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Question Number 93733 by liki last updated on 14/May/20

Commented by liki last updated on 14/May/20

$$..\mathrm{please}\mathrm{help}\mathrm{qn}\mathrm{no}6\mathrm{a},\mathrm{b},\mathrm{c}$$

Commented by Rio Michael last updated on 14/May/20

$$\mathrm{Given}\mathrm{that}f\left(t\right)\mathrm{is}\mathrm{a}\mathrm{function}$$$$\left(\mathrm{i}\right)f\left(t\right)\mathrm{is}\mathrm{even}\mathrm{if}f(-t)=f\left(t\right)\mathrm{for}\mathrm{example}f\left(t\right)=t^2$$$$f(-t)={(-t)}^2=t^2=f\left(t\right).$$$$\left(\mathrm{ii}\right)f\left(t\right)\mathrm{is}\mathrm{odd}\mathrm{if}f(-t)=-f\left(t\right)\mathrm{for}\mathrm{example},f\left(t\right)=t$$$$f(-t)=-t=-f\left(t\right)$$

Commented by Ritu Jain last updated on 14/May/20

$$\mathrm{hi}\mathrm{liki}$$$$\mathrm{can}\mathrm{you}\mathrm{please}\mathrm{use}\mathrm{smaller}\mathrm{font}\mathrm{size}.$$

Commented by mathmax by abdo last updated on 14/May/20

$$c)f\left(x\right)=2+\sqrt{4-x}andg\left(x\right)=x^2+1\Rightarrow fog\left(x\right)=f\left(g\left(x\right)\right)$$$$=2+\sqrt{4-g\left(x\right)}=2+\sqrt{4-x^2-1}=2+\sqrt{3-x^2}$$$$x\in D_{f0g}\iff 3-x^2⩾0\iff x^2⩽3\iff -\sqrt{3}⩽x⩽\sqrt{3}butx<0\Rightarrow$$$$D_{fog}=[-\sqrt{3},0[$$

Commented by liki last updated on 14/May/20

$$..\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{i}\mathrm{appriciate}\mathrm{your}\mathrm{time}.$$

Commented by mathmax by abdo last updated on 14/May/20

$$8)thelinex=3isdirctedbyvectorv=j(0,1)wehave$$$$2y-x=-2\Rightarrow x-2y=2\Rightarrow x-2y-2=0\Rightarrow thislineisdirectedby$$$$u(2,1)\Rightarrow (D,\mathrm{\Delta })=(u,v)cos(u,v)=\frac{u.v}{\mid \mid u\mid \mid .\mid \mid v\mid \mid }=\frac{1}{\sqrt{5}}$$$$sin(u,v)=\frac{det(u,v)}{\mid \mid u\mid \mid .\mid \mid v\mid \mid }=\frac{\left|\begin{array}{c}20\\ 11\end{array}\right|}{\sqrt{5}}=\frac{2}{\sqrt{5}}\Rightarrow tan(u,v)=2\Rightarrow$$$$(u,v)=arctan\left(2\right)\left[\pi \right]$$

Commented by abdomathmax last updated on 15/May/20

$$youarewelcome.$$

Commented by liki last updated on 15/May/20

$$..\mathrm{sory}\mathrm{sir}\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{add}\mathrm{more}\mathrm{than}256\mathrm{in}\mathrm{whatsap}\mathrm{group}??$$

Commented by mathmax by abdo last updated on 15/May/20

$$A={\int }_0^{\pi }\sqrt{sinx}dx\left(itsnotsimpsonmethod\right)$$$$A={\int }_0^{\frac{\pi }{2}}\sqrt{sinx}dx+{\int }_{\frac{\pi }{2}}^{\pi }\sqrt{sinx}dxbut{\int }_{\frac{\pi }{2}}^{\pi }\sqrt{sinx}dx{=}_{x=\frac{\pi }{2}+t}$$$$={\int }_0^{\frac{\pi }{2}}\sqrt{cost}dt\Rightarrow A={\int }_0^{\frac{\pi }{2}}\sqrt{sinx}dx+{\int }_0^{\frac{\pi }{2}}\sqrt{cosx}dx=A_1+A_2$$$$wehavex-\frac{x^3}{6}⩽sinx⩽xforx\in [0,\frac{\pi }{2}]\Rightarrow$$$$\sqrt{x-\frac{x^3}{6}}⩽\sqrt{sinx}⩽\sqrt{x}\Rightarrow {\int }_0^{\frac{\pi }{2}}\sqrt{x-\frac{x^3}{6}}dx⩽A_1⩽{\int }_0^{\frac{\pi }{2}}\sqrt{x}dx$$$$but{\int }_0^{\frac{\pi }{2}}\sqrt{x}dx{=}_{\sqrt{x}=t}{\int }_0^{\sqrt{\frac{\pi }{2}}}t\left(2t\right)dt=2{\left[\frac{t^3}{3}\right]}_0^{\sqrt{\frac{\pi }{2}}}$$$$=\frac{2}{3}\times \frac{\pi }{2}\sqrt{\frac{\pi }{2}}=\frac{\pi }{3}\sqrt{\frac{\pi }{2}}$$$${\int }_0^{\frac{\pi }{2}}\sqrt{x-\frac{x^3}{6}}dx={\int }_0^{\frac{\pi }{2}}\sqrt{x}\sqrt{1-\frac{x^2}{6}}dx{=}_{\sqrt{x}=t}{\int }_0^{\sqrt{\frac{\pi }{2}}}t\sqrt{1-\frac{t^4}{6}}\left(2t\right)dt$$$$=2{\int }_0^{\sqrt{\frac{\pi }{2}}}{t}^2\sqrt{1-\frac{t^4}{6}}dtbutweknow$$$${(1+u)}^{\frac{1}{2}}=1+\frac{u}{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2}{u}^2+...=1+\frac{u}{2}-\frac{u^2}{8}\Rightarrow$$$$1+\frac{u}{2}-\frac{u^2}{8}⩽\sqrt{1+u}⩽1+\frac{u}{2}\Rightarrow 1-\frac{u}{2}-\frac{u^2}{8}⩽\sqrt{1-u}⩽1-\frac{u}{2}$$$$\Rightarrow 1-\frac{t^4}{12}-\frac{t^8}{36\times 8}⩽\sqrt{1-\frac{t^4}{6}}⩽1-\frac{t^4}{12}\Rightarrow$$$$t^2-\frac{t^6}{12}-\frac{t^{10}}{36\times 8}⩽t^2\sqrt{1-\frac{t^4}{6}}⩽t^2-\frac{t^6}{12}\Rightarrow$$$${\int }_0^{\sqrt{\frac{\pi }{2}}}(2t^2-\frac{t^6}{6}-\frac{t^{10}}{144})dt⩽{\int }_0^{\frac{\pi }{2}}\sqrt{x-\frac{x^3}{6}}dx⩽{\int }_0^{\sqrt{\frac{\pi }{2}}}(2t^2-\frac{t^6}{6})dt$$$${\int }_0^{\sqrt{\frac{\pi }{2}}}(2t^2-\frac{t^6}{6})dt={[\frac{2}{3}{t}^3-\frac{t^7}{42}]}_0^{\sqrt{\frac{\pi }{2}}}=\frac{2}{3}\times \frac{\pi }{2}\sqrt{\frac{\pi }{2}}-\frac{1}{42}{\left(\sqrt{\frac{\pi }{2}}\right)}^7$$$${\int }_0^{\sqrt{\frac{\pi }{2}}}(2t^2-\frac{t^6}{6}-\frac{t^{10}}{144})dt={[\frac{2}{3}{t}^3-\frac{t^7}{42}-\frac{t^{11}}{11\times 144}]}_0^{\sqrt{\frac{\pi }{2}}}$$$$....becontinued....$$$$$$

Commented by mathmax by abdo last updated on 15/May/20

$$I={\int }_0^{\pi }\sqrt{sinx}dxletusesimpsonmethod$$$${\int }_a^{b}f\left(x\right)dx\sim \frac{b-a}{6}\{f\left(a\right)+f\left(\frac{a+b}{2}\right)+f\left(b\right)\}withf\left(x\right)=\sqrt{sinx}weget$$$${\int }_0^{\pi }\sqrt{sinx}dx\sim \frac{\pi }{6}\{\sqrt{sin0}+\sqrt{sin\left(\frac{\pi }{2}\right)}+\sqrt{sin\pi }\}$$$$\Rightarrow {\int }_0^{\pi }\sqrt{sinx}dx\sim \frac{\pi }{6}$$

Answered by Rio Michael last updated on 14/May/20

$$6.\left(\mathrm{b}\right)f\left(x\right)=(x-2)(8-x)\mathrm{for}2⩽x⩽8$$$$\left(\mathrm{i}\right)f(-1)=(-1-2)(8+1)=-27$$$$\left(\mathrm{ii}\right)f(1-2t)=(1-2t-2)(8-1+2t)=(-1-2t)(9-2t){\mathrm{D}}_f:t\in \mathbb{R}$$$$$$

Commented by Rasheed.Sindhi last updated on 15/May/20

$$\mathcal{T}hefunctionisdefinedininterval$$$$2⩽x⩽8.\mathcal{H}owcanyoucalculate$$$$f(-1)?$$$$\therefore f(-1)isundefined!$$

Answered by Rio Michael last updated on 14/May/20

$$f\left(t\right)=\frac{t}{2}\left(\frac{e^t+1}{e^t-1}\right)$$$$f(-t)=\frac{-t}{2}\left(\frac{e^{-t}+1}{e^{-t}-1}\right)=-\frac{t}{2}\left(\frac{1+e^t}{1-e^t}\right)=\frac{t}{2}\left(\frac{e^t+1}{e^t-1}\right)$$$$\mathrm{hence}f\left(t\right)\mathrm{is}\mathrm{even}\mathrm{as}f(-t)=f\left(t\right)$$

Answered by Rio Michael last updated on 14/May/20

$$6\left(\mathrm{c}\right)f\left(x\right)=2+\sqrt{4-x}\mathrm{and}\mathrm{g}\left(x\right)=x^2+1$$$$fog=f\left[\mathrm{g}\left(x\right)\right]=f(x^2+1)=2+\sqrt{4-x^2-1}=2+\sqrt{3-x^2}{\mathrm{D}}_{fg}:x\in \mathbb{R}-\{-\sqrt{3},\sqrt{3}\}$$$$\mathrm{as}\mathrm{an}\mathrm{exercise}\mathrm{find}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{your}\mathrm{function}$$

Commented by liki last updated on 14/May/20

$$..\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}.$$

Commented by mr W last updated on 14/May/20

$$youarerequestedtouseasmaller$$$$fontforyourtextsir!$$

Commented by Tinku Tara last updated on 14/May/20

$$\mathrm{Another}\mathrm{request}\mathrm{is}\mathrm{to}\mathrm{please}\mathrm{crop}$$$$\mathrm{image}\mathrm{so}\mathrm{that}\mathrm{only}\mathrm{useful}\mathrm{portion}$$$$\mathrm{is}\mathrm{posted}.\mathrm{Latest}\mathrm{verion}\mathrm{provides}$$$$\mathrm{crop}\mathrm{function}\mathrm{which}\mathrm{uploading}$$$$\mathrm{images}.$$

Commented by Ar Brandon last updated on 14/May/20

OK Sir��

Commented by liki last updated on 15/May/20

$$..\mathrm{ok}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 14/May/20

$$6\left(a\right)$$$$\mathcal{E}venfunction:f(-t)=f\left(t\right)$$$$\mathcal{O}ddfunction:f(-t)=-f\left(t\right)$$$$f\left(t\right)=\frac{t}{2}\left(\frac{e^t+1}{e^t-1}\right):Given$$$$Now,$$$$f(-t)=\frac{-t}{2}\left(\frac{e^{-t}+1}{e^{-t}-1}\right)$$$$=\frac{-t}{2}\left(\frac{\frac{1}{e^t}+1}{\frac{1}{e^t}-1}\right)$$$$=\frac{-t}{2}\left(\frac{\frac{1+e^t}{e^t}}{\frac{1-{\mathrm{e}}^{\mathrm{t}}}{e^t}}\right)$$$$=\frac{-t}{2}\left(\frac{\frac{e^t+1}{e^t}}{-\frac{{\mathrm{e}}^{\mathrm{t}}-1}{e^t}}\right)$$$$=\frac{t}{2}\left(\frac{e^t+1}{e^t-1}\right)=f\left(t\right)$$$$\because f(-t)=f\left(t\right)$$$$\therefore f\left(t\right)isanevenfunction.$$

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