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If-a-lt-0-2pi-1-10-3-cos-x-dx-lt-b-then-the-ordered-pair-a-b-is-




Question Number 53118 by gunawan last updated on 18/Jan/19
If a<∫_0 ^(2π) (1/(10+3 cos x)) dx<b, then the  ordered pair (a, b) is
$$\mathrm{If}\:{a}<\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{10}+\mathrm{3}\:\mathrm{cos}\:{x}}\:{dx}<{b},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{ordered}\:\mathrm{pair}\:\left({a},\:{b}\right)\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19
i am using simple logic..  then i[shall solve in details  we know  cosx max value=+1  cosx minimum value=−1  so   10+3(1)>10+3cosx>10+3(−1)  13>10+3cosx>7  (1/(13))<(1/(10+3cosx))<(1/7)  ∫_0 ^(2π) (dx/(13))<∫_0 ^(2π) (dx/(10+3cosx))<∫_0 ^(2π) (dx/7)  ((2π)/(13))<∫_0 ^(2π) (dx/(10+3cosx))<((2π)/7)  so a=((2π)/(13))   b=((2π)/7)    (((2π)/(13)),((2π)/7))
$${i}\:{am}\:{using}\:{simple}\:{logic}.. \\ $$$${then}\:{i}\left[{shall}\:{solve}\:{in}\:{details}\right. \\ $$$${we}\:{know}\:\:{cosx}\:{max}\:{value}=+\mathrm{1} \\ $$$${cosx}\:{minimum}\:{value}=−\mathrm{1} \\ $$$${so} \\ $$$$\:\mathrm{10}+\mathrm{3}\left(\mathrm{1}\right)>\mathrm{10}+\mathrm{3}{cosx}>\mathrm{10}+\mathrm{3}\left(−\mathrm{1}\right) \\ $$$$\mathrm{13}>\mathrm{10}+\mathrm{3}{cosx}>\mathrm{7} \\ $$$$\frac{\mathrm{1}}{\mathrm{13}}<\frac{\mathrm{1}}{\mathrm{10}+\mathrm{3}{cosx}}<\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{13}}<\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{10}+\mathrm{3}{cosx}}<\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{7}} \\ $$$$\frac{\mathrm{2}\pi}{\mathrm{13}}<\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{10}+\mathrm{3}{cosx}}<\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$${so}\:{a}=\frac{\mathrm{2}\pi}{\mathrm{13}}\:\:\:{b}=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:\:\:\left(\frac{\mathrm{2}\pi}{\mathrm{13}},\frac{\mathrm{2}\pi}{\mathrm{7}}\right) \\ $$
Commented by gunawan last updated on 19/Jan/19
wow thank you Sir
$$\mathrm{wow}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
most welcome...
$${most}\:{welcome}… \\ $$

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