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Question Number 113682 by mohammad17 last updated on 14/Sep/20

Commented by mohammad17 last updated on 14/Sep/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 14/Sep/20

$$\underset{x\to 4}{\lim }\frac{4-x}{5-\sqrt{x^2+9}}$$$$\underset{x\to 4}{\lim }\frac{4-x}{25-x^2-9}.(5+\sqrt{x^2+9})=\frac{4-x}{(4-x)(4+x)}.(5+5)=\frac{5}{4}$$

Commented by mohammad17 last updated on 15/Sep/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 14/Sep/20

$${(y+x)}^7={sin}^{7}x{e}^{3y}$$$$7log(y+x)=7log\left(sinx\right)+3y$$$$\frac{7}{y+x}.(\frac{dy}{dx}+1)=7cotx+3\frac{dy}{dx}$$$$(\frac{7}{y+x}-3)\frac{dy}{dx}=7(cotx-\frac{1}{y+x})$$$$\frac{dy}{dx}=\frac{7(cotx-\frac{1}{y+x})}{(\frac{7}{y+x}-3)}$$$$$$

Commented by mohammad17 last updated on 15/Sep/20

$$thankyousir$$

Answered by Mr.D.N. last updated on 14/Sep/20

$$$$$$\mathrm{A})\mathrm{Prove}\mathrm{that}{\cos }^2\theta +{\sin }^2\theta =1$$$${\mathrm{Sol}}^{\mathrm{n}}:$$$$\mathrm{We}\mathrm{know},\mathrm{trigonometric}\mathrm{ratio}:$$$$\sin \theta =\frac{\mathrm{p}}{\mathrm{h}}\mathrm{and}\cos \theta =\frac{\mathrm{b}}{\mathrm{h}}$$$$\mathrm{or},\mathrm{p}=\mathrm{h}\sin \theta \mathrm{and}\mathrm{b}=\mathrm{h}\cos \theta$$$$\mathrm{We}\mathrm{know}\mathrm{Pythagorous}\mathrm{theorem},$$$${\mathrm{p}}^2+{\mathrm{b}}^2={\mathrm{h}}^2$$$$\mathrm{or},{\left(\mathrm{h}\sin \theta \right)}^2+{\left(\mathrm{h}\cos \theta \right)}^2={\mathrm{h}}^2$$$$\mathrm{or},{\mathrm{h}}^2{\sin }^2\theta +{\mathrm{h}}^2{\cos }^2\theta ={\mathrm{h}}^2$$$$\mathrm{or},{\mathrm{h}}^2({\sin }^2\theta +{\cos }^2\theta )={\mathrm{h}}^2$$$$\mathrm{or},{\sin }^2\theta +{\cos }^2\theta =\frac{{\mathrm{h}}^2}{{\mathrm{h}}^2}$$$$\therefore {\sin }^2\theta +{\cos }^2\theta =1\mathrm{which}\mathrm{is}\mathrm{required}\mathrm{proof}.$$$$$$$$$$

Commented by mohammad17 last updated on 15/Sep/20

$$thankyousir$$

Answered by Mr.D.N. last updated on 14/Sep/20

$$\mathrm{Q}5$$$$1-\mathrm{By}\mathrm{using}\mathrm{first}\mathrm{principle}\mathrm{derivative}\mathrm{prove}$$$$\mathrm{that}\mathrm{the}\mathrm{derivative}\mathrm{of}$$$$\tan \mathrm{x}={\sec }^2\mathrm{x}$$$${\mathrm{Sol}}^{\mathrm{n}}:$$$$\mathrm{We}\mathrm{first}\mathrm{principle}\mathrm{derivative}\mathrm{rules}:$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}$$$$\mathrm{Let},\mathrm{f}\left(\mathrm{x}\right)=\tan \mathrm{x}$$$$\mathrm{f}(\mathrm{x}+\mathrm{h})=\tan (\mathrm{x}+\mathrm{h})$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan \mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\tan (\mathrm{x}+\mathrm{h})-\tan \mathrm{x}}{\mathrm{h}}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\frac{\sin (\mathrm{x}+\mathrm{h})}{\cos (\mathrm{x}+\mathrm{h})}-\frac{\sin \mathrm{x}}{\cos \mathrm{x}}}{\mathrm{h}}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\frac{\cos \mathrm{x}\sin (\mathrm{x}+\mathrm{h})-\mathrm{sinx}\cos (\mathrm{x}+\mathrm{h})}{\cos \mathrm{x}\cos (\mathrm{x}+\mathrm{h})}}{\mathrm{h}}$$$$=\underset{\mathbf{h}\to 0}{\stackrel{\lim }{}}\frac{\cos \mathrm{x}(\sin \mathrm{x}\cos \mathrm{h}+\mathrm{cosx}\sin \mathrm{h})-\mathrm{sinx}(\cos \mathrm{x}\cos \mathrm{h}-\mathrm{sinx}\sin \mathrm{h})}{\mathrm{h}\cos \mathrm{x}\cos (\mathrm{x}+\mathrm{h})}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\cos \mathrm{x}\sin \mathrm{x}\cos \mathrm{h}+{\cos }^2\mathrm{x}\sin \mathrm{h}-\sin \mathrm{x}\cos \mathrm{x}\cos \mathrm{h}+{\sin }^2\mathrm{x}\sin \mathrm{h}}{\mathrm{h}\cos \mathrm{x}\cos (\mathrm{x}+\mathrm{h})}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{{\cos }^2\mathrm{x}\sin \mathrm{h}+{\sin }^2\mathrm{x}\sin \mathrm{h}}{\mathrm{h}\cos \mathrm{x}\cos (\mathrm{x}+\mathrm{h})}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\sin \mathrm{h}({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x})}{\mathrm{h}\cos \mathrm{x}\cos (\mathrm{x}+\mathrm{h})}\{\because {\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}=1\}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\sin \mathrm{h}}{\mathrm{h}}\times \underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{1}{\cos \mathrm{x}\cos (\mathrm{x}+\mathrm{h})}\{\because \underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\sin \mathrm{h}}{\mathrm{h}}=1\}$$$$=1\times \frac{1}{\cos \mathrm{x}.\cos (\mathrm{x}+0)}=\frac{1}{\cos \mathrm{x}.\cos \mathrm{x}}=\frac{1}{{\cos }^2\mathrm{x}}$$$$={\sec }^2\mathrm{x}$$$$\therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\tan \mathrm{x}\right)={\sec }^2\mathrm{x}\mathrm{proved}//.$$

Commented by mohammad17 last updated on 15/Sep/20

$$thankyousir$$

Answered by Mr.D.N. last updated on 14/Sep/20

$$2-\left(\mathrm{b}\right)iff\left(x\right)={tan}^{-1}\frac{x}{a}showthat\frac{dx}{dy}=\frac{a^2+x^2}{a}$$$${\mathrm{Sol}}^{\mathrm{n}}:$$$$\mathrm{given}\mathrm{function}:\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$$$$\mathrm{D}.\mathrm{w}.\mathrm{r}.\mathrm{to}\mathrm{x}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^2}.\left(\frac{1}{\mathrm{a}}\right)$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\frac{{\mathrm{a}}^2+{\mathrm{x}}^2}{{\mathrm{a}}^2}}.\left(\frac{1}{\mathrm{a}}\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2+{\mathrm{x}}^2}.\frac{1}{\mathrm{a}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{{\mathrm{a}}^2+{\mathrm{x}}^2}$$$$\mathrm{a}\mathrm{dx}=({\mathrm{a}}^2+{\mathrm{x}}^2)\mathrm{dy}$$$$\therefore \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{{\mathrm{a}}^2+{\mathrm{x}}^2}{\mathrm{a}}\mathrm{proved}//.$$$$$$$$$$

Commented by mohammad17 last updated on 15/Sep/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 14/Sep/20

$$\mathrm{A}){\cos }^2\theta +{\sin }^2\theta ={\left(\frac{{\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }}{2}\right)}^2+{\left(\frac{{\mathrm{e}}^{\mathrm{i}\theta }-{\mathrm{e}}^{-\mathrm{i}\theta }}{2\mathrm{i}}\right)}^2$$$$=\frac{1}{4}({\mathrm{e}}^{2\mathrm{i}\theta }+2+{\mathrm{e}}^{-2\mathrm{i}\theta })-\frac{1}{4}({\mathrm{e}}^{2\mathrm{i}\theta }-2+{\mathrm{e}}^{-2\mathrm{i}\theta })$$$$=\frac{1}{4}\{{\mathrm{e}}^{2\mathrm{i}\theta }+2+{\mathrm{e}}^{-2\mathrm{i}\theta }-{\mathrm{e}}^{2\mathrm{i}\theta }+2-{\mathrm{e}}^{-2\mathrm{i}\theta }\}=\frac{4}{4}=1$$

Commented by mohammad17 last updated on 15/Sep/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by mathmax by abdo last updated on 15/Sep/20

$$\mathrm{b})\mathrm{we}\mathrm{have}\sqrt{5-{2\mathrm{x}}^2}⩽\mathrm{f}\left(\mathrm{x}\right)⩽\sqrt{5-{\mathrm{x}}^2}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{5}$$$${\lim }_{\mathrm{x}\to 4}\frac{4-\mathrm{x}}{5-\sqrt{{\mathrm{x}}^2+9}}={\lim }_{\mathrm{x}\to 4}\frac{(4-\mathrm{x})(5+\sqrt{{\mathrm{x}}^2+9})}{25-{\mathrm{x}}^2-9}$$$$={\lim }_{\mathrm{x}\to 4}\frac{(4-\mathrm{x})(5+\sqrt{{\mathrm{x}}^2+9})}{(4-\mathrm{x})(4+\mathrm{x})}={\lim }_{\mathrm{x}\to 4}\frac{5+\sqrt{{\mathrm{x}}^2+9}}{4+\mathrm{x}}$$$$=\frac{5+5}{4+4}=\frac{10}{8}=\frac{5}{4}$$

Answered by mathmax by abdo last updated on 15/Sep/20

$$5)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{tanx}\right)={\lim }_{\mathrm{h}\to 0}\frac{\tan (\mathrm{x}+\mathrm{h})-\mathrm{tanx}}{\mathrm{h}}$$$$={\lim }_{\mathrm{h}\to 0}\frac{\frac{\mathrm{tanx}+\tanh }{1-\mathrm{tanx}.\tanh }-\mathrm{tanx}}{\mathrm{h}}$$$$={\lim }_{\mathrm{h}\to 0}\frac{\mathrm{tanx}+\tanh -\mathrm{tanx}+{\tan }^2\mathrm{x}\tanh }{\mathrm{h}(1-\mathrm{tanx}.\tanh )}$$$$={\lim }_{\mathrm{h}\to 0}\frac{\tanh (1+{\tan }^2\mathrm{x})}{\mathrm{h}(1-\mathrm{tanx}.\tanh })=1+{\tan }^2\mathrm{x}=\frac{1}{{\cos }^2\mathrm{x}}({\lim }_{\mathrm{h}\to 0}\frac{\tanh }{\mathrm{h}}=1)$$

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