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Question Number 55908 by gunawan last updated on 06/Mar/19 | ||
$$\mathrm{known}\:{a}\:<\:\frac{\pi}{\mathrm{2}}\:. \\ $$ $$\mathrm{If}\:\:\mathrm{M}<\mathrm{1}\:\mathrm{with}\:\mid\mathrm{cos}\:{x}−\mathrm{cos}\:{y}\mid\leqslant\mathrm{M}\:\mid{x}−{y}\mid \\ $$ $$\mathrm{for}\:\mathrm{every}\:{x},\:{y}\:\in\:\left[\mathrm{0},{a}\right],\:\mathrm{then}\:\mathrm{M}=.. \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Mar/19 | ||
$${let}\:{y}>{x} \\ $$ $${so}\:{cosx}>{cosy} \\ $$ $$={cosx}−{cosy} \\ $$ $$=+{ve} \\ $$ $${but}\:{x}−{y}=−{ve} \\ $$ $$\mid{x}−{y}\mid=+{ve} \\ $$ $${from}\:{attached}\:{graph} \\ $$ $${tan}\theta=\frac{{y}_{\mathrm{2}} −{y}_{\mathrm{1}} }{{x}_{\mathrm{2}} −{x}_{\mathrm{1}} }=\frac{{cosy}−{cosx}}{{y}−{x}} \\ $$ $${tan}\theta={M}=\frac{{cosy}−{cosx}}{{y}−{x}}\:{but}\:{cosx}>{cosy} \\ $$ $${y}>{x}\:\:{so}\:{tan}\theta={M}\:\:{is}\:−{ve} \\ $$ $${wait}... \\ $$ | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 07/Mar/19 | ||
Commented bygunawan last updated on 07/Mar/19 | ||
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:+{ve}\:\mathrm{Sir}? \\ $$ $$ \\ $$ | ||