Let 

\(\large y=\Large(\frac{1}{x})\large ^x\\\large y=x^{-x}\)

Taking log both side,

\(\large\log y=-x\log x\)

diff. w.r.t x

\(\Large\frac{1}{y}\frac{dy}{dx}\large=-1-\log x\\\large \frac{dy}{dx}=-x^{-x}(1+\log x)\)

For max value, 

\(\Large \frac{dy}{dx}\large=0\\\large -x^{-x}(1+\log x)=0\\\large 1+\log x=0\\\large \log x=-1\\\large x=e^{-1}\)

Therefore the max value attaind at \(\large x=e^{-1}\) and max value is \(\Large(\frac{1}{e^{-1}})^{e^{-1}}=\large e^{\Large\frac{1}{e}}\)

The contrapositive of the converse of the statement is "if x is not s prime number, then x is not odd"

\(\large i^n+i^{n+1}+i^{n+2}+i^{n+3}=i^n(1+i+i^2+i^3)\)

                                            \(\large= i^n(1+i-1-i)=i^n.0=0\)

\(\large CV=\Large\frac{\sigma}{\overline x}\large\times 100\)

Hence

\(\large 60=\Large\frac{21}{\overline x_1}\large\times 100\\\large \overline x_1=35\)

and 

\(\large 70=\Large\frac{16}{\overline x_2}\large\times 100\\\large \overline x_2=22.85\)

Hence their mean is 35 and 22.85

\(\Large\frac{dx}{dt}\large=2t+3, \Large\frac{dy}{dt}\large=4t-2\)

So \(\Large\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t-2}{2t+3}\)

Now put x=2, y=-1 in the given equation 

\(\large 2=t^2+3t-8\\\large t^2+3t-10=0\\\large(t+5)(t-2)\\\large t=-5,2\)

\(\large -1=2t^2-2t-5\\\large 2t^2-2t-4=0\\\large(t-2)(t+1)\\\large t=2,-1\)

Therefore common point is \(\large t=2\)

Hence \(\large t=2\), \(\Large\frac{dy}{dx}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}\)

\(\large \vec a+\vec b+\vec c=0\\\large \vec a+\vec b=-\vec c\\\large |\vec a+\vec b|=|-\vec c|\\\large (\vec a+\vec b).(\vec a+ \vec b)=49\\\large |\vec a|^2+|\vec b|^2+2|\vec a|\vec b|\cos\theta =49\\\large 9+25+2.3.5\cos\theta=49\\\large 30\cos\theta=15\\\large \cos\theta=\Large\frac{1}{2}\\\large\theta=\Large\frac{\pi}{3}\)

\(\large a*b=\Large\frac{a+b}{4}=\Large\frac{b+a}{4}\large=b*a\)

\(\large\therefore *\) is commutative.

Now \(\large a*(b*c)=a*\Large \Large\frac{b+c}{4}=\Large\frac{a+\frac{b+c}{4}}{4}\)

and \(\large (a*b)*c=\Large \Large\frac{a+b}{4}\large *c=\Large\frac{\frac{a+b}{4}+c}{4}\)

Hence \(\large a*(b*c)\neq (a*b)*c\)

\(\large\therefore *\) is not associative.

\(\large x^my^n=(x+y)^{m+n}\)

taking log both side

\(\large m\log x+n\log y=(m+n)\log(x+y)\)

diff. w.r.t x

\(\Large\frac{m}{x}+\Large\frac{n}{y}\Large\frac{dy}{dx}=\Large\frac{m+n}{x+y}\large(1+\Large\frac{dy}{dx})\\\large (\Large\frac{nx+ny-my-ny}{y(x+y)})\Large\frac{dy}{dx}=\Large\frac{mx+nx-mx-my}{x(x+y)}\\\large \Large\frac{dy}{dx}\large=\Large\frac{y}{x}\)

\(\large xy=e^{\sin^{-1}(t^2-1)}e^{\sec^{-1}\Large(\frac{1}{t^2-1})}\)

     \(\large =e^{\sin^{-1}(t^2-1)}e^{\cos^{-1}(t^2-1)}\\\large=e^{\sin^{-1}(t^2-1)+\cos^{-1}(t^2-1)}\\\large=e^{\Large\frac{\pi}{2}}\)

diff. w.r.t x

\(\large y+x\Large\frac{dy}{dx}\large=0\\\large \Large\frac{dy}{dx}=\frac{-y}{x}\)

\(\large1+\sin\theta+\sin^2\theta+...\infty=2\sqrt3+4\\\Large \frac{1}{1-\sin\theta}\large=2\sqrt3+4 (\because 1+r+r^2+...\infty=\Large\frac{1}{1-r})\\\large \sin\theta=1-\Large\frac{1}{2\sqrt3+4}\\\large \sin\theta=\Large\frac{2\sqrt3+3}{2(\sqrt3+2)}\\\large\sin\theta=\Large\frac{\sqrt3}{2}\\\large\theta=\Large\frac{\pi}{3}\)

\(\large[1+\Large(\frac{dy}{dx})^2+\large \sin\Large (\frac{dy}{dx})]^{\frac{3}{4}}=\frac{d^2y}{dx^2}\\\large[1+\Large(\frac{dy}{dx})^2+\large \sin\Large (\frac{dy}{dx})]^3=(\frac{d^2y}{dx^2})^4\)

Therefore order of diff. equation is of order 2 and degree not defined.

\(\large \sin^{-1}(\cos \Large\frac{53\pi}{4})\large=\sin^{-1}(\cos(10\pi+\Large\frac{3\pi}{5}))\)

                             \(\large =\sin^{-1}(\cos \Large\frac{3\pi}{5})\\\large=\sin^{-1}(\sin\Large(\frac{\pi}{2}-\frac{3\pi}{5})\\\large=\sin^{-1}(\sin\Large(\frac{-\pi}{10}))=\Large\frac{-\pi}{10}\)

\(\large \vec a, \vec b, \vec c\) are perpendicular to the sum of the remaining.

Hence \(\large \vec a.\vec b+\vec b.\vec c+\vec c.\vec a=0\)

Now \(\large |\vec a+\vec b+\vec c|^2=(\vec a+\vec b+\vec c).(\vec a+\vec b+\vec c)\)

                               \(\large =|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)\\\large =9+16+25+2.0\\\large =50\)

Hence \(|\vec a+\vec b+\vec c|=5\sqrt2\)

\(\large (1-\cos \theta+i\sin\theta)^{-1}=\Large\frac{1}{1-\cos\theta+i\sin\theta}\)

                                         \(\Large=\frac{1-\cos\theta-i\sin\theta}{(1-\cos\theta)^2+\sin^2\theta}\\\Large=\frac{1-\cos\theta-i\sin\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}\\\Large=\frac{1-\cos\theta-i\sin\theta}{2-2\cos\theta}\\\Large=\frac{1}{2}\large-i\frac{\sin\theta}{2(1-\cos\theta)}\)

Hence real part is \(\Large=\frac{1}{2}\).

\(\large 2x+3y-3=0 , x+ky+7=0\) are perpendicular .

Hence 

\(\large 2.1+3.k=0\\\large 2+3k=0\\\large k=-\Large\frac{2}{3}\)

\(\large A-B=\Large\frac{1}{\pi}\large\begin{bmatrix}\sin^{-1}(\pi x)+\cos^{-1}(\pi x)&\tan^{-1}\frac{x}{\pi}-\tan^{-1}\frac{x}{\pi}\\\sin^{-1}\frac{x}{\pi}-\sin^{-1}\frac{x}{\pi}&\cot^{-1}(\pi x)+\tan^{-1}(\pi x)\end{bmatrix}\)

              \(\large= \Large\frac{1}{\pi}\begin{bmatrix}\Large\frac{\pi}{2}\large&0\\0&\Large\frac{\pi}{2}\end{bmatrix}\\\large=\Large\frac{1}{2}\large\begin{bmatrix}1&0\\0&1\end{bmatrix}\\\large=\Large\frac{1}{2}\large I\)

Number of one-one mapping from A to B is \(\large ^5P_4=\Large\frac{5!}{1!}\large=5.4.3.2.1=120\)

\(\large x\Large\frac{dy}{dx}\large-y=x^4-3x\\\large \Large\frac{dy}{dx}-\frac{y}{x}\large=x^3-3\)

which is linear in y.

so \(\large IF=e^{-\int \Large\frac{1}{x}\large dx}=e^{-\log x}=\Large\frac{1}{x}\)

\(\large A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)

\(\large 5A=\begin{bmatrix}15&5\\-5&10\end{bmatrix}\)

Hence \(\large A^2-5A=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}=-7\begin{bmatrix}1&0\\0&1\end{bmatrix}=-7I\)

Probability of the card being aces

 \(\large=\Large\frac{^4c_2}{^{52c_2}}\\\Large=\frac{\frac{4!}{2!2!}}{\frac{52!}{2!50!}}\\\Large=\frac{4.3}{52.51}=\frac{1}{221}\)

let 

\(\large f(x)=\sin^{103}x\cos^{101}x\\\large \therefore \large f(-x)=\sin^{103}(-x)\cos^{101}(-x)\)

                 \(\large =-\sin^{103}x\cos^{101}x\\\large=-f(x)\)

Hence f(x) is odd function.

So \(\large\int _{-\frac{\pi}{4}}^{\frac{\pi}{4}}\large \sin^{103}x\cos^{101}x dx=0\)

  \(\large\lim_{x\to 0}\Large\frac{xe^x-\sin x}{x} [\frac{0}{0}]\)

\(\large =\lim_{x\to0}\Large\frac{xe^x+e^x-\cos x}{1}\\\large=0+1-1=0\)[Using L'hospital rule]

\(\large x=2+3\cos\theta, y=1-3\sin\theta\\\large x-2=3cos\theta...(1)\\\large y-1=-3\sin\theta....(2)\)

Squaring (1) and (2) and adding

\(\large (x-2)^2+(y-1)^2=9cos^2\theta+9\sin^2\theta\\\large (x-2)^2+(y-1)^2=3^2\)

Therefore centre of teh circle is \(\large (2,1) \) and radius is 3.

\(\large \sin^{-1}x+\sin^{-1}y=\Large\frac{\pi}{2}\\\large \sin^{-1}x=\Large\frac{\pi}{2}\large-\sin^{-1}y\\\large \sin^{-1}x=\cos^{-1}y\\\large \sin^{-1}x=\sin^{-1}\sqrt{1-y^2}\\\large x=\sqrt{1-y^2}\\\large x^2=1-y^2\)

\(\large \sin1^{\circ}+ \sin2^{\circ}+..+ \sin359^{\circ}=( \sin1^{\circ}+ \sin359^{\circ})+( \sin2^{\circ}+ \sin358^{\circ})+....+ \sin180^{\circ}\)

                                                   \(\large =2 \sin180 ^{\circ}. \cos179^{\circ}+2\sin180 ^{\circ}.\cos178 ^{\circ}+...+\sin180 ^{\circ}\) (use the formula \(\large \sin c+\sin d=2\sin \Large\frac{c+d}{2}\large\cos\Large\frac{c-d}{2} \))

                                                   \(\large =0+0+...+0(\because \sin 180^{\circ}=0)\\\large =0 \)

The 11 th term is 

\(\large T_{11}=^{14}c_{10}x^{14-10}\Large(\frac{1}{\sqrt x})^{10}\)

       \(\Large=\frac{14!}{10!.4!}\frac{x^4}{x^5}\\\Large=\frac{14.13.12.11}{4.3.2.1}\frac{1}{x}\\\Large= \frac{1001}{x}\)

order of matroix A is \(\large m\times n\)

Let order od B' is \(\large x\times y\)

Since AB' is defined, so column of A=row of B' hence \(\large n=x\)

Also B'A is defined, so column of B'=row of A hence \(\large y=m\)

Hence order of B' is \(\large n\times m\)

Let \(\large u=\log_{10}x, v=log_x10\)

So \(\large u=\Large\frac{\log x}{\log 10}, v=\Large\frac{\log 10}{\log x}\)

\(\Large\frac{du}{dx}=\Large\frac{1}{x\log 10}, \frac{dv}{dx}=\Large\frac{-\log 10}{x(\log x)^2}\)

\(\Large\frac{du}{dv}=\frac{du/dx}{dv/dx}\)

       \(\Large=\frac{1}{x\log10}\div(\Large\frac{-\log 10}{x(\log x)^2})\\\Large =\frac{-(\log x)^2}{(\log10)^2}\\\large=-(\log_{10}x)^2\)

\(\large x^3-3xy^2+2=0\)

diff. w.r.t x

\(\large 3x^2-3y^2-6xy\Large\frac{dy}{dx}\large=0\\\Large \Large\frac{dy}{dx}=\frac{x^2-y^2}{2xy}\)

Therefore slope at \(\large P(h,k)\) is

\(\large m_1=\Large \frac{h^2-k^2}{2hk}\)

\(\large 3x^2y-y^3=2\)

diff. w.r.t x

\(\large 6xy+3x^2\Large\frac{dy}{dx}\large-3y^2\Large\frac{dy}{dx}\large=0\\\large \Large\frac{dy}{dx}\large=\Large\frac{-2xy}{x^2-y^2}\large\)

So slope at \(\large P(h,k)\)is

\(\large m_2=\Large\frac{-2hk}{h^2-k^2}\large\)

Here \(\large m_1.m_2=-1\)

Therefore both the curves cut at right angle.

\(\large\tan^{-1}(x^2+y^2)=\alpha\\\large x^2+y^2=\tan\alpha\)

diff. w.r.t x

\(\large 2x+2y\Large\frac{dy}{dx}\large=0\\\Large\frac{dy}{dx} =\frac{-x}{y}\)

Area of the circle is \(\large A=\pi r^2\)

diff. w.r.t r

\(\Large\frac{dA}{dr}\large=2\pi r\)

at r=2, \(\Large\frac{dA}{dr}\large=2\pi \times 2=4\pi\)

Therefore the rate of change of area with respect to radius r=2 is \(\large =4\pi\)

\(\large I=\int _0^{\Large\frac{\pi}{2}}\frac{\sin^{1000}x}{\sin^{1000}x+\cos^{1000}x}\large dx....(1)\)

\(\large I=\int _0^{\Large\frac{\pi}{2}}\frac{\sin^{1000}(\Large\frac{\pi}{2}-x)}{\sin^{1000}(\Large\frac{\pi}{2}-x)+\cos^{1000}(\Large\frac{\pi}{2}-x)}\large dx \\\large=\int _0^{\Large\frac{\pi}{2}}\frac{\sin^{1000}x}{\cos^{1000}x+\sin^{1000}x}\large dx....(2)\)

Adding (1) and (2)

\(\large 2I=\int _0^{\Large\frac{\pi}{2}}\frac{\sin^{1000}x+\cos^{1000}x}{\sin^{1000}x+\cos^{1000}x}\large dx\\\large 2I=\int _0^{\Large\frac{\pi}{2}}dx\\\large 2I=\Large\frac{\pi}{2}\\\large I=\Large\frac{\pi}{4}\)

\(\large \tan\theta=\Large\frac{\sin2\theta}{1+\cos2\theta}\)

So \(\large \tan\Large\frac{\pi}{8}=\Large\frac{\sin2(\frac{\pi}{8})}{1+\cos2(\frac{\pi}{8})}\)

                 \(\Large=\frac{\sin\frac{\pi}{4}}{1+\cos \frac{\pi}{4}}\\\Large=\frac{\frac{1}{\sqrt2}}{1+\frac{1}{\sqrt2}}\\\Large=\frac{1}{\sqrt2+1}\)

\(\Large\frac{dy}{y}+\frac{dx}{x}\large=0\)

\(\Large\int\frac{dy}{y}+\int\frac{dx}{x}\large=ln c\\\large \ln y+\ln x=ln c\\\large xy=c\)integrating both side

Let (x,y,z) be the foot of the perpendicular drawn from origin to the palne 5y+8=0

Hence

\(\Large\frac{x-0}{0}=\frac{y-0}{5}=\frac{z-0}{0}=\frac{-(0+0+0+8)}{0^2+5^2+0^2}\\\large x=0, y=\Large\frac{-5.8}{25}\large ,z=0\\\large x=0, y=\Large\frac{-8}{5}\large, z=0\)

Therefore the coordinate of the foot of the perpendicular is \(\large (0,\Large\frac{-8}{5}\large,0)\)

\(\large\tan^{-1}\Large\frac{x}{y}\large-\tan^{-1}\frac{x-y}{x+y}=\large\tan^{-1}\Large\frac{x}{y}\large-\tan^{-1}\frac{\frac{x}{y}-1}{\frac{x}{y}+1}\\\large =\tan^{-1}\Large\frac{x}{y}\large -(\tan^{-1}\Large\frac{x}{y}\large-\tan^{-1}1)\\\large=\tan^{-1}1=\Large\frac{\pi}{4}\)

\(\large A+A^T=I\\\large \begin{bmatrix}\cos2\theta&-\sin2\theta\\\sin2\theta&\cos2\theta\end{bmatrix}+\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\\\large \begin{bmatrix}2\cos2\theta&0\\0&2\cos2\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

comparing both side 

\(\large 2\cos2\theta=1\\\large cos2\theta=\Large\frac{1}{2}\\\large 2\theta=\Large\frac{\pi}{3}\\\large \theta=\Large\frac{\pi}{6}\)

\(\large\int\Large\frac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}\large\mathrm{d}x=\int \Large\frac{x^6-x^5}{x^4-x^3}\large\mathrm{d}x\)

                                   \(\large=\int \Large\frac{x^5(x-1)}{x^3(x-1)}\large\mathrm{d}x\\\large=\int x^2 \mathrm{d}x =\Large\frac{x^3}{3}\)

\(\large3\tan^{-1}x+\cot^{-1}x=\pi\\\large 2\tan^{-1}x+\tan^{-1}x+\cot^{-1}x=\pi\\\large 2\tan^{-1}x+\Large\frac{\pi}{2}\large=\pi\\\large 2\tan^{-1}x=\Large\frac{\pi}{2}\\\large \tan^{-1}x=\Large\frac{\pi}{4}\\\large x=1\)

\(\large I=\int \Large\frac{e^x(x^2\tan^{-1}x+\tan^{-1}x+1)}{x^2+1}\large\mathrm{d}x\)

   \(\large=\int e^x(\tan^{-1}x+ \Large\frac{1}{x^2+1})\large\mathrm{d}x\\\large=e^x\tan^{-1}x-\int e^x \Large\frac{1}{x^2+1}\large\mathrm{d}x+\int e^x \Large\frac{1}{x^2+1}\large\mathrm{d}x+c\\\large=e^x\tan^{-1}x+c\)

By given condition

\(\large |x.(\hat i+\hat j+\hat k)|=1\\\large \sqrt{x^2+x^2+x^2}=1\\\large 3x^2=1\\\large x=\pm\Large\frac{1}{\sqrt3}\)

\(\large \cos\alpha,\cos\beta,\cos\gamma\) are direcion cosine of a vector a.

\(\large \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\\\Large\frac{1}{2}\large(1+\large \cos2\alpha+1+\cos2\beta+1+\cos2\gamma)=1\\\large \large \cos2\alpha+\cos2\beta+\cos2\gamma+3=2\\\large \cos2\alpha+\cos2\beta+\cos2\gamma=-1\)

If A is a matrix of order n then \(\large |kA|=k^n|A|\)

\(\large |3A|=3^3|A|=27|A|\)

\(\large\begin{vmatrix}\log x &\log y&\log z\\\log2x&\log2y&\log2z\\\log3x&\log3y&\log3z\end{vmatrix}=\large\begin{vmatrix}\log x &\log y&\log z\\\log2&\log2&\log2\\\log3&\log3&\log3\end{vmatrix}\) (applying the operation \(\large R_2\rightarrow R_2-R_1, R_3\rightarrow R_3-R_1\))

                                             \(\large=\log2.\log3\begin{vmatrix}\log x &\log y&\log z\\1&1&1\\1&1&1\end{vmatrix}\\\large=0\)

(as two rows are same)

\(\large f(x)=[x]\) is conmtinuous at only non-integral values. Hence here f(x) is continuous at 1.5.

\(\large 2 \vec a.\vec b=|\vec a|.|\vec b|\\\large 2|\vec a|.|\vec b|\cos \theta=|\vec a|.|\vec b|\\\large \cos\theta =\Large\frac{1}{2}\\\large \theta=60^{\circ}\)

\(\large 4y^2+3x+3y+1=0\\\large y^2+\Large\frac{3y}{4}=-\frac{3x}{4}-\frac{1}{4}\\\large y^2+\Large\frac{3y}{4}+\frac{3^2}{8^2}=-\frac{3x}{4}-\frac{1}{4}+\frac{9}{64}\\\large (y+\frac{3}{8})^2=-\frac{3}{4}(x+\frac{7}{48})\)

Hence the length of latus recum is \(\Large \frac{3}{4}\)

\(\large P(A|B)=\Large\frac{P(A\cap B)}{P(B)}\)

               \(\Large=\frac{7/10}{17/20}=\frac{14}{17}\)

\(\large\begin{vmatrix}1+x&1&1\\1&1+y&1\\1&1&1+z\end{vmatrix}=0\\\large (1+x)(1+y+z+yz-1)+1(1-1-z)+1(1-1-y)=0\\\large y+z+yz +xy+xz+xyz-z-y=0\\\large xy+yz+zx=-xyz\\\large \Large\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=\large-1\\\large x^{-1}+y^{-1}+z^{-1}=-1\)

Let
 \(\large xe^x=t\\\large (xe^x+e^x)dx=dt\\\large e^x(x+1)dx=dt\)

Hence \(\large I=\int \Large\frac{dt}{\cos^2t}\)

              \(\large=\int\sec^2t dt\\\large= \tan t+c\\\large=\tan(xe^x)+c\)

When two dice are thrown, then total number of outcomes is 36.

And the sample space such that total number is 5 is \(\large\{(1,4), (2,3),(3,2),(4,1)\}\)

Hence the required probability \(\Large=\frac{4}{36}=\frac{1}{9}\)

\(\large |\vec a|=|\vec b|=| \sqrt3\vec a-\vec b|=1\)

\(\large |\sqrt3\vec a-\vec b|^2=(\sqrt3\vec a-\vec b).(\sqrt3\vec a - \vec b)\\\large 1=3|\vec a|^2+|\vec b|^2-2\sqrt3|\vec a||\vec b|\cos\theta\\\large 1=3+1-2\sqrt3.1.1\cos\theta\\\large 2\sqrt3\cos\theta=3\\\large \cos\theta=\Large\frac{\sqrt3}{2}\\\large\theta=30^{\circ}\)

\(\large \cot\theta+\tan\theta=2\\\Large \frac{\cos\theta}{\sin\theta}+ \frac{\sin\theta}{\cos\theta}\large=2\\\Large \frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}\large=2\\\large 2\sin\theta\cos\theta=1\\\large \sin2\theta=1\\\large 2\theta=n\pi+(-1)^n\Large\frac{\pi}{2}\\\large\theta=\Large \frac{n\pi}{2}\large+(-1)^n\Large\frac{\pi}{4}\)

Vector equation of the plane is 

\(\large \vec r.\hat n=d\\\large \vec r.\Large \frac{2\hat i-3\hat j+\hat k}{\sqrt{2^2+(-3)^2+1^2}}=\frac{3}{\sqrt{14}}\\\large \vec r.(2\hat i-3\hat j+\hat k)=3\)

\(\large I=\int _2^8\Large\frac{\sqrt{10-x}}{\sqrt x+\sqrt{10-x}}\large dx.....(1)\)

\(\large I=\int _2^8\Large\frac{\sqrt{x}}{\sqrt {10-x}+\sqrt{x}}\large dx.....(2) [\because \int f(x) dx=\int f(a+b-x)dx]\)

adding (1) and (2)

\(\large 2I=\int _2^8\Large\frac{\sqrt{10-x}+\sqrt x}{\sqrt x\sqrt{10-x}}\large dx\\\large 2I=\int _2^8 dx\\\large 2I=6\\\large I=3\)

n th term \(\large a_n=\Large \frac{1^2+2^2+3^2+...}{1+2+3+...}\)

                     \(\Large=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}\\\Large=\frac{2n+1}{3}\)

Hence thesum of the series is 

\(\large S=\sum _{n=1}^{\infty}\Large\frac{2n+1}{3}\)

   \(\Large=\frac{1}{3}\large (\sum 2n+\sum 1)\\\Large=\frac{1}{3} (\frac{2n(n+1)}{2}\large +n)\\\Large=\frac{n(n+2)}{3}\)

\(\large x^y=e^{x-y}\) 

Taking log both side

\(\large y\log x=x-y\\\large y=\Large \frac{x}{1+\log x}\)

diff. w.r.t x

\(\Large \frac{dy}{dx}=\Large \frac{1+\log x-1}{(1+\log x)^2}=\Large \frac{\log x}{(1+\log x)^2}\)

\(\large f(x)=2x+6\)

let

 \(\large y=2x+6\\\large 2x=y-6\\\large x=\Large\frac{y-6}{2} \\\large x=\Large\frac{y}{2}\large-3\)

Hence \(\large f^{-1}(x)=\Large\frac{x}{2}\large-3\)