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Question Number 112128 by mohammad17 last updated on 06/Sep/20

Commented by mohammad17 last updated on 06/Sep/20

$p l e a s e s i r h e l p m e$
Commented by bobhans last updated on 06/Sep/20

$(Q 4 B) y^{3} = 16 - x^{3}; y = {(16 - x^{3})}^{\frac{1}{3}}$ $\frac{dy}{dx} = - x^{2} {(16 - x^{3})}^{- \frac{2}{3}}$ $\frac{d^{2} y}{{dx}^{2}} = - (2 x {(16 - x^{3})}^{- \frac{2}{3}} + x^{2} ({2 x}^{2} {(16 - x^{3})}^{- \frac{5}{3}})$ $\frac{d^{2} y}{{dx}^{2}} = - 2 x {(16 - x^{3})}^{- \frac{2}{3}} - {2 x}^{4} {(16 - x^{3})}^{- \frac{5}{3}}$ $\frac{d^{2} y}{{dx}^{2}} ∣_{x = 2} = - 4 {(8)}^{- \frac{2}{3}} - 32 {(8)}^{- \frac{5}{3}}$ $= \frac{- 4}{4} - \frac{32}{32} = - 2$
	Answered by bemath last updated on 06/Sep/20
	$(Q1) y−xy = 2x, y = ((2x)/(−x+1)) ⇒ y′=(2/((1−x)^2 )) , m_(normal ) =−(1/(y′)) m_(normal) =−(((1−x)^2 )/2) parallel to line 2x+y=0 ⇒ −(((1−x)^2 )/2) = −2 ⇒(1−x)^2 = 4 → { ((1−x=2, x=−1)),((1−x=−2,x=3)) :} for { ((x=−1, y=((−2)/2)=−1)),((x=3,y=(6/(−2))=−3)) :} → { ((normal at point(−1,−1) is 2x+y=−3)),((normal at point(3,−3) is 2x+y=3)) :}$
	$(Q 1) y - x y = 2 x, y = \frac{2 x}{- x + 1}$ $\Rightarrow y^{'} = \frac{2}{{(1 - x)}^{2}}, m_{n o r m a l} = - \frac{1}{y^{'}}$ $m_{n o r m a l} = - \frac{{(1 - x)}^{2}}{2} p a r a l l e l t o$ $l i n e 2 x + y = 0 \Rightarrow - \frac{{(1 - x)}^{2}}{2} = - 2$ $\Rightarrow {(1 - x)}^{2} = 4 \to {\begin{cases} 1 - x = 2, x = - 1 \\ 1 - x = - 2, x = 3 \end{cases}$ $f o r {\begin{cases} x = - 1, y = \frac{- 2}{2} = - 1 \\ x = 3, y = \frac{6}{- 2} = - 3 \end{cases}$ $\to {\begin{cases} n o r m a l a t p o i n t (- 1, - 1) i s 2 x + y = - 3 \\ n o r m a l a t p o i n t (3, - 3) i s 2 x + y = 3 \end{cases}$
	Commented by mohammad17 last updated on 06/Sep/20

	$t h a n k y o u s i r c a n y o u h e l p m e ?$
	Commented by bemath last updated on 06/Sep/20

	$o k$
	Answered by bemath last updated on 06/Sep/20

	$(Q 5 B) \lim_{x \to 0} \frac{x \sin x}{1 - \cos x} = \lim_{x \to 0} \frac{\sin x + x \cos x}{\sin x} = \lim_{x \to 0} (1 + \frac{x \cos x}{\sin x})$ $1 + \lim_{x \to 0} \frac{\cos x - x \sin x}{\cos x} = 2$
	Answered by bemath last updated on 06/Sep/20

	$(Q 2 B) \lim_{x \to \infty} \frac{3 x^{2} - 6 x}{4 x - 8} = \lim_{x \to 0} \frac{x^{2} (3 - \frac{6}{x})}{x^{2} (\frac{4}{x} - \frac{8}{x^{2}})} = \infty$
	Answered by mathmax by abdo last updated on 06/Sep/20
	$Q_3 ) y(x)=((x^3 −1)/(x^2 −1)) y is not conutinue at x_0 =1 but lim_(x→1) y(x) =lim_(x→1) (((x−1)(x^2 +x+1))/((x−1)(x+1))) =lim_(x→1) ((x^2 +x+1)/(x+1))=(3/2) but if we put { ((y(x)=((x^3 −1)/(x^2 −1)) if x≠1)),((y(1)=(3/2))) :} this function become continue on R 2) y(x)=((∣x∣)/x) lim_(x→0^+ ) y(x) =lim_(x→0^+ ) (x/x) =1 (continue at right) lim_(x→0^− ) y(x)=lim_(x→0^− ) ((−x)/x) =−1 (continue at left) but y is not continue at$
	$Q_{3}) y (x) = \frac{x^{3} - 1}{x^{2} - 1} y is not conutinue at x_{0} = 1 but$ $\lim_{x \to 1} y (x) = \lim_{x \to 1} \frac{(x - 1) (x^{2} + x + 1)}{(x - 1) (x + 1)} = \lim_{x \to 1} \frac{x^{2} + x + 1}{x + 1} = \frac{3}{2}$ $but if we put {\begin{cases} y (x) = \frac{x^{3} - 1}{x^{2} - 1} if x \neq 1 \\ y (1) = \frac{3}{2} \end{cases}$ $this function become continue on R$ $2) y (x) = \frac{∣ x ∣}{x} \lim_{x \to 0^{+}} y (x) = \lim_{x \to 0^{+}} \frac{x}{x} = 1 (continue at right)$ $\lim_{x \to 0^{-}} y (x) = \lim_{x \to 0^{-}} \frac{- x}{x} = - 1 (continue at left)$ $but y is not continue at$
	Commented by mathmax by abdo last updated on 06/Sep/20

	$at 0$
	Answered by mathmax by abdo last updated on 06/Sep/20

	${q_{4})}_{b} y^{2} = 7 - x^{2} - xy so x = 0 \Rightarrow y = \bar{+} \sqrt{7}$ $the curve crosses the x axis at x_{1} (\sqrt{7}, 0) and x_{2} = (- \sqrt{7}, o)$ $we have by derivation 2 x + y + {xy}^{^{'}} + {2 yy}^{^{'}} = 0 \Rightarrow$ $(x + 2 y) y^{^{'}} = - (2 x + y) \Rightarrow y^{^{'}} = - \frac{2 x + y}{x + 2 y}$ $the dirctor coefficient of tanvent at x_{1} is y^{^{'}} (\sqrt{7})$ $= - \frac{2 \sqrt{7} + y (\sqrt{7})}{\sqrt{7} + 2 y (\sqrt{7})} = - \frac{2 \sqrt{7} + 0}{\sqrt{7}} = - 2$ $the director coefficient of tangent at x_{2} is y^{^{'}} (- \sqrt{7})$ $= - \frac{2 (- \sqrt{7}) + y (- \sqrt{7})}{- \sqrt{7} + 2 y (- \sqrt{7})} = - 2 \Rightarrow the tangentes are paralleles$
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