Class 12th

From structure it is clear [CO₂ (CO)₂] has bridging carbonyl ligand.

In presence of light allylic substitution occur.

In presence of CCl₄, addition reaction will occur.

KMnO₄ decomposes upon heating at 513 K and forms K₂MnO₄ and MnO₂.

2KMnO₄  K₂MnO₄ + O₂ + MnO₂

  $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  

⇒   $\overrightarrow{a}+5\overrightarrow{b}$ = $\overrightarrow{c}$           …(1)

$\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$  

⇒   $\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}$               …(2)

From (1) and (2)

  $\overrightarrow{b}+6\overrightarrow{c}=\mu \left(\lambda \overrightarrow{c}-5\overrightarrow{b}\right)$          

-> $\left(1+5\mu \right)\overrightarrow{b}+\left(6-\lambda \mu \right)\overrightarrow{c}=0$

$?\overrightarrow{b}$ and $\overrightarrow{c}$  are non-collinear

-> 1 + 5m = 0 $\mu =\frac{-1}{5}$  and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

$\overrightarrow{b}=6\overrightarrow{c}=\frac{-1}{5}\overrightarrow{a}$

$5\overrightarrow{b}+30\overrightarrow{c}=-\overrightarrow{a}$

$\begin{array}{cc}\hfill \overrightarrow{a}+5\overrightarrow{b}+30\overrightarrow{c}=0\hfill & \hfill \overrightarrow{a}+\alpha \overrightarrow{b}+\beta \overrightarrow{c}=0\hfill \end{array}]$

On comparing

α = 5, β = 30 ⇒ α + β = 35

y (x) = 2^x – x²

y? (x) = 2x log 2 – 2x

M = 3

N = 2

M + N = 5

${\displaystyle \int \frac{\left(sinx-cosx\right)si{n}^{2}x}{tanx\left(si{n}^{3}x+co{s}^{3}x\right)}dx}$

${\displaystyle \int \frac{\left(sinx-cosx\right)sinxcosx}{si{n}^{3}x+co{s}^{3}x}dx}$ , put sin³x + cos³x = t(3 sin²x*cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3}{\displaystyle \int \frac{dt}{t}}$

$=\frac{lnt}{3}+c$

$=\frac{ln\left|si{n}^{3}x+co{s}^{3}x\right|}{3}+c$

$f\left(x\right)=\frac{\left(2^x+2^{-x}\right)tanx\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}}{{\left(7x^2-3x+1\right)}^3}$

$f\left(x\right)=\left(2^x+2^{-x}\right).tanx.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}$

$f\text{'}\left(x\right)=\left(2^x+2^{-x}\right).se{c}^{2}x.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}+tanx.\left(Q\left(x\right)\right)$

$f\text{'}\left(0\right)=2.1\sqrt[]{\frac{\pi }{4}}.1$

$=\sqrt[]{\pi }$

$Area=\frac{\pi {\left(13\right)}^2}{2}-\left[\frac{1}{2}*25*5+{\displaystyle \underset{12}{\overset{13}{\int }}\sqrt[]{\left(169-y^2\right)}}\cdot dy\right]$

$=\frac{169\pi }{2}-\left[\frac{125}{2}+{\left[\frac{y}{2}\sqrt[]{169-y^2}+\frac{169}{2}si{n}^{-1}\frac{y}{13}\right]}_{12}^{13}\right]$

$=\frac{169}{2}\pi -\frac{125}{2}-\left[\frac{169}{2}*\frac{\pi }{2}-6*5-\frac{169}{2}si{n}^{-1}\frac{12}{13}\right]$

$=\frac{169\pi }{4}-\frac{65}{2}+\frac{169}{2}si{n}^{-1}\frac{12}{13}$

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{x^{2}cosx}{1+{\pi }^2}+\frac{1+si{n}^{2}x}{1+e^{{\left(sinx\right)}^{2023}}}\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{x^{2}cosx}{1+{\pi }^x}+\frac{1+si{n}^{2}x}{1+e^{{\left(sinx\right)}^{2023}}}\right)+\left(\frac{x^{2}cosx}{1+{\pi }^{-x}}+\frac{1+si{n}^{2}x}{1+e^{-{\left(sinx\right)}^{2023}}}\right)\right\}dx}$

$=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(x^{2}cosx+1+si{n}^{2}x\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}cosxdx}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$ .(1)

Let $I_1={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx}$

$I_1={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}1\cdot dx}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{1-cos2x}{2}\right)dx}$

$I_1=\frac{\pi }{2}+\frac{1}{2}\left[\frac{\pi }{2}+0\right]$

$I_1=\frac{3\pi }{4}$

Let $I_2={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}cosxdx}$

$I_2={\left[x^2\left(sinx\right)-{\displaystyle \int 2x}{\displaystyle \int cosxdx}\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^2\left(sinx\right)-2{\displaystyle \int xsinx}\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^{2}sinx-2\left(x\left(-cosx\right)+{\displaystyle \int cosx}\right)\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^{2}sinx-2\left(-xcosx+sinx\right)\right]}_0^{\frac{\pi }{2}}$

$I_2=\left(\frac{{\pi }^2}{4}-2\right)$

Put l₁ and l₂ in (1)

$\frac{{\pi }^2}{4}-2+\frac{3\pi }{4}$

$\frac{{\pi }^2}{4}+\frac{3\pi }{4}-2$

$\frac{\pi }{4}\left(\pi +3\right)-2$

α = 3

  $\frac{dy}{dx}-\left(\frac{sin2x}{1+co{s}^{2}x}\right)y=\frac{sinx}{1+co{s}^{2}x}$

IF = $e^{-{\displaystyle \int \frac{sin2xdx}{1+co{s}^{2}x}}}$  

$=e^{ln\left(1+co{s}^{2}x\right)}=\left(1+co{s}^{2}x\right)$        

So, y(1 + cos² x) = ${\displaystyle \int \frac{sinx}{\left(1+co{s}^{2}x\right)}\cdot \left(1+co{s}^{2}x\right)dx}$  

y(1 + cos² x) = – cos x + c

$?$      y(0) = 0

0 = – 1 + c

-> c = 1

$y=\frac{1-cosx}{1+co{s}^{2}x}$   

Now, $y\left(\frac{\pi }{2}\right)=1$