Question Number 316 by 123456 last updated on 25/Jan/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=\begin{vmatrix}{{x}}&{{g}\left({x}\right)}\\{{g}\left(−{x}\right)}&{−{x}}\end{vmatrix} \\ $$$${g}\left({x}\right)=\begin{vmatrix}{{f}\left({x}\right)}&{{x}}\\{−{x}}&{{f}\left(−{x}\right)}\end{vmatrix} \\ $$ Answered by prakash jain last updated on…
Question Number 303 by 123456 last updated on 25/Jan/15 $${f}\left({x},{y},{z}\right)=\frac{{y}^{\mathrm{2}} {z}^{\mathrm{3}} }{\mathrm{1}−{x}}+\frac{{xz}^{\mathrm{3}} }{\mathrm{1}−{y}^{\mathrm{2}} }+\frac{{xy}^{\mathrm{2}} }{\mathrm{1}−{z}^{\mathrm{3}} } \\ $$$$\mathrm{find} \\ $$$$\frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{y}}+\frac{\partial{f}}{\partial{z}} \\ $$ Answered by prakash…
Question Number 298 by 123456 last updated on 25/Jan/15 $${a}\left(\mathrm{n},{m}\right)=\begin{cases}{{n}+{m}}&{{n}\leqslant\mathrm{0}}\\{{b}\left({n}\right)+{m}}&{{n}>\mathrm{0}\wedge{m}\leqslant\mathrm{0}}\\{{b}\left({n}\right)+{b}\left({m}\right)}&{{n}>\mathrm{0}\wedge{m}>\mathrm{0}}\end{cases} \\ $$$${b}\left({n}\right)=\begin{cases}{\mathrm{0}}&{{n}\leqslant\mathrm{0}}\\{{b}\left({n}−\mathrm{1}\right)}&{\mathrm{0}<{n}<\mathrm{5}}\\{{b}\left({n}−\mathrm{1}\right)+{b}\left({n}+\mathrm{1}\right)}&{{n}=\mathrm{5}}\\{{b}\left({n}+\mathrm{1}\right)}&{\mathrm{5}<{n}\leqslant\mathrm{10}}\\{\mathrm{1}}&{{n}>\mathrm{10}}\end{cases} \\ $$$$\mathrm{find} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)+{a}\left(\mathrm{4},\mathrm{6}\right) \\ $$ Answered by prakash jain last updated on…
Question Number 291 by 123456 last updated on 25/Jan/15 $${a}\left({n},{m}\right)=\begin{cases}{{m}}&{{n}\leqslant\mathrm{0}}\\{{a}\left({n}−\mathrm{1},{m}+\mathrm{2}\right)}&{{n}>\mathrm{0}\wedge{n}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right)}\\{{a}\left({n}−\mathrm{2},{m}−\mathrm{1}\right)+{nn}}&{{n}>\mathrm{0}\wedge{n}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right)\wedge{m}\leqslant\mathrm{0}}\\{{a}\left({m}−\mathrm{1},{n}−\mathrm{1}\right)+{a}\left({n}−\mathrm{2},{m}−\mathrm{2}\right)}&{{n}>\mathrm{0}\wedge{n}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right)\wedge{m}>\mathrm{0}}\end{cases} \\ $$$$\mathrm{evaluate}\:{a}\left(\mathrm{7},\mathrm{5}\right) \\ $$ Answered by prakash jain last updated on 19/Dec/14 $${a}\left(\mathrm{7},\mathrm{5}\right)={a}\left(\mathrm{4},\mathrm{6}\right)+{a}\left(\mathrm{5},\mathrm{3}\right) \\ $$$$={a}\left(\mathrm{3},\mathrm{8}\right)+{a}\left(\mathrm{2},\mathrm{4}\right)+{a}\left(\mathrm{3},\mathrm{1}\right)…
Question Number 272 by 123456 last updated on 25/Jan/15 $$\mathrm{given}\:\mathrm{f}\left(\mathrm{x},{y}\right)={ax}+{bxy}+{cy}\:\mathrm{for}\:\left(\mathrm{x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{proof}\:\mathrm{that}\:\mathrm{if}\:\mid{a}−{c}\mid\leqslant\mathrm{1}\:\mathrm{then}\:\mid{f}\left({x},{y}\right)−{f}\left({y},{x}\right)\mid\leqslant\mid{x}−{y}\mid \\ $$$$\mathrm{is}\:\mathrm{f}\left(\mathrm{x},{y}\right)\:\mathrm{bounced}? \\ $$ Answered by prakash jain last updated on 18/Dec/14 $${f}\left({x},{y}\right)−{f}\left({y},{x}\right)={ax}+{bxy}+{cy}−{ay}−{bxy}−{cx}…
Question Number 257 by abcd last updated on 25/Jan/15 $$\mathrm{If}\:\mathrm{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x},\mathrm{y}\in\mathbb{N}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{3}\:\mathrm{and}\:\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{120}.\:\mathrm{Then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{n}. \\ $$ Answered by 123456 last updated…
Question Number 65778 by mathmax by abdo last updated on 03/Aug/19 $${find}\:{lim}_{{n}\rightarrow+\infty} \:{e}^{−{n}^{\mathrm{2}} } \left({n}+\mathrm{1}\right)^{{n}!} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 65779 by mathmax by abdo last updated on 03/Aug/19 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)−{xtan}\left({x}\right)}{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)} \\ $$ Commented by mathmax by abdo last updated on 04/Aug/19…
Question Number 65777 by mathmax by abdo last updated on 03/Aug/19 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\:{e}^{−{n}} \left(\left({n}+\mathrm{1}\right)!\right)^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 239 by 123456 last updated on 25/Jan/15 $$\mathrm{find}\:\mathrm{f}\left(\mathrm{10}\right)\:\mathrm{given} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{2}\right),\mathrm{x}>\mathrm{1},\mathrm{x}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)\mathrm{f}\left(\mathrm{x}−\mathrm{2}\right),\mathrm{x}>\mathrm{1},\mathrm{x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$ Answered by prakash jain last…