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Question Number 106604 by mohammad17 last updated on 06/Aug/20

Commented by bemath last updated on 06/Aug/20

$$\mathrm{Q}4\left(5\right)\underset{x\to 1/2}{\lim }\frac{6\sqrt{\mathrm{x}}-6\sqrt{2\mathrm{x}-1}}{\mathrm{x}-1}=\frac{\frac{6}{\sqrt{2}}-0}{-\frac{1}{2}}$$$$=-\frac{12}{\sqrt{2}}=-6\sqrt{2}@\mathrm{bemath}@$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$y=2\sqrt{x+2}-4$$$$(x+2)⩾0$$$$Domainx\in [-2,\mathrm{\infty })$$$$\frac{y+4}{2}=\sqrt{x+2}$$$$x=\frac{{(y+4)}^2}{4}-2$$$$-2⩽x<\mathrm{\infty }$$$$-2⩽\frac{{(y+4)}^2-8}{4}<\mathrm{\infty }$$$$-8⩽{(y+4)}^2-8<\mathrm{\infty }$$$$0⩽{(y+4)}^2<\mathrm{\infty }$$$$-4⩽y<\mathrm{\infty }$$$$$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$4)\underset{x\to 1}{\lim }\frac{sin(x-1)}{(x-1)(x+3)}=\underset{x\to 1}{\lim }\frac{(x-1)}{(x-1)(x+3)}=\frac{1}{4}$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$4)\int (x^2+2x+3)sin4xdx=$$$$\int x^{2}sin4x+2\int xsin4x+3\int sin4x$$$$=-\frac{1}{4}{x}^{2}cos4x+\frac{1}{2}\int xcos4x-\frac{1}{2}xcos4x+\frac{1}{2}\int cos4x-\frac{3}{4}sin4x$$$$=-\frac{1}{4}{x}^{2}cos4x+\frac{1}{8}xsin4x-\frac{1}{8}\int sin4x+\frac{1}{2}xcos4x+\frac{1}{8}sin4x-\frac{3}{4}sin4x$$$$=-\frac{1}{4}{x}^{2}cos4x+sin4x(\frac{x}{8}-\frac{5}{8})+\frac{1}{32}cos4x+\frac{1}{2}cos4x+C$$$$$$

Answered by bemath last updated on 06/Aug/20

$$\mathrm{Q}4\left(1\right)\underset{x\to 0^{+}}{\lim }\frac{\mid \mathrm{x}+2\mid -2}{\mid \mathrm{x}\mid }=\underset{x\to 0^{+}}{\lim }\frac{\mathrm{x}+2-2}{\mathrm{x}}=1$$$$\underset{x\to 0^{-}}{\lim }\frac{\mid \mathrm{x}+2\mid -2}{\mid \mathrm{x}\mid }=\underset{x\to 0^{-}}{\lim }\frac{\mathrm{x}+2-2}{-\mathrm{x}}=-1$$$$\mathrm{since}\underset{x\to 0^{+}}{\lim f}\left(\mathrm{x}\right)\ne \underset{x\to 0^{-}}{\lim f}\left(\mathrm{x}\right)$$$$\mathrm{then}\underset{x\to 0}{\lim }\frac{\mid \mathrm{x}+2\mid -2}{\mid \mathrm{x}\mid }{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$$$@\mathrm{bemath}@$$

Commented by bemath last updated on 06/Aug/20

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$3)\underset{x\to \frac{1}{2}}{\lim }\frac{\frac{1}{x}-2}{x-\frac{1}{2}}=\frac{1-2x}{2x-1}.\frac{2}{x}=-1.4=-4$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$2)\underset{x\to -2}{\lim }\frac{\frac{x+2}{2x}}{x^3+8}=\underset{x\to -2}{\lim }\frac{\frac{1}{2x}}{x^2-2x+4}=-\frac{1}{4}.\frac{1}{12}=-\frac{1}{48}$$

Answered by 1549442205PVT last updated on 06/Aug/20

$$\mathrm{Q}1\mathrm{a})\mathrm{y}=2\sqrt{\mathrm{x}+2}-4.\mathrm{y}\mathrm{is}\mathrm{definited}\mathrm{for}$$$$\mathrm{\forall }\mathrm{x}\in [-2;+\mathrm{\infty })\Rightarrow \mathrm{D}\left(\mathrm{x}\right)=[-2;+\mathrm{\infty })$$$$\mathrm{We}\mathrm{have}\mathrm{y}⩾2.0-4=-4(\mathrm{since}\sqrt{\mathrm{x}+2}⩾0)$$$$\mathrm{Thus},\mathrm{the}\mathrm{domain}\mathrm{of}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{y}\mathrm{be}$$$$\mathrm{D}\left(\mathrm{y}\right)=[-4;+\mathrm{\infty })$$$$\mathrm{Q}2\mathrm{a})\mathrm{Solve}\mathrm{the}\mathrm{equations}:$$$${\mathrm{e}}^{2\ln \left(\cos \mathrm{x}\right)}={\left({\mathrm{lne}}^{\mathrm{sinx}}\right)}^2\iff {\mathrm{e}}^{\ln {\left(\mathrm{cosx}\right)}^2}+{\left(\mathrm{sinx}\right)}^2$$$$\iff {\cos }^2\mathrm{x}={\sin }^2\mathrm{x}(\mathrm{since}{\mathrm{e}}^{\mathrm{lnN}}=\mathrm{N},{\mathrm{lne}}^{\mathrm{m}}=\mathrm{m})$$$$\iff {\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}=0\iff \cos 2\mathrm{x}=0$$$$\iff 2\mathbf{x}=\frac{\mathit{\pi }}{2}+\mathbf{k}\mathit{\pi }\iff \mathbf{x}=\frac{\mathit{\pi }}{4}+\frac{\mathbf{k}\mathit{\pi }}{2}$$$$\mathrm{b})2^{\mathrm{x}+1}=3^{\mathrm{x}+1}\iff {\left(\frac{2}{3}\right)}^{\mathrm{x}+1}=1\iff \mathrm{x}+1=0$$$$\iff \mathbf{x}=-1$$$$\mathrm{c})4^{\mathrm{x}}=2^{{\mathrm{x}}^2-3}\iff 2^{2\mathrm{x}}=2^{{\mathrm{x}}^2-3}\iff {\mathrm{x}}^2-3=2\mathrm{x}$$$$\iff {\mathrm{x}}^2-2\mathrm{x}-3=0\mathrm{\Delta }{;}^{\prime }=1+3=2^2$$$$\Rightarrow \mathrm{x}=1\pm 2\Rightarrow \mathbf{x}\in \{-1,3\}$$$$\mathrm{d}){\log }_4(\mathrm{x}+1)-{2\log }_4(\mathrm{x}+1)=\frac{1}{2}\left(1\right)$$$$\mathrm{Condition}\mathrm{for}\mathrm{the}\mathrm{eqn}.\mathrm{is}\mathrm{deined}\mathrm{as}\mathrm{x}⩾-1$$$$\left(1\right)\iff {\log }_4(\mathrm{x}+1)=-\frac{1}{2}\iff \mathrm{x}+1=4^{-\frac{1}{2}}$$$$\iff \mathrm{x}+1=\frac{1}{\sqrt{4}}=\frac{1}{2}\iff \mathbf{x}=-\frac{1}{2}$$$$\mathrm{Q}3.\mathrm{a})\mathrm{y}=\sqrt{-3\mathrm{x}-5}\Rightarrow {\mathrm{y}}^{\prime }=\frac{-3}{2\sqrt{-3\mathrm{x}-5}}$$$$\mathrm{b})\sin 3\mathrm{x}=\sin (\mathrm{x}+2\mathrm{x})=\mathrm{sinxcos}2\mathrm{x}+\mathrm{cosxsin}2\mathrm{x}$$$$=\mathrm{sinx}(1-{2\sin }^2\mathrm{x})+\mathrm{cosx}.2\mathrm{sinxcosx}$$$$=\mathrm{sinx}-{2\sin }^3\mathrm{x}+2\mathrm{sinx}(1-{\sin }^2\mathrm{x})$$$$=3\mathrm{sinx}-{4\sin }^3\mathrm{x}(\mathrm{q}.\mathrm{e}.\mathrm{d})$$$$\mathrm{Q}4.\mathrm{a})\underset{\mathrm{x}\to 0}{\mathrm{Lim}}\frac{\mid \mathrm{x}+2\mid -2}{\mid \mathrm{x}\mid }=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}+2-2}{\mid \mathrm{x}\mid }$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{\mid \mathrm{x}\mid }=\{\begin{array}{l}\underset{\mathrm{x}\to -0}{\lim }\frac{\mathrm{x}}{-\mathrm{x}}=-1\\ \underset{\mathrm{x}\to +0}{\lim }\frac{\mathrm{x}}{\mathrm{x}}=1\end{array}$$$$\mathrm{Since}\mathrm{left}-\mathrm{limit}\ne \mathrm{right}-\mathrm{limit}$$$$\mathrm{the}\mathrm{given}\mathrm{limit}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{exist}$$$$\mathrm{b})\underset{\mathrm{x}\to -2}{\lim }\frac{\frac{1}{\mathrm{x}}+\frac{1}{2}}{{\mathrm{x}}^3+8}\underset{\mathrm{x}\to -2}{=\lim }\frac{\frac{\mathrm{x}+2}{2\mathrm{x}}}{(\mathrm{x}+2)({\mathrm{x}}^2-\mathrm{x}+4)}=$$$$=\underset{\mathrm{x}\to -2}{\lim }\frac{1}{2\mathrm{x}({\mathrm{x}}^2-\mathrm{x}+4)}=\frac{1}{-4(4+2+4)}=\frac{-1}{40}$$$$\mathrm{c})\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{{\mathrm{x}}^{-1}-2}{\mathrm{x}-\frac{1}{2}}=\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{\frac{1}{\mathrm{x}}-2}{\frac{2\mathrm{x}-1}{2}}=\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{\frac{1-2\mathrm{x}}{\mathrm{x}}}{\frac{2\mathrm{x}-1}{2}}$$$$=\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{-2}{\mathrm{x}}=-4$$$$\mathrm{d})\underset{\mathrm{x}\to 1}{\mathrm{li}\mathrm{m}}\frac{\sin (\mathrm{x}-1)}{{\mathrm{x}}^2+2\mathrm{x}-3}\underset{\mathrm{x}-1=\mathrm{t}}{=}\underset{\mathrm{x}\to 1}{\lim }\frac{\sin (\mathrm{x}-1)}{(\mathrm{x}-1)(\mathrm{x}+3)}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\mathrm{sint}}{\mathrm{t}(\mathrm{t}+4)}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\mathrm{sint}}{\mathrm{t}}\times \frac{1}{\mathrm{t}+4}=1\times \frac{1}{4}=\frac{1}{4}$$$$\mathrm{e})\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{6\sqrt{\mathrm{x}}-6\sqrt{2\mathrm{x}-1}}{\mathrm{x}-1}=\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{6[\mathrm{x}-(2\mathrm{x}-1)}{(\mathrm{x}-1)(\sqrt{\mathrm{x}}+\sqrt{2\mathrm{x}-1})}$$$$=\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{-6(\mathrm{x}-1)}{(\mathrm{x}-1)(\sqrt{\mathrm{x}}+\sqrt{2\mathrm{x}-1})}=\underset{\mathrm{x}\to \frac{1}{2}}{\lim }\frac{-6}{\sqrt{\mathrm{x}}+\sqrt{2\mathrm{x}-1}}$$$$=\frac{-6}{\sqrt{\frac{1}{2}}}=-6\sqrt{2}$$

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