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Question Number 113587 by mohammad17 last updated on 14/Sep/20

Commented by mohammad17 last updated on 14/Sep/20

$h e l p m e s i r$
	Answered by 1549442205PVT last updated on 14/Sep/20
	$Q1 1)((2x+1)/(x+2))≤1⇔((2x+1)/(x+2))−1≤0⇔((x−1)/(x+2))≤0 ⇔x∈(−2;1] 2)(2y+1)^2 >9⇔4y^2 +4y−8>0 ⇔y^2 +y−2>0⇔(y−1)(y+2)>0 ⇔y∈(−∞;−2)∪(1;+∞) 3)∣3x−4∣>2(1) determinant ((x,,(4/3),),((∣3x−4∣),(4−3x),0,(3x−4)),((∣3x−4∣−2),(2−3x),(−2),(3x−6))) From above tablet we have two cases i) { ((x<4/3)),((2−3x>0)) :}⇔ { ((x<4/3)),((3x<2)) :}⇔ { ((x<4/3)),((x<2/3)) :}⇔x<2/3 ii) { ((x≥4/3)),((3x−6>0)) :}⇔ { ((x≥4/3)),((x>2)) :}⇔x>2 Combining two cases we get x∈(−∞;2/3)∪(2;+∞) 4)y^2 +2y−3>0⇔(y−1)(y+3)>0 ⇔y∈(−∞;−3)∪(1;+∞) Q2. 1)The equation of the line pass through the point (0;5) and (4;2) is ((y−5)/(x−0))=((2−5)/(4−0))=((−3)/4)⇒y=((−3)/4)x+5 The equation of the line pass through the point (0;−2) and (3;6) is ((y+2)/(x−0))=((6+2)/(3−0))=(8/3)⇒y=(8/3)x−2 The acute angle between two lines is θ then tanθ=∣((k_1 −k_2 )/(1+k_1 k_2 ))∣ (1) where k_1 =((−3)/4),k_2 =(8/3).Replace into (1) we get tanθ=∣(((8/3)+(3/4))/(1+(((−3)/4))(8/3)))∣=∣((41)/(−12))∣=((41)/(12)) ⇒θ≈73°41′ 2)The intersection point of two graphs of two functions y=1−x^2 and y=(√(1−x^2 )) is roots of the system { ((y=1−x^2 (1))),((y=(√(1−x^2 )) (2))) :} ⇒1−x^2 =(√(1−x^2 )) ⇔ { ((x^2 ≤1)),(((1−x^2 )^2 =(1−x^2 ))) :} ⇔ { ((∣x∣≤1(3))),((1−x^2 )[(1−x^2 )−1]=0(4))) :} i)1−x^2 =0⇔x^2 =1⇔x=±1⇒y=0 ii)1−x^2 −1=0⇔x^2 =0⇔x=0⇒y=1 Thus,two graphs have 3 intersecrion points are:A(1,0),B(−1,0),C(0,1) 3)Suppose P_1 (x_1 ,y_1 ),P_2 (x_2 ,y_2 ) and M is the midpoint of P_1 P_2 .Then OP_1 ^(→) =(x_1 ,y_1 ),OP_2 ^(→) =(x_2 ,y_2 )and We have OM^(→) =OP_1 ^(→) +P_1 M^(→) =OP_2 ^(→) +P_2 M^(→) ⇒OM^(→) =((OP_1 ^(→) +P_1 M^(→) +OP_2 ^(→) +P_2 M^(→) )/2) =((OP_1 ^(→) +OP_2 ^(→) )/2) (since P_1 M^(→) =−P_2 M^(→) ) Therefore,if denote M(x_M ,y_M )then OM^(→) =(x_M ,y_M ) and by adding rule for the vectors we get x_M =((x_1 +x_2 )/2),y_M =((y_1 +y_2 )/2).That shows that the point with coordintes (((x_1 +x_2 )/2),((y_1 +y_2 )/2)) is the midpoint of P_1 P_2 .Q3. A) a)F(x)=1−2x−x^2 The domain of F(x) is D_F =(−∞;+∞) Range:we have 1−2x−x^2 =2−(x+1)^2 ≤2 ⇒−∞<F(x)≤2.Hence,R_F =(−∞;2] b)y=(√(∣x∣)) D_y =(−∞;+∞).Since y=(√(∣x∣))≥0∀x R_y =[0;+∞) c)y=(√((1/x)−1)) . The domain:we need must have x≠0 (1/x)−1≥0⇔((1−x)/x)≥0⇔0<x≤1.Hence D_y =(0;1] Range:(√((1/x)−1)) ≥0∀x∈D_y ⇒R_y =[0;+∞) Hence,x is impossible (I)x<0 (II)x=0 (III)x>1 B)We need find the values x so that (x/2)>1+(4/x)⇔(x/2)−1−(4/x)⇔((x^2 −2x−8)/(2x))>0 (((x+2)(x−4))/(2x))o>0⇔x∈(−2;0)∪(4;+∞)$
	$Q 1$ $1) \frac{2 x + 1}{x + 2} ⩽ 1 \Leftrightarrow \frac{2 x + 1}{x + 2} - 1 ⩽ 0 \Leftrightarrow \frac{x - 1}{x + 2} ⩽ 0$ $\Leftrightarrow x \in (- 2; 1]$ $2) {(2 y + 1)}^{2} > 9 \Leftrightarrow {4 y}^{2} + 4 y - 8 > 0$ $\Leftrightarrow y^{2} + y - 2 > 0 \Leftrightarrow (y - 1) (y + 2) > 0$ $\Leftrightarrow y \in (- \infty; - 2) \cup (1; + \infty)$ $3) ∣ 3 x - 4 ∣> 2 (1)$ $\| \begin{matrix} x & \frac{4}{3} \\ ∣ 3 x - 4 ∣ & 4 - 3 x & 0 & 3 x - 4 \\ ∣ 3 x - 4 ∣ - 2 & 2 - 3 x & - 2 & 3 x - 6 \end{matrix} \|$ $From above tablet we have two cases$ $i) {\begin{cases} x < 4 / 3 \\ 2 - 3 x > 0 \end{cases} \Leftrightarrow {\begin{cases} x < 4 / 3 \\ 3 x < 2 \end{cases} \Leftrightarrow {\begin{cases} x < 4 / 3 \\ x < 2 / 3 \end{cases} \Leftrightarrow x < 2 / 3$ $ii) {\begin{cases} x ⩾ 4 / 3 \\ 3 x - 6 > 0 \end{cases} \Leftrightarrow {\begin{cases} x ⩾ 4 / 3 \\ x > 2 \end{cases} \Leftrightarrow x > 2$ $Combining two cases we get$ $x \in (- \infty; 2 / 3) \cup (2; + \infty)$ $4) y^{2} + 2 y - 3 > 0 \Leftrightarrow (y - 1) (y + 3) > 0$ $\Leftrightarrow y \in (- \infty; - 3) \cup (1; + \infty)$ $Q 2 .$ $1) The equation of the line pass through$ $the point (0; 5) and (4; 2) is$ $\frac{y - 5}{x - 0} = \frac{2 - 5}{4 - 0} = \frac{- 3}{4} \Rightarrow y = \frac{- 3}{4} x + 5$ $The equation of the line pass through$ $the point (0; - 2) and (3; 6) is$ $\frac{y + 2}{x - 0} = \frac{6 + 2}{3 - 0} = \frac{8}{3} \Rightarrow y = \frac{8}{3} x - 2$ $The acute angle between two lines$ $is θ then \tan θ =∣ \frac{k_{1} - k_{2}}{1 + k_{1} k_{2}} ∣ (1) where$ $k_{1} = \frac{- 3}{4}, k_{2} = \frac{8}{3} . Replace into (1) we get$ $\tan θ =∣ \frac{\frac{8}{3} + \frac{3}{4}}{1 + (\frac{- 3}{4}) \frac{8}{3}} ∣=∣ \frac{41}{- 12} ∣= \frac{41}{12}$ $\Rightarrow θ \approx 73 ° 41^{'}$ $2) The intersection point of two$ $graphs of two functions y = 1 - x^{2}$ $and y = \sqrt{1 - x^{2}} is roots of the system$ ${\begin{cases} y = 1 - x^{2} (1) \\ y = \sqrt{1 - x^{2}} (2) \end{cases}$ $\Rightarrow 1 - x^{2} = \sqrt{1 - x^{2}} \Leftrightarrow {\begin{cases} x^{2} ⩽ 1 \\ {(1 - x^{2})}^{2} = (1 - x^{2}) \end{cases}$ $\Leftrightarrow {\begin{cases} ∣ x ∣⩽ 1 (3) \\ 1 - x^{2}) [(1 - x^{2}) - 1] = 0 (4) \end{cases}$ $i) 1 - x^{2} = 0 \Leftrightarrow x^{2} = 1 \Leftrightarrow x = \pm 1 \Rightarrow y = 0$ $ii) 1 - x^{2} - 1 = 0 \Leftrightarrow x^{2} = 0 \Leftrightarrow x = 0 \Rightarrow y = 1$ $Thus, two graphs have 3 intersecrion$ $points are : A (1, 0), B (- 1, 0), C (0, 1)$ $3) Suppose P_{1} (x_{1}, y_{1}), P_{2} (x_{2}, y_{2}) and$ $M is the midpoint of P_{1} P_{2} . Then$ $\vec{{OP}_{1}} = (x_{1}, y_{1}), \vec{{OP}_{2}} = (x_{2}, y_{2}) and$ $We have \vec{OM} = \vec{{OP}_{1}} + \vec{P_{1} M} = \vec{{OP}_{2}} + \vec{P_{2} M}$ $\Rightarrow \vec{OM} = \frac{\vec{{OP}_{1}} + \vec{P_{1} M} + \vec{{OP}_{2}} + \vec{P_{2} M}}{2}$ $= \frac{\vec{{OP}_{1}} + \vec{{OP}_{2}}}{2} (since \vec{P_{1} M} = - \vec{P_{2} M})$ $Therefore, if denote M (x_{M}, y_{M}) then$ $\vec{OM} = (x_{M}, y_{M}) and by adding rule$ $for the vectors we get$ $x_{M} = \frac{x_{1} + x_{2}}{2}, y_{M} = \frac{y_{1} + y_{2}}{2} . That shows$ $that the point with coordintes$ $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) is the midpoint of$ $P_{1} P_{2}$ $. Q 3 .$ $A)$ $a) F (x) = 1 - 2 x - x^{2}$ $The domain of F (x) is D_{F} = (- \infty; + \infty)$ $Range : we have 1 - 2 x - x^{2} = 2 - {(x + 1)}^{2} ⩽ 2$ $\Rightarrow - \infty < F (x) ⩽ 2 . Hence, R_{F} = (- \infty; 2]$ $b) y = \sqrt{∣ x ∣}$ $D_{y} = (- \infty; + \infty) . Since y = \sqrt{∣ x ∣} ⩾ 0 \forall x$ $R_{y} = [0; + \infty)$ $c) y = \sqrt{\frac{1}{x} - 1} .$ $The domain : we need must have x \neq 0$ $\frac{1}{x} - 1 ⩾ 0 \Leftrightarrow \frac{1 - x}{x} ⩾ 0 \Leftrightarrow 0 < x ⩽ 1 . Hence$ $D_{y} = (0; 1]$ $Range : \sqrt{\frac{1}{x} - 1} ⩾ 0 \forall x \in D_{y} \Rightarrow R_{y} = [0; + \infty)$ $Hence, x is impossible (I) x < 0 (II) x = 0$ $(III) x > 1$ $B) We need find the values x so that$ $\frac{x}{2} > 1 + \frac{4}{x} \Leftrightarrow \frac{x}{2} - 1 - \frac{4}{x} \Leftrightarrow \frac{x^{2} - 2 x - 8}{2 x} > 0$ $\frac{(x + 2) (x - 4)}{2 x} o > 0 \Leftrightarrow x \in (- 2; 0) \cup (4; + \infty)$
	Commented by mohammad17 last updated on 14/Sep/20

	$t h a n k y o u s i r$
	Commented by 1549442205PVT last updated on 15/Sep/20

	$You are welcome .$
	Answered by Dwaipayan Shikari last updated on 14/Sep/20

	$f (x) = 1 - 2 x - x^{2}$ $f (x) = 2 - {(1 + x)}^{2}$ $- \infty < x < \infty (D o m a i n)$ $R a n g e 2 ⩽ f (x) < - \infty$
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