(2)

$f\left(x\right)$$=\{\begin{array}{cc}-2,& -2\le x\le 0\\ x-2,& 0<x\le 2\end{array}$

$h\left(x\right)=f\left(\right|x\left|\right)+\left|f\right(x\left)\right|$

$f\left(\right|x\left|\right)$$=\{\begin{array}{cc}-x-2,& -2\le x\le 0\\ x-2,& 0<x\le 2\end{array}$

$\left|f\right(x\left)\right|$$=\{\begin{array}{cc}2,& -2\le x\le 0\\ -(x-2),& 0<x\le 2\end{array}$

So, $h\left(x\right)$$=\{\begin{array}{cc}-x-2+2=-x,& -2\le x\le 0\\ x-2-(x-2)=0,& 0<x\le 2\end{array}$

$\therefore$     ${\int }_{-2}^{2}h\left(x\right) dx={\int }_{-2}^0-x dx+{\int }_0^{2}0 dx$

        $=-{\left[\frac{{\displaystyle x^2}}{{\displaystyle 2}}\right]}_{-2}^0=-[0-\frac{{\displaystyle 4}}{{\displaystyle 2}}]=2$

(3)

Let $I={\int }_{-1}^1\frac{\cos \alpha x}{1+3^x} dx$                                              ....(i)

$={\int }_{-1}^1\frac{\cos \alpha (1-1-x)}{1+3^{1-1-x}} dx$

$\left(\text{using property }{\int }_a^{b}f\right(x) dx={\int }_a^{b}f(a+b-x) dx)$

$\Rightarrow I={\int }_{-1}^1\frac{\cos \alpha x}{1+3^{-x}} dx={\int }_{-1}^1\frac{3^x\cos \alpha x}{1+3^x} dx$              ...(ii)

On adding (i) and (ii), we get

$2I={\int }_{-1}^1\cos \left(\alpha x\right) dx={\left(\frac{{\displaystyle \sin \left(\alpha x\right)}}{{\displaystyle \alpha }}\right)}_{-1}^1$

$\Rightarrow 2\left(\frac{{\displaystyle 2}}{{\displaystyle \pi }}\right)=\frac{2\sin \alpha }{\alpha } [\because I={\int }_{-1}^1\frac{{\displaystyle \cos \alpha x}}{{\displaystyle 1+3^x}} dx=\frac{{\displaystyle 2}}{{\displaystyle \pi }}\text{ (Given)}]$

$\Rightarrow \frac{\sin \alpha }{\alpha }=\frac{2}{\pi }\Rightarrow \alpha =\frac{\pi }{2}.$

(2)

Given $f\left(x\right)={\int }_0^x(t+\sin (1-e^t\left)\right)dt$

$\text{Now, }\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^3}\left(\frac{{\displaystyle 0}}{{\displaystyle 0}}\text{ form}\right)=\underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)}{3x^2}$

$=\underset{x\to 0}{\lim }\frac{x+\sin (1-e^x)}{3x^2} \left(\frac{{\displaystyle 0}}{{\displaystyle 0}}\text{ form}\right)$

$=\underset{x\to 0}{\lim }\frac{1+\cos (1-e^x)(-e^x)}{6x} \left(\frac{{\displaystyle 0}}{{\displaystyle 0}}\text{ form}\right)$

$=\underset{x\to 0}{\lim }\frac{-\sin (1-e^x){\left(e^x\right)}^2+\cos (1-e^x)(-e^x)}{6}=-\frac{1}{6}$

(2)

Let $I={\int }_0^{\pi /4}\frac{136\sin x}{3\sin x+5\cos x} dx$

Let $\sin x=\lambda (3\sin x+5\cos x)+\mu (3\cos x-5\sin x)$

For $x=0,$ we have $0=5\lambda +3\mu$

and for $x=\pi /2,$ we have $1=3\lambda -5\mu$

Solving these two equations, we get $\lambda =\frac{3}{34}$ and $\mu =\frac{-5}{34}$

$\therefore$     $I=136{\int }_0^{\pi /4}\frac{3}{34}\left(\frac{3\sin x+5\cos x}{3\sin x+5\cos x}\right)dx-\frac{136\times 5}{34}{\int }_0^{\pi /4}\frac{3\cos x-5\sin x}{3\sin x+5\cos x} dx$

$=\frac{136\times 3}{34}{\left[x\right]}_0^{\pi /4}-\frac{136\times 5}{34}\ln |3\sin x+5\cos x{|}_0^{\pi /4}$

$=3\pi -20[\ln \frac{8}{\sqrt{2}}-\ln 5]=3\pi -20\ln 4\sqrt{2}+20\ln 5$

$=3\pi -20\ln {\left(2\right)}^{\frac{5}{2}}+20\ln 5=3\pi -50\ln 2+20\ln 5$

(1)

$\text{Let }I={\int }_{-\pi }^{\pi }\frac{2y(1+\sin y)}{1+{\cos }^{2}y} dy$

$I=\underset{\mathrm{Odd} \mathrm{function}}{\underbrace{{\int }_{-\pi }^{\pi }\frac{{\displaystyle 2y}}{{\displaystyle 1+{\cos }^{2}y}} dy}}+\underset{\mathrm{Even} \mathrm{function}}{\underbrace{{\int }_{-\pi }^{\pi }\frac{{\displaystyle 2y\sin y}}{{\displaystyle 1+{\cos }^{2}y}} d}}$

$\therefore$     $I=2{\int }_0^{\pi }\frac{2y\sin y}{1+{\cos }^{2}y} dy$                                        ...(i)

$\text{Now, }I=4{\int }_0^{\pi }\frac{(\pi -y)\sin (\pi -y)}{1+{\cos }^2(\pi -y)} dy$

$\Rightarrow I=4{\int }_0^{\pi }\frac{(\pi -y)\sin y}{1+{\cos }^{2}y} dy$                                          ...(ii)

$\text{Adding (i) and (ii), we get}$

$2I={\int }_0^{\pi }\frac{4\pi \sin y}{1+{\cos }^{2}y} dy$

$\text{Let }\cos y=t\Rightarrow \sin y dy=-dt$

$\text{When }y=0, t=1$

$\text{when }y=\pi , t=-1$

$\therefore I=-2\pi {\int }_1^{-1}\frac{1}{1+t^2} dt$

$=-2\pi {\left[{\tan }^{-1}t\right]}_1^{-1}=-2\pi [\frac{{\displaystyle -\pi }}{{\displaystyle 4}}-\frac{{\displaystyle \pi }}{{\displaystyle 4}}]$

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

(2)

Let $I={\int }_0^1{(1-x^{10})}^{20} dx$

Put $x^{10}=t$

$\Rightarrow x=t^{1/10}$

$\Rightarrow dx=\frac{1}{10}{t}^{-9/10} dt$

$\therefore$    $I={\int }_0^1{(1-t)}^{20}\frac{1}{10}{t}^{-9/10} dt=\frac{1}{10}{\int }_0^1{t}^{-9/10}{(1-t)}^{21-1} dt$

$=\frac{1}{10}\beta (\frac{{\displaystyle 1}}{{\displaystyle 10}},21)\Rightarrow a=\frac{1}{10}, b=\frac{1}{10}, c=21$

$\therefore 100(a+b+c)=100(\frac{{\displaystyle 2}}{{\displaystyle 10}}+21)=2120$

(1)

Let $I={\int }_0^{\pi /4}\frac{{\cos }^{2}x{\sin }^{2}x}{{({\cos }^{3}x+{\sin }^{3}x)}^2} dx$

On dividing Nr. and Dr. by ${\cos }^{6}x,$ we get

        $I={\int }_0^{\pi /4}\frac{{\tan }^{2}x{\sec }^{2}x}{{(1+{\tan }^{3}x)}^2} dx$

Put $1+{\tan }^{3}x=t$

$\Rightarrow 3{\tan }^{2}x{\sec }^{2}x dx=dt$

when $x=0,$ $t=1$ and when $x=\pi /4,$ $t=2$

$\therefore$     $I=\frac{1}{3}{\int }_1^2\frac{dt}{t^2}=\frac{1}{3}{\left[\frac{{\displaystyle -1}}{{\displaystyle t}}\right]}_1^2=\frac{1}{3}[\frac{{\displaystyle -1}}{{\displaystyle 2}}+1]=\frac{1}{6}$

(2)

$I_n={\int }_0^1{(1-x^k)}^{n}1 dx$

By ILATE rule, we have

$I_n={\left[{(1-x^k)}^{n}x\right]}_0^1+nk{\int }_0^1{(1-x^k)}^{n-1}{x}^k dx$

$=-nk{\int }_0^1{(1-x^k)}^{n-1}(1-x^k-1) dx$

$=-nk[{\int }_0^1{(1-x^k)}^n dx-{\int }_0^1{(1-x^k)}^{n-1} dx]$

$=-nk$ $I_n+nk{I}_{n-1}\Rightarrow I_n(1+nk)=nk$ $I_{n-1}$

$\Rightarrow \frac{I_n}{I_{n-1}}=\frac{nk}{1+nk}\Rightarrow \frac{I_{21}}{I_{20}}=\frac{21k}{1+21k}=\frac{147}{148}$

$\Rightarrow 21k=147\Rightarrow k=7$

(3)

We have, ${\int }_{\alpha }^{{\log }_{e}4}\frac{dx}{\sqrt{e^x-1}}=\frac{\pi }{6}$                           ...(i)

Put $\sqrt{e^x-1}=t\Rightarrow e^x=1+t^2$

$\Rightarrow e^{x}dx=2t dt \Rightarrow dx=\frac{2t dt}{1+t^2}$

When $x=\alpha ,$ $t=\sqrt{e^{\alpha }-1},$ when $x={\log }_{e}4$

          $t=\sqrt{e^{\log 4}-1}=\sqrt{3}$

$\therefore$      From (i), ${\int }_{\sqrt{e^{\alpha }-1}}^{\sqrt{3}}\frac{2t dt}{t(1+t^2)}=\frac{\pi }{6}\Rightarrow 2{\left[{\tan }^{-1}t\right]}_{\sqrt{e^{\alpha }-1}}^{\sqrt{3}}=\frac{\pi }{6}$

$\Rightarrow 2({\tan }^{-1}\sqrt{3}-{\tan }^{-1}\sqrt{e^{\alpha }-1})=\frac{\pi }{6}$

$\Rightarrow \frac{\pi }{3}-{\tan }^{-1}\left(\sqrt{e^{\alpha }-1}\right)=\frac{\pi }{12} \Rightarrow {\tan }^{-1}\left(\sqrt{e^{\alpha }-1}\right)=\frac{\pi }{4}$

$\Rightarrow e^{\alpha }-1=1$                     $(\because \tan \frac{\pi }{4}=1)$         

$\Rightarrow e^{\alpha }=2 \Rightarrow e^{-\alpha }=\frac{1}{2}$

$\therefore \text{ Quadratic equation whose roots are }{e}^{\alpha }\text{ and }{e}^{-\alpha }\text{ is}$

       $x^2-(e^{\alpha }+e^{-\alpha })x+e^{\alpha }\times e^{-\alpha }=0$

$\Rightarrow x^2-(2+\frac{{\displaystyle 1}}{{\displaystyle 2}})x+1=0 \Rightarrow 2x^2-5x+2=0$

(3)

Let $I={\int }_{-1}^2{\log }_e(x+\sqrt{x^2+1})dx$

$={\left[{\log }_e\right(x+\sqrt{x^2+1}\left)x\right]}_{-1}^2-{\int }_{-1}^2\frac{1}{(x+\sqrt{x^2+1})}(1+\frac{{\displaystyle x}}{{\displaystyle \sqrt{x^2+1}}})·x dx$

$=2{\log }_e(2+\sqrt{5})+{\log }_e(\sqrt{2}-1)-{\int }_{-1}^2\frac{x}{\sqrt{x^2+1}}dx$

$=\log \left[{(2+\sqrt{5})}^2\right(\sqrt{2}-1\left)\right]-{\left[\sqrt{x^2+1}\right]}_{-1}^2$

$=\log \left[\frac{{\displaystyle 9+4\sqrt{5}}}{{\displaystyle 1+\sqrt{2}}}\right]-\sqrt{5}+\sqrt{2}$

(4)

Let $L=\underset{x\to \pi /2}{\lim }\left[\frac{{\int }_{x^3}^{{(\pi /2)}^3}\left(\sin \right(2t^{1/3})+\cos (t^{1/3}\left)\right) dt}{{(x-\frac{\pi }{2})}^2}\right]$ $\frac{0}{0}\text{ form, so by L'Hospital rule}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\displaystyle \frac{d}{dx}}{\int }_{x^3}^{{(\pi /2)}^3}\left(\sin \right(2t^{1/3})+\cos (t^{1/3}\left)\right)dt}{2(x-\frac{\pi }{2})}$ $\sin (2\times \frac{\pi }{2})·\frac{d}{dx}{\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^3-\sin \left(2x\right)·\frac{d}{dx}{x}^3$

$=\underset{x\to \pi /2}{\lim }\frac{{\displaystyle +(\cos \frac{{\displaystyle \pi }}{{\displaystyle 2}}\frac{{\displaystyle d}}{{\displaystyle dx}}{\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^3-\cos x·\frac{{\displaystyle d}}{{\displaystyle dx}}{x}^3)}}{2(x-\frac{\pi }{2})}$    (By Leibnitz rule of integration)

$=\underset{x\to \pi /2}{\lim }\frac{-3x^2\sin 2x-3x^2\cos x}{2(x-\frac{\pi }{2})} \left(\frac{{\displaystyle 0}}{{\displaystyle 0}}\text{ form}\right)$

$=\underset{x\to \pi /2}{\lim }\frac{-6x\sin 2x-6x^2\cos 2x-6x\cos x+3x^2\sin x}{2}$

$=\frac{{\displaystyle 6}\frac{{\pi }^2}{4}+\frac{3{\pi }^2}{4}}{2}=\frac{9{\pi }^2}{8}$

(2)

$I={\int }_{1/4}^{3/4}\cos \left(2{\cot }^{-1}\sqrt{\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}}\right)dx$

$={\int }_{1/4}^{3/4}\cos \left(2{\tan }^{-1}\sqrt{\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}}\right)dx$

$={\int }_{1/4}^{3/4} \frac{1-{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right)}{ 1+{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right)}dx$

$={\int }_{1/4}^{3/4} \frac{{\displaystyle 1}{\displaystyle -}\frac{1+x}{1-x}}{{\displaystyle 1}{\displaystyle +}\frac{1+x}{1-x}}dx$

$={\int }_{1/4}^{3/4}\frac{-2x}{2} dx={(-\frac{{\displaystyle x^2}}{{\displaystyle 2}})}_{1/4}^{3/4}=\frac{-9}{32}+\frac{1}{32}=\frac{-8}{32}=\frac{-1}{4}$

(4)

Let $I={\int }_0^{\pi /4}\frac{xdx}{{\sin }^4\left(2x\right)+{\cos }^4\left(2x\right)}$

Put $2x=t$

$\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}$    $\therefore I={\int }_0^{\pi /2}\frac{t/2·dt/2}{{\sin }^{4}t+{\cos }^{4}t}$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\pi /2}\frac{t dt}{{\sin }^{4}t+{\cos }^{4}t}$                                                 ...(i)

$\Rightarrow I=\frac{1}{4}{\int }_0^{\pi /2}\frac{(\frac{\pi }{2}-t) dt}{{\sin }^{4}t+{\cos }^{4}t}$                                                 ...(ii)

Adding (i) and (ii), we get

$2I=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{dt}{{\sin }^{4}t+{\cos }^{4}t}\Rightarrow 2I=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{{\sec }^{4}t dt}{{\tan }^{4}t+1}$

Let $\tan t=y\Rightarrow {\sec }^{2}t dt=dy$

$\therefore 2I=\frac{\pi }{8}{\int }_0^{\infty }\frac{(1+y^2) dy}{1+y^4}$

$\Rightarrow 2I=\frac{\pi }{8}{\int }_0^{\infty }\frac{(\frac{1}{y^2}+1) dy}{y^2+\frac{1}{y^2}-2+2}=\frac{\pi }{8}{\int }_0^{\infty }\frac{(\frac{1}{y^2}+1) dy}{{{\displaystyle (}{\displaystyle y}{\displaystyle -}{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle y}}}{\displaystyle )}}^2+2}$

Let $y-\frac{1}{y}=u\Rightarrow (1+\frac{1}{y^2})dy=du,$ we get $2I=\frac{\pi }{8}{\int }_{-\infty }^{\infty }\frac{du}{u^2+2}$

$\Rightarrow 2I=\frac{\pi }{8}{\left[\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}{\tan }^{-1}\right(\frac{{\displaystyle u}}{{\displaystyle \sqrt{2}}}\left)\right]}_{-\infty }^{\infty }$

$\Rightarrow 2I=\frac{\pi }{8}\left[\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}{\tan }^{-1}\right(\infty )-\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}{\tan }^{-1}(-\infty \left)\right]$

$\Rightarrow I=\frac{\pi }{16\sqrt{2}}(\frac{{\displaystyle \pi }}{{\displaystyle 2}}+\frac{{\displaystyle \pi }}{{\displaystyle 2}})=\frac{{\pi }^2}{16\sqrt{2}}=\frac{\sqrt{2}{\pi }^2}{32}$

(1)

Let  $I={\int }_0^{\pi /3}{\cos }^{4}x dx={\int }_0^{\pi /3}{\left(\frac{{\displaystyle 1+\cos 2x}}{{\displaystyle 2}}\right)}^2 dx$

$=\frac{1}{4}{\int }_0^{\pi /3}(1+{\cos }^{2}2x+2\cos 2x) dx$

$=\frac{1}{4}{\int }_0^{\pi /3}(1+2\cos 2x+(\frac{{\displaystyle 1+\cos 4x}}{{\displaystyle 2}}\left)\right) dx$

$=\frac{1}{4}{\int }_0^{\pi /3}(\frac{{\displaystyle 3}}{{\displaystyle 2}}+2\cos 2x+\frac{{\displaystyle \cos 4x}}{{\displaystyle 2}}) dx$

$=\frac{1}{4}{[\frac{{\displaystyle 3}}{{\displaystyle 2}}x+\sin 2x+\frac{{\displaystyle \sin 4x}}{{\displaystyle 8}}]}_0^{\pi /3}$

$=\frac{1}{4}[\frac{{\displaystyle \pi }}{{\displaystyle 2}}+\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}-\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 16}}]=\frac{\pi }{8}+\frac{7\sqrt{3}}{64}\Rightarrow a=\frac{1}{8}, b=\frac{7}{64}$

$\therefore 9a+8b=\frac{9}{8}+\frac{7}{8}=2$

(3)

Let $I={\int }_0^1{(2x^3-3x^2-x+1)}^{1/3}dx$

$\Rightarrow I={\int }_0^1{[2{(1-x)}^3-3{(1-x)}^2-(1-x)+1]}^{1/3}dx$

$\Rightarrow I={\int }_0^1(-2x^3+3x^2+x-1)dx$

$\Rightarrow I={\int }_0^1-(2x^3-3x^2-x+1)dx$

$\Rightarrow I=-I\Rightarrow 2I=0\Rightarrow I=0$

(4)

${\int }_0^1\frac{1}{\sqrt{3+x}+\sqrt{1+x}} dx$

$={\int }_0^1\frac{1}{\sqrt{3+x}+\sqrt{1+x}}\times \frac{\sqrt{3+x}-\sqrt{1+x}}{\sqrt{3+x}-\sqrt{1+x}} dx$        [$\because$Rationalising the denominator]

$={\int }_0^1\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} dx=\frac{1}{2}{\int }_0^1(\sqrt{3+x}-\sqrt{1+x}) dx$

$=\frac{1}{2}[$$\frac{{\displaystyle 2{(3+x)}^{3/2}}}{{\displaystyle 3}}$${|}_0^1$$-$$\frac{{\displaystyle 2}}{{\displaystyle 3}}{(1+x)}^{3/2}$${|}_0^1$$]$

$=3-\frac{2\sqrt{2}}{3}-\sqrt{3}=a+b\sqrt{2}+c\sqrt{3}\Rightarrow a=3, b=\frac{-2}{3}, c=-1$

$\therefore 2a+3b-4c=2\times 3+3\times \left(\frac{{\displaystyle -2}}{{\displaystyle 3}}\right)-4(-1)=8$

(1)

Let $I={\int }_0^{\pi }\frac{dx}{1+a^2-2a\cos x}$

$={\int }_0^{\pi }\frac{dx}{1+a^2-2a\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$

$={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}dx}{{\displaystyle (}{\displaystyle 1}{\displaystyle +}{\displaystyle {{\displaystyle a}}^2}{\displaystyle )}{\displaystyle (}{\displaystyle 1}{\displaystyle +}{\displaystyle {{\displaystyle \tan }}^2}{\displaystyle \frac{{\displaystyle x}}{{\displaystyle 2}}}{\displaystyle )}-2a(1-{\tan }^2\frac{x}{2})}$

Let $\tan \frac{x}{2}=t \Rightarrow {\sec }^2\frac{x}{2} dx=2 dt$

$I=2{\int }_0^{\infty }\frac{dt}{(1+a^2)(1+t^2)-2a+2a{t}^2}$

$=2{\int }_0^{\infty }\frac{dt}{1+a^2+t^2+a^2{t}^2-2a+2a{t}^2}$

$=2{\int }_0^{\infty }\frac{dt}{(1+a^2-2a)+(t^2+a^2{t}^2+2a{t}^2)}$

$I=2{\int }_0^{\infty }\frac{dt}{{(1-a)}^2+{(t+at)}^2}$

Let $t+at=u \Rightarrow dt+a dt=du$

$\Rightarrow I=\frac{2}{a+1}{\int }_0^{\infty }\frac{du}{{(1-a)}^2+u^2}=\frac{2}{(a+1)}\times \frac{1}{1-a}\times {\left[{\tan }^{-1}\right(\frac{{\displaystyle u}}{{\displaystyle 1-a}}\left)\right]}_0^{\infty }$

$=\frac{2}{1-a^2}\times \frac{\pi }{2}=\frac{\pi }{1-a^2}$

(3)

Let $I={\int }_{-\pi /2}^{\pi /2}(\frac{{\displaystyle x^2\cos x}}{{\displaystyle 1+{\pi }^x}}+\frac{{\displaystyle 1+{\sin }^{2}x}}{{\displaystyle 1+e^{{\left(\sin x\right)}^{2023}}}})dx$                  ...(i)

$\Rightarrow I={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+{\pi }^{-x}} +\frac{1+{\sin }^{2}x}{1+e^{-\sin x^{2023}}} dx$                 ...(ii)

Adding (i) and (ii), we get

$2I={\int }_{-\pi /2}^{\pi /2}(x^2\cos x+1+{\sin }^{2}x)dx$

$\Rightarrow I={\int }_0^{\pi /2}(x^2\cos x+1+{\sin }^{2}x)dx$

$={\int }_0^{\pi /2}{x}^2\cos x dx+{\int }_0^{\pi /2}(1+{\sin }^{2}x)dx$                     ...(iii)