Question-7413

Question Number 7413 by Tawakalitu. last updated on 28/Aug/16

Answered by sandy_suhendra last updated on 28/Aug/16

9) t h e c u r v e t h r o u g h t h e p o i n t (- 1, 0), (3, 0) a n d (- 2, - 10)

f (x) = a (x - x_{1}) (x - x_{2}) w h i c h x_{1} = - 1 a n d x_{2} = 3

- 10 = a (- 2 + 1) (- 2 - 3)

- 10 = 5 a \Rightarrow a = - 2

f (x) = - 2 (x + 1) (x - 3)

= - 2 x^{2} + 4 x + 6

Commented by Tawakalitu. last updated on 28/Aug/16

T h a n k s s o m u c h s i r

Answered by sandy_suhendra last updated on 28/Aug/16

10 a) \vec{A B} = \vec{O B} - \vec{O A} = (3 i + 4 j) - (2 i - 3 j) = i + 7 j

b) ∣ \vec{A B} ∣= \sqrt{1^{2} + 7^{2}} = \sqrt{50} = 5 \sqrt{2}

c) ∣ \vec{O A} ∣= \sqrt{2^{2} + {(- 3)}^{2}} = \sqrt{13}

∣ \vec{O B} ∣= \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5

c o s ∠ A O B = \frac{\vec{O A} \cdot \vec{O B}}{∣ \vec{O A} ∣ ∣ \vec{O B} ∣} = \frac{2 \times 3 + (- 3) \times 4}{(\sqrt{13}) \times 5} = \frac{- 6}{5 \sqrt{13}}

∠ A O B = a r c c o s \frac{- 6}{5 \sqrt{13}}

Commented by Tawakalitu. last updated on 28/Aug/16

T h a n k s s o m u c h s i r

Answered by sandy_suhendra last updated on 28/Aug/16

8) f (x) = x^{2} + 2 k x + 5 \Rightarrow a = 1; b = 2 k; c = 5

a l w a y s p o s i t i v e \Rightarrow a > 0 a n d D < 0

a = 1 s o a > 0

D = b^{2} - 4 a c < 0

(2 k^{2}) - 4 \times 1 \times 5 < 0

4 k^{2} - 20 < 0

k^{2} - 5 < 0

(k + \sqrt{5}) (k - \sqrt{5}) < 0

- \sqrt{5} < k < \sqrt{5}

Commented by nume1114 last updated on 28/Aug/16

I t h i n k w h e n a > 0 a n d D < 0,

f (x) i s a l w a y s p o s i t i v e .

S o,

D = b^{2} - 4 a c < 0

{(2 k)}^{2} - 4 \times 1 \times 5 < 0

(k + \sqrt{5}) (k - \sqrt{5}) < 0

- \sqrt{5} < k < \sqrt{5}

Commented by Rasheed Soomro last updated on 28/Aug/16

f (x) = x^{2} + 6 x + 8

f (- 3) = {(- 3)}^{2} + 6 (- 3) + 8 = 9 - 18 + 8 = - 1 < 0

D = 4 > 0 a n d a = 1 > 0

Commented by Tawakalitu. last updated on 28/Aug/16

T h a n k s s o m u c h s i r

Commented by sandy_suhendra last updated on 28/Aug/16

{T h a t}^{'} s r i g h t . S o r r y I h a v e m a d e a m i s t a k e . I^{'} v e c o r r e c t e d i t