (x-\alpha)| \sin 2 \alpha+\mathrm{c} \\ & \text { But } \int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} \mathrm{d} x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\mathrm{c} \quad \text{.... [Given]}\\ & \Rightarrow \mathrm{A}(x)=x-\alpha, \mathrm{B}(x)=\log |\sin (x-\alpha)|+\mathrm{c} \\ \end{aligned}$$

$\log (x+y)=2 x y$

Differentiating w.r.t. $x$, we get

$$\begin{aligned} & \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \\ & \frac{1}{x+y}+\frac{1}{(x+y)} \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \\ & \left(\frac{1}{x+y}-2 x\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y-\frac{1}{x+y} \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}\left(\frac{1}{x+y}-2 x\right)=2 y-\frac{1}{x+y} \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left(2 y-\frac{1}{x+y}\right)}{\left(\frac{1}{x+y}-2 x\right)} \\ & \text { For } x=0, \log (y)=0 \\ & \Rightarrow y=1 \\ & \left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{(0,1)}=\frac{\left(2-\frac{1}{0+1}\right)}{\left(\frac{1}{0+1}-0\right)}=1 \end{aligned}$$

$$\begin{aligned} & \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C(1-\tan A \tan C)}{\cos A \cos C(1+\tan A \tan C)} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{(1-\tan A \tan C)}{(1+\tan A \tan C)} \end{aligned}$$

$$\begin{aligned} & \left(1-\tan ^2 \mathrm{~B}\right)(1+\tan \mathrm{A} \tan \mathrm{C}) \\ & =(1-\tan \mathrm{A} \tan \mathrm{C})\left(1+\tan ^2 \mathrm{~B}\right) \\ & 1+\tan \mathrm{A} \tan \mathrm{C}-\tan ^2 \mathrm{~B}-\tan ^2 \mathrm{~B} \tan \mathrm{A} \tan \mathrm{C} \\ & =1+\tan ^2 \mathrm{~B}-\tan \mathrm{A} \tan \mathrm{C}-\tan \mathrm{A} \tan \mathrm{C} \tan ^2 \mathrm{~B} \\ & 2 \tan \mathrm{A} \tan \mathrm{C}=2 \tan ^2 \mathrm{~B} \\ & \tan ^2 B=\tan A \cdot \tan C \\ \therefore \quad & \tan A, \tan B, \tan C \text { are in G.P. } \\ \end{aligned}$$

$$\begin{aligned} & P(A)=\frac{1}{2} \\ \therefore \quad & P\left(A^{\prime}\right)=1-\frac{1}{2}=\frac{1}{2} \\ P(B) & =\frac{1}{3} \end{aligned}$$

$$\begin{array}{ll} \therefore \quad & \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3} \\ & \mathrm{P}(\mathrm{C})=\frac{1}{4} \\ \therefore \quad & \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{1}{4}=\frac{3}{4} \\ \therefore \quad & \mathrm{P}\left(\mathrm{Problem}^{\prime} \text { is not solve }\right) \\ & =\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right) \\ & =\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right) \quad \cdots\left[\because \mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}^{\prime} \text { are independent }\right] \\ & =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\\ & =\frac{1}{4} \\ \therefore \quad & \mathrm{P}(\text { the problem will be solve }) \\ = & 1-\mathrm{P}(\text { Problem is not solved }) \\ = & 1-\frac{1}{4} \\ = & \frac{3}{4} \end{array}$$

$$\begin{array}{ll} & \text { Given equation } x^3+x-1=0 \\ & \text { Let } \mathrm{f}(x)=x^3+x-1 \\ \therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+1 \\ & \Rightarrow \mathrm{f}^{\prime}(x)>0 \forall x \in \mathrm{R} \\ & \Rightarrow \mathrm{f}(x) \text { is increasing } \\ \therefore \quad & \text { for } x_2>x_1, \mathrm{f}\left(x_2\right)>\mathrm{f}\left(x_1\right) \\ & \text { Now, } \mathrm{f}(0)=-1 \text { and } \mathrm{f}(1)=1 \\ \therefore \quad & \mathrm{f}(x)=0 \text { for some } x \in(0,1) \\ \therefore \quad & \text { Equation has one real root. } \end{array}$$

Total number of cards $=52$

Total number of queens $=4$

Probability of getting a queen

P(queen) $=\frac{4}{52}=\frac{1}{13}$

Probability of not getting a queen

$\mathrm{P}(\text { non queen })=\frac{48}{52}=\frac{12}{13}$

Let $\mathrm{X}$ be a random variable such that $X=$ number of queens in 2 draws

Case I: No queens are drawn $(X=0)$

$\mathrm{P}(\mathrm{X}=0)=\mathrm{P}$(non queen) $\times \mathrm{P}$(non queen)

$=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}$

Case II: One queen is drawn $(X=1)$

$P(X=1)=P($ non queen and queen $)$ or $\mathrm{P}($ queen and non queen)

$$\begin{aligned} & =\frac{12}{13} \times \frac{1}{13}+\frac{1}{13} \times \frac{12}{13} \\ & =\frac{24}{169} \end{aligned}$$

Case III: Two queens are drawn $(\mathrm{X}=2)$

$$\begin{aligned} \mathrm{P}(\mathrm{X}=2) & =\mathrm{P}(\text { queen }) \times \mathrm{P}(\text { queen }) \\ & =\frac{1}{13} \times \frac{1}{13} \\ & =\frac{1}{169} \end{aligned}$$

$$\begin{aligned} \begin{aligned} & \therefore \quad \text { Required Mean is } \\ & \begin{aligned} \mathrm{E}(\mathrm{X}) & =\Sigma x \cdot \mathrm{P}(x) \\ & =0 \times \frac{144}{169}+1 \times \frac{24}{169}+2 \times \frac{1}{169} \\ & =\frac{26}{169} \end{aligned} \\ & \mathrm{E}(\mathrm{X})=\frac{2}{13} \end{aligned} \end{aligned}$$

Circle passes through origin and centre lie on Y-axis.

Let $(0, k)$ be centre and '$k$' be radius

$\therefore \quad$ Equation of circle is

$$\begin{aligned} & (x-0)^2+(y-\mathrm{k})^2=\mathrm{k}^2 \\ & x^2+y^2-2 y \mathrm{k}+\mathrm{k}^2=\mathrm{k}^2 \\ & x^2+y^2-2 \mathrm{k} y=0 \\ & x^2+y^2=2 \mathrm{k} y \quad \text{.... (i)}\\ & \frac{x^2+y^2}{2 y}=\mathrm{k} \quad \text{.... (ii)} \end{aligned}$$

Differentiating equation (i) with respect to $x$, we get

$$\begin{aligned} & 2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{k} \frac{\mathrm{d} y}{\mathrm{~d} x} \\ & 2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}-2 \mathrm{k} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ & 2 x+2(y-\mathrm{k}) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ & 2 x+2\left[y-\left(\frac{x^2+y^2}{2 y}\right)\right] \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \quad \text{...[From (ii)]}\\ & 2 x+2\left[\frac{2 y^2-x^2-y^2}{2 y}\right] \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ & 2 x+\left(\frac{y^2-x^2}{y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ & 2 x y+\left(y^2-x^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ & \text { i.e. }\left(x^2-y^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}-2 x y=0 \end{aligned}$$