Question-66602

Question Number 66602 by aliesam last updated on 17/Aug/19

Commented by Rio Michael last updated on 17/Aug/19

16) l e t y = x^{3}

\Rightarrow \frac{d y}{d x} = 3 x^{2}

\frac{d y}{d x} ∣_{x = - 1} = 3

w h e n x = - 1, y = - 1 h e n c e p t (- 1, - 1)

e q u a t i o n

y = m x + c

- 1 = 3 (- 1) + c

c = 2

e q u a t i o n; y = 3 x + 2 a n s w e r C

17) \frac{9 x^{2} - 4 y^{2}}{3 x + 2 y} = \frac{(3 x - 2 y) (3 x + 2 y)}{3 x + 2 y}

= 3 x - 2 y

= 3 (2) - 2 (3)

= 0 D

Commented by aliesam last updated on 17/Aug/19

p e r f e c t

Commented by mathmax by abdo last updated on 17/Aug/19

1 5) t h e c o r r c t a n s w e r i s a)

16) e q u a t i o n o f t a n g e n t e a t x = - 1 i s y = f^{^{'}} (- 1) (x + 1) + f (- 1)

f (x) = x^{3} \Rightarrow f^{^{'}} (x) = 3 x^{2} \Rightarrow f^{^{'}} (- 1) = 3 a n d f (- 1) = - 1 \Rightarrow

y = 3 (x + 1) - 1 \Rightarrow y = 3 x + 2

18) {\begin{cases} x - y + 1 = 0 \Rightarrow {\begin{cases} x - y = - 1 \\ 7 x + y = 17 \end{cases} \\ 7 x + y = 17 \end{cases}

Δ = 8 \neq 0 \Rightarrow t h e s y s t e m e h a s o n e s o l u t i o n (a)

Answered by Rio Michael last updated on 17/Aug/19

15) A

Commented by aliesam last updated on 17/Aug/19

t h a n k y o u s i r

Answered by Tanmay chaudhury last updated on 17/Aug/19

15) a e d i t e d

16) (y + 1) = 3 {(- 1)}^{2} (x + 1) \to y = 3 x + 2 s o C

17) \frac{36 - 36}{6 + 6} = 0 d

18) y = x + 1 y + 7 x = 17 \to x = 2 y = 3 s o a

18) s i n x + c o s x + 3

3 + \frac{1}{\sqrt{2}} ⩾ \frac{1}{\sqrt{2}} s i n (\frac{π}{4} + x) + 3 ⩾ 3 - \frac{1}{\sqrt{2}} a l w Y s m o r e t h a n 2

s o b

19) y = x^{3} - 3 x^{2} + 3 x - 1

w a i t f o r 19