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Question Number 34739 by chakraborty ankit last updated on 10/May/18
Commented by abdo mathsup 649 cc last updated on 10/May/18
5) we have the formula Σ_(k=0) ^(n−1) [x +(k/n)] =[nx] ⇒ Σ_(k=0) ^(98) [ (2/3) +(k/(99))]=[99.(2/3)]=[66]=66 answer c.
$$\left.\mathrm{5}\right)\:{we}\:{have}\:{the}\:{formula} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left[{x}\:+\frac{{k}}{{n}}\right]\:=\left[{nx}\right]\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{\mathrm{98}} \left[\:\frac{\mathrm{2}}{\mathrm{3}}\:+\frac{{k}}{\mathrm{99}}\right]=\left[\mathrm{99}.\frac{\mathrm{2}}{\mathrm{3}}\right]=\left[\mathrm{66}\right]=\mathrm{66}\:\:\:{answer}\:{c}. \\ $$
Commented by math khazana by abdo last updated on 11/May/18
7) 2f(x) +f(1−x)=x^2 f is a polynom let put f(x) =ax^2 +bx +c ⇔ 2ax^2 +2bx +2c +a(1−x)^2 +b(1−x) +c =x^2 ⇔ 2ax^2 x^2 +2bx +2c +a(x^2 −2x+1) +b −bx +c =x^2 ⇔3ax^2 +(2b −2a −b)x +2c +a +b+c =x^2 ⇔ 3a=1 ,b−2a=0 , 3c +a+b=0 ⇔a=(1/3) , b=(2/3) , c =−((a+b)/3) =−(1/3) ⇒ f(x)= (1/3)x^2 +(2/3)x −(1/3) = ((x^2 +2x −1)/3) answer B.
$$\left.\mathrm{7}\right)\:\mathrm{2}{f}\left({x}\right)\:+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} \:\:\:{f}\:{is}\:{a}\:{polynom}\: \\ $$$${let}\:{put}\:{f}\left({x}\right)\:={ax}^{\mathrm{2}} \:+{bx}\:+{c} \\ $$$$\Leftrightarrow\:\mathrm{2}{ax}^{\mathrm{2}} \:+\mathrm{2}{bx}\:+\mathrm{2}{c}\:+{a}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \:+{b}\left(\mathrm{1}−{x}\right)\:+{c}\:={x}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{2}{ax}^{\mathrm{2}} \:{x}^{\mathrm{2}} \:+\mathrm{2}{bx}\:+\mathrm{2}{c}\:+{a}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)\:+{b}\:−{bx}\:+{c}\:={x}^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{3}{ax}^{\mathrm{2}} \:\:\:+\left(\mathrm{2}{b}\:−\mathrm{2}{a}\:−{b}\right){x}\:\:+\mathrm{2}{c}\:+{a}\:\:+{b}+{c}\:={x}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{3}{a}=\mathrm{1}\:,{b}−\mathrm{2}{a}=\mathrm{0}\:,\:\mathrm{3}{c}\:+{a}+{b}=\mathrm{0} \\ $$$$\Leftrightarrow{a}=\frac{\mathrm{1}}{\mathrm{3}}\:\:,\:{b}=\frac{\mathrm{2}}{\mathrm{3}}\:,\:{c}\:=−\frac{{a}+{b}}{\mathrm{3}}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} \:\:+\frac{\mathrm{2}}{\mathrm{3}}{x}\:−\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:−\mathrm{1}}{\mathrm{3}}\:\:{answer}\:{B}. \\ $$
Commented by chakraborty ankit last updated on 13/May/18
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 13/May/18
1. A={1,2,3,4},B={2,4,6} A∩B ⊇ C ⊆ A∪B A∩B={2,4} , A∪B={1,2,3,4,6} C={2,4}∪X , ∀ X∈P( (A∪B)−(A∩B) ) P({1,2,3,4,6}−{2,4}) =P({1,3,6} Number of X′s =2^3 =8 Number of C′s=8 [∵ C={2,4}∪X] (C) 8
$$\mathrm{1}. \\ $$$$\mathrm{A}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\},\mathrm{B}=\left\{\mathrm{2},\mathrm{4},\mathrm{6}\right\} \\ $$$$\mathrm{A}\cap\mathrm{B}\:\supseteq\:\mathrm{C}\:\subseteq\:\mathrm{A}\cup\mathrm{B} \\ $$$$\mathrm{A}\cap\mathrm{B}=\left\{\mathrm{2},\mathrm{4}\right\}\:\:,\:\:\mathrm{A}\cup\mathrm{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6}\right\} \\ $$$$\:\:\:\:\mathrm{C}=\left\{\mathrm{2},\mathrm{4}\right\}\cup\mathrm{X}\:,\:\:\forall\:\mathrm{X}\in\mathbb{P}\left(\:\left(\mathrm{A}\cup\mathrm{B}\right)−\left(\mathrm{A}\cap\mathrm{B}\right)\:\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{P}\left(\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6}\right\}−\left\{\mathrm{2},\mathrm{4}\right\}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathbb{P}\left(\left\{\mathrm{1},\mathrm{3},\mathrm{6}\right\}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Number}\:\mathrm{of}\:\mathrm{X}'\mathrm{s}\:=\mathrm{2}^{\mathrm{3}} =\mathrm{8} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{C}'\mathrm{s}=\mathrm{8}\:\:\:\:\:\:\:\:\:\:\left[\because\:\mathrm{C}=\left\{\mathrm{2},\mathrm{4}\right\}\cup\mathrm{X}\right] \\ $$$$\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{8} \\ $$
Answered by rahul 19 last updated on 12/May/18
3) (A) 4) (B) (4,∞). 6) (B)
$$\left.\mathrm{3}\right)\:\left({A}\right) \\ $$$$\left.\mathrm{4}\right)\:\left({B}\right)\:\left(\mathrm{4},\infty\right). \\ $$$$\left.\mathrm{6}\right)\:\left({B}\right)\: \\ $$
Answered by Rasheed.Sindhi last updated on 13/May/18
2. Let the number of families in the town is y. 25% of y=y/4 having cell phone. 15% of y=3y/20 having scooter. 1500 families having both. −−−− (y/4) −1500 having cell phones only. (3y/20) −1500 having scooters only. Families having a cell phone or a scooter: [(y/4) −1500]+1500+[(3y/20) −1500] (y/4)+((3y)/(20))−1500=((5y+3y−30000)/(20)) Having neither cell phone nor scooter y−((8y−30000)/(20))=65% of y ((20y−8y+30000)/(20))=((13y)/(20)) 13y−12y=30000 y=30000 Total number of families in the town (C) 30,000
$$\mathrm{2}. \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{families}\:\mathrm{in}\:\mathrm{the}\:\mathrm{town}\:\mathrm{is}\:\mathrm{y}. \\ $$$$\:\:\:\:\:\:\mathrm{25\%}\:\mathrm{of}\:\:\mathrm{y}=\mathrm{y}/\mathrm{4}\:\mathrm{having}\:\mathrm{cell}\:\mathrm{phone}. \\ $$$$\:\:\:\:\:\:\:\mathrm{15\%}\:\mathrm{of}\:\:\mathrm{y}=\mathrm{3y}/\mathrm{20}\:\mathrm{having}\:\mathrm{scooter}. \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1500}\:\mathrm{families}\:\mathrm{having}\:\mathrm{both}. \\ $$$$−−−− \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{y}/\mathrm{4}\right)\:−\mathrm{1500}\:\mathrm{having}\:\mathrm{cell}\:\mathrm{phones}\:\mathrm{only}. \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{3y}/\mathrm{20}\right)\:−\mathrm{1500}\:\mathrm{having}\:\mathrm{scooters}\:\mathrm{only}. \\ $$$$\mathrm{Families}\:\mathrm{having}\:\mathrm{a}\:\mathrm{cell}\:\mathrm{phone}\:\mathrm{or}\:\mathrm{a}\:\mathrm{scooter}: \\ $$$$\:\:\left[\left(\mathrm{y}/\mathrm{4}\right)\:−\mathrm{1500}\right]+\mathrm{1500}+\left[\left(\mathrm{3y}/\mathrm{20}\right)\:−\mathrm{1500}\right] \\ $$$$\:\:\:\:\:\:\frac{\mathrm{y}}{\mathrm{4}}+\frac{\mathrm{3y}}{\mathrm{20}}−\mathrm{1500}=\frac{\mathrm{5y}+\mathrm{3y}−\mathrm{30000}}{\mathrm{20}} \\ $$$$\mathrm{Having}\:\mathrm{neither}\:\mathrm{cell}\:\mathrm{phone}\:\mathrm{nor}\:\mathrm{scooter} \\ $$$$\:\:\:\:\:\:\mathrm{y}−\frac{\mathrm{8y}−\mathrm{30000}}{\mathrm{20}}=\mathrm{65\%}\:\mathrm{of}\:\mathrm{y} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{20y}−\mathrm{8y}+\mathrm{30000}}{\mathrm{20}}=\frac{\mathrm{13y}}{\mathrm{20}} \\ $$$$\:\:\:\:\:\:\:\mathrm{13y}−\mathrm{12y}=\mathrm{30000} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}=\mathrm{30000} \\ $$$$\mathcal{T}{otal}\:{number}\:{of}\:{families}\:{in}\:{the}\:{town} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\:\:\:\mathrm{30},\mathrm{000} \\ $$
Commented by chakraborty ankit last updated on 13/May/18
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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