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Question Number 59114 by kelly33 last updated on 04/May/19

Commented by kelly33 last updated on 04/May/19

$P l e a s e h e l p m e .$
	Answered by tanmay last updated on 04/May/19

	$5) t r y i n g p = 4 x + y^{2}$ $p = 4 x + 4 - 2 x^{2}$ $\frac{d p}{d x} = 4 - 4 x f o r m a x / m i n \frac{d p}{d x} = 0 \to x = 1$ $\frac{d^{2} p}{{d x}^{2}} = - 4 \frac{d^{2} p}{{d x}^{2}} < 0 s o m a x i m u m$ $y^{2} = 4 - 2 x^{2} \to y^{2} = 4 - 2 {(1)}^{2}$ $y = \pm \sqrt{2}$ $p = 4 x + y^{2}$ $= 4 (1) + 2 = 6$ ${(4 x + y^{2})}_{m a x} = 6$ $a n o t h e r w a y$ $p = 4 x + 4 - 2 x^{2}$ $p = - 2 (x^{2} - 2 x + 1) + 2 + 4$ $p = - 2 {(x - 1)}^{2} + 6$ $p = 6 - 2 {(x - 1)}^{2}$ $s i n c e {(x - 1)}^{2} ⩾ 0$ $s o p_{m a x} = 6$ $p_{m i n} = n o t f e a s i b l e$
	Answered by tanmay last updated on 04/May/19

	$4) f (x) =∣ x - 1 ∣ + ∣ x^{2} - 2 x ∣$ $f (x) =∣ x - 1 ∣ + ∣ x (x - 2) ∣$ $c r i t i c a l v a l u e o f x a r e 0, 1 s n d 2$ $f (x) = 1 w h e n x = 2$ $f (x) = (x - 1) - x (x - 2) w h e n 2 > x > 1$ $= - x^{2} + 2 x + x - 1$ $= - x^{2} + 3 x - 1$ $= 1 w h e n x = 1$ $= - (x - 1) - x (x - 2) w h e n 1 > x > 0$ $= - x + 1 - x^{2} + 2 x$ $= - x^{2} + 3 x + 1$ $f (x) = 1 w h e n x = 0$ $n o w f (x) = - x^{2} + 3 x - 1 w h e n 2 > x > 1$ $\frac{d f}{d x} = - 2 x + 3 f o r m a x / m i n \frac{d f}{d x} = 0 s o x = \frac{3}{2} = 1.5$ $\frac{d^{2} f}{{d x}^{2}} = - 2 < 0$ $s o a t x = \frac{3}{2} f (x) m a x i m u m$ $f {(x)}_{m a x} = - x^{2} + 3 x - 1$ $= - {(\frac{3}{2})}^{2} + 3 \times \frac{3}{2} - 1$ $= \frac{- 9}{4} + \frac{9}{2} - 1$ $= \frac{- 9 + 18 - 4}{4} = \frac{5}{4}$ $f (x) = - x^{2} + 3 x + 1 w h e n 1 > x > 0$ $\frac{d f}{d x} = - 2 x + 3 f o r m a x / m i n \frac{d f}{d x} = 0 s o x = \frac{3}{2}$ $\frac{d^{2} f}{{d x}^{2}} = - 2 < 0 m a x i m u m a t x = \frac{3}{2} b u t x = \frac{3}{2} d o e s n o t$ $l i e i n t h e i n t e r v a l [t h a t i s 1 > x > 0$ $c o n c l u s i o n f {(x)}_{m a x} = \frac{5}{4} a t x = \frac{3}{2}$ $f {(x)}_{m i n} = 1 w h e n x = 2$ $f {(x)}_{m i n} = 1 w h e n x = 1$ $f {(x)}_{m i n} = 1 w h e n x = 0$ $f {(x)}_{m a x} = \frac{5}{4} w h e n x = \frac{3}{2}$
	Answered by tanmay last updated on 05/May/19

	$3) y = {a x}^{3} + {b x}^{2} + c x + d$ $(0, 1) (- 1, - 2), (1, 2) (2, 9)$ $1 = d$ $- 2 = - a + b - c + 1$ $- a + b - c = - 3$ $2 = a + b + c + 1$ $- a + b - c = - 3$ $a + b + c = 1$ $2 b = - 2 b = - 1$ $s o$ $- a - 1 - c = - 3$ $a + c = 2 . . . .$ $a - 1 + c = 1$ $a + c = 2$ $9 = a {(2)}^{3} + b {(2)}^{2} + c (2) + d$ $9 = 8 a - 4 + 2 c + 1$ $8 a + 2 c = 12$ $4 a + c = 6$ $a + c = 2$ $3 a = 4$ $a = \frac{4}{3}$ $c = 2 - \frac{4}{3} \to c = \frac{2}{3}$ $y = \frac{4}{3} x^{3} - x^{2} + \frac{2}{3} x + 1$
	Answered by tanmay last updated on 05/May/19

	$2) f o r m u l a$ $X = (x - h) c o s θ + (y - k) s i n θ$ $Y = - (x - h) s i n θ + (y - k) c o s θ$ $h e r e (h, k) \to (2, 1) θ = 45^{o}$ $(x, y) = (3, 3)$ $X = (3 - 2) c o s 45^{o} + (3 - 1) s i n 45^{o}$ $X = \frac{1}{\sqrt{2}} (1 + 2) = \frac{3}{\sqrt{2}}$ $Y = - (3 - 2) s i n 45^{o} + (3 - 1) c o s 45^{o}$ $Y = \frac{1}{\sqrt{2}} (- 1 + 2) = \frac{1}{\sqrt{2}}$
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