Menu Close

Question-21526

Tinku Tara 0 Comments
Pin




Question Number 21526 by tawa tawa last updated on 26/Sep/17
Commented by tawa tawa last updated on 26/Sep/17
Please help me. Thanks in advance.
$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}.\:\mathrm{Thanks}\:\mathrm{in}\:\mathrm{advance}. \\ $$
Commented by $@ty@m last updated on 26/Sep/17
4. (i) The general term in a Binomial expansion is given by: t_(r+1) =^n C_r x^(n−r) a^r Here t_(r+1) =^6 C_r (2x)^(6−r) (1)^r To find coefficient of x^3 , we put r=3 ∴ t_4 =^6 C_3 (2x)^3 =20×8x^3 ⇒coefficient of x^3 =160 Ans. (ii) t_5 =^6 C_4 (2x)^2 (1)^4 =15×4x^2 =60x^2 ⇒coefficient of x^3 in (1−3x)(1+2x)^6 =160−3×60 =160−180 =−20 Ans.
$$\mathrm{4}.\:\left({i}\right)\:{The}\:{general}\:{term}\:{in}\:{a}\:{Binomial} \\ $$$${expansion}\:{is}\:{given}\:{by}: \\ $$$${t}_{{r}+\mathrm{1}} =\:^{{n}} {C}_{{r}} {x}^{{n}−{r}} {a}^{{r}} \\ $$$$\:{Here}\:{t}_{{r}+\mathrm{1}} =\:^{\mathrm{6}} {C}_{{r}} \left(\mathrm{2}{x}\right)^{\mathrm{6}−{r}} \left(\mathrm{1}\right)^{{r}} \\ $$$${To}\:{find}\:{coefficient}\:{of}\:{x}^{\mathrm{3}} ,\:{we}\:{put} \\ $$$${r}=\mathrm{3} \\ $$$$\therefore\:{t}_{\mathrm{4}} =\:^{\mathrm{6}} {C}_{\mathrm{3}} \left(\mathrm{2}{x}\right)^{\mathrm{3}} =\mathrm{20}×\mathrm{8}{x}^{\mathrm{3}} \\ $$$$\Rightarrow{coefficient}\:{of}\:{x}^{\mathrm{3}} =\mathrm{160}\:{Ans}. \\ $$$$\left({ii}\right)\:{t}_{\mathrm{5}} =\:^{\mathrm{6}} {C}_{\mathrm{4}} \left(\mathrm{2}{x}\right)^{\mathrm{2}} \left(\mathrm{1}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{15}×\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{60}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{coefficient}\:{of}\:{x}^{\mathrm{3}} \:{in}\:\left(\mathrm{1}−\mathrm{3}{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{160}−\mathrm{3}×\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{160}−\mathrm{180} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{20}\:{Ans}. \\ $$
Commented by tawa tawa last updated on 26/Sep/17
wow. God bless you sir
$$\mathrm{wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by $@ty@m last updated on 26/Sep/17
5. (i) ar(OCD)=(1/2)×OC×OD×sin ∠COD =(1/2)×10×10×sin(0.8) =50×0.7174 =35.87 cm^2 −−−(1) ar(OAB)=((0.8)/(2π))×π×6^2 =0.4×36 =14.4 cm^2 −−−(2) ∴ Area of shaded region=35.87−14.40 =21.47 cm^2 Ans. (ii) We have l=rθ ∴AB^(⌢) =6×0.8=4.8 −−(3) and CD^2 =OC^2 +OD^2 −2.OC.OD.cos∠COD =10^2 +10^2 −2×10×10×cos(0.8) =200−200×0.6967 =60.66 CD=7.79 cm −−(4) ∴ Perimeter of shaded region =4+4.8+4+7.79 =20.59 cm Ans.
$$\mathrm{5}.\:\left({i}\right)\:{ar}\left({OCD}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{OC}×{OD}×\mathrm{sin}\:\angle{COD} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{10}×\mathrm{sin}\left(\mathrm{0}.\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{50}×\mathrm{0}.\mathrm{7174} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{35}.\mathrm{87}\:{cm}^{\mathrm{2}} \:\:−−−\left(\mathrm{1}\right) \\ $$$${ar}\left({OAB}\right)=\frac{\mathrm{0}.\mathrm{8}}{\mathrm{2}\pi}×\pi×\mathrm{6}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{4}×\mathrm{36}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{14}.\mathrm{4}\:{cm}^{\mathrm{2}} \:−−−\left(\mathrm{2}\right) \\ $$$$\therefore\:{Area}\:{of}\:{shaded}\:{region}=\mathrm{35}.\mathrm{87}−\mathrm{14}.\mathrm{40} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{21}.\mathrm{47}\:{cm}^{\mathrm{2}} \:{Ans}. \\ $$$$\left({ii}\right)\:{We}\:{have} \\ $$$${l}={r}\theta \\ $$$$\therefore\overset{\frown} {{AB}}=\mathrm{6}×\mathrm{0}.\mathrm{8}=\mathrm{4}.\mathrm{8}\:−−\left(\mathrm{3}\right) \\ $$$${and}\:{CD}^{\mathrm{2}} ={OC}^{\mathrm{2}} +{OD}^{\mathrm{2}} −\mathrm{2}.{OC}.{OD}.\mathrm{cos}\angle{COD} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{10}×\mathrm{10}×\mathrm{cos}\left(\mathrm{0}.\mathrm{8}\right) \\ $$$$\:\:\:=\mathrm{200}−\mathrm{200}×\mathrm{0}.\mathrm{6967} \\ $$$$\:\:\:=\mathrm{60}.\mathrm{66} \\ $$$${CD}=\mathrm{7}.\mathrm{79}\:{cm}\:−−\left(\mathrm{4}\right) \\ $$$$\therefore\:{Perimeter}\:{of}\:{shaded}\:{region} \\ $$$$=\mathrm{4}+\mathrm{4}.\mathrm{8}+\mathrm{4}+\mathrm{7}.\mathrm{79} \\ $$$$=\mathrm{20}.\mathrm{59}\:{cm}\:{Ans}.\: \\ $$
Commented by tawa tawa last updated on 26/Sep/17
God bless you sir. I really appreciate
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *