Maths

Required area is

= ${\displaystyle \underset{e-e^2}{\overset{0}{\int }}\mathcal{l}n\left(x+e^2\right)-1dx+{\displaystyle \underset{0}{\overset{\mathcal{l}n2}{\int }}2e^{-x}-1dx}}$

$=1+e-\mathcal{l}n2$

 $l_n\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{dt}{{\left(t^2+s\right)}^n}}$

Applying integral by parts

$l_n\left(x\right)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-{\displaystyle \underset{0}{\overset{x}{\int }}n{\left(t^2+5\right)}^{-n-1}.2t^2}$

$10n{l}_{n+1}\left(x\right)+\left(1-2n\right)l_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}$

Put n = 5

 $x=2\sqrt[]{2}cost\sqrt[]{sin2t}$

$\frac{dx}{dt}=\frac{2\sqrt[]{2}cos3t}{\sqrt[]{sin2t}}$

$y\left(t\right)=2\sqrt[]{2}sint\sqrt[]{sin2t}$

$\frac{dy}{dx}=\frac{2\sqrt[]{2}sin3t}{\sqrt[]{sin2t}}$

$\therefore \frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{1+1}{-3}=\frac{-2}{3}$

 $f\left(x\right)=ta{n}^{-1}\left(sinx-cosx\right)$

$f\text{'}\left(x\right)=\frac{cosx+sinx}{{\left(sinx-cosx\right)}^2+1}$ = 0

$\therefore x=\frac{3\pi }{4}$

Sum = tan^-1 $\sqrt[]{2}-\frac{\pi }{4}$

= $co{s}^{-1}\frac{1}{\sqrt[]{3}}-\frac{\pi }{4}$

 $f\left(x\right)=x{e}^{x\left(1-x\right)}$

$f\text{'}\left(x\right)=-e^{x\left(1-x\right)}\left(2x+1\right)\left(x-1\right)$

$f\left(x\right)is\uparrow in\left(\frac{-1}{2},-1\right)$

Note : n should be given as a natural number:

$f\left(x\right)=\{\begin{array}{l}\begin{array}{ccc}\hfill \frac{-sin\left(x-1\right)}{x-1}\hfill & \hfill ,\hfill & \hfill x<-1\hfill \\ \hfill -\left(sin2+1\right)\hfill & \hfill ,\hfill & \hfill x=-1\hfill \\ \hfill cos2\pi x\hfill & \hfill ,\hfill & \hfill -1<x<1\hfill \end{array}\\ 1,x=1\end{array}$

$\frac{-sin\left(x-1\right)}{x-1},x>1$

f (x) is discontinuous at x = 1 and x = 1

 $f\left(0\right)+3+\lambda +4=14$

$\therefore f\left(0\right)=7-\lambda =c$

$f\left(1\right)=a+b+c=3$ ……. (i)

$f\left(3\right)=9a+3b+c=4$ …… (ii)

$f\left(-2\right)=4a-2b+c=\lambda$ …. (iii)

(ii) – (iii)

$a+b=\frac{4-\lambda }{5},$ put in equation (i)

$6\lambda =24\lambda =4$

Given that $A^T=A,B^T=-B$

(A) $C=A^4-B^4$

$=A^4-B^4=C$

(B)C = AB – BA

$=B^T{A}^T-A^T{B}^T=-BA+AB=C$

(C) $C=B^5-A^5$

$C^T={\left(B^5-A^5\right)}^T={\left(B^5\right)}^T-{\left(A^5\right)}^T=-B^5-A^5$

(D)C = AB + BA

= BA – AB = C

$\therefore$ option is true

 $a=\frac{-1}{{\alpha }^2}-\frac{1}{{\beta }^2}-2,b=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+1+\frac{1}{{\alpha }^2{\beta }^2}$

$6x^2+17x+7=0,x=\frac{-7}{3},x=\frac{-1}{2}$ are roots

Both roots, real and negative