Maths

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$  

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

-> $C=+\frac{\pi }{2}$

now at x =  $\mathcal{l}n\sqrt[]{3}$  

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$  

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

 Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$  

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$  

So, f(x) = ax - b

Now,  $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$  

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$  

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$  

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$  

f(x) = ax – b

$=\frac{18x-4}{13}$

option (D) satisfies

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$  

option (C) is incorrect, there will be minima.

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

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

=2

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$  

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$  

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$  

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$  

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$  

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$  

$\frac{25S}{36}=1+\frac{3/5}{1}$  

$S=\frac{288}{125}$

Kindly go through the solution

B = (I – adjA)⁵

$\left|z\right|=3\Rightarrow$ circle with radius = 3

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4},$ part of a circle (with radius $\sqrt[]{2}$ ). no common points

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

$\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

$x=\pm \omega ,?{\omega }^2$

Now, =  ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$

=  ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$

= 1 + 1 – 1 = 1