Test 1. Solve equation (k^2 −1)x^2 +(k−1)x+(k+1)=0 k∈R (30) 2. Prove ((sin(x))/(1+cos(x)))=((1−cos(x))/(sin(x))) (35) 3.P(x)=−2x^3 −2x^2 −x+2409 Find P(−11) (35) Evaluate other answers and give marks I want to see how math teachers evaluate in other countries Sorry foy my english


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Question Number 13491 by 433 last updated on 20/May/17

$T e s t$ $1 . S o l v e e q u a t i o n$ $(k^{2} - 1) x^{2} + (k - 1) x + (k + 1) = 0 k \in R$ $(30)$ $2 . P r o v e$ $\frac{s i n (x)}{1 + c o s (x)} = \frac{1 - c o s (x)}{s i n (x)}$ $(35)$ $3 . P (x) = - 2 x^{3} - 2 x^{2} - x + 2409$ $F i n d P (- 11)$ $(35)$ $E v a l u a t e o t h e r a n s w e r s a n d g i v e m a r k s$ $I w a n t t o s e e h o w m a t h t e a c h e r s e v a l u a t e i n o t h e r c o u n t r i e s$ $S o r r y f o y m y e n g l i s h$
	Answered by 433 last updated on 20/May/17

	$1 . I f k = 1$ $2 = 0$ $I f k = - 1$ $- 2 x = 0 \Rightarrow x = 0$ $I f k \neq \pm 1$ $Δ = {(k - 1)}^{2} - 4 (k^{2} - 1) (k + 1)$ $= (k - 1) ((k - 1) - 4 {(k + 1)}^{2})$ $= (k - 1) (k - 1 - 4 k^{2} - 8 k - 4)$ $(k - 1) (- 4 k^{2} - 7 k - 5)$ $Δ^{'} = 7^{2} - 20 \times 4 = 49 - 80 < 0$ $- 4 k^{2} - 7 k - 5 < 0 \forall k$ $I f k > 1 \Rightarrow Δ < 0$ $I f k \in (- \infty, - 1) \cup (- 1, 1) \Rightarrow Δ > 0$ $x_{1, 2} = \frac{1 - k \pm \sqrt{(k - 1) (- 4 k^{2} - 7 k - 5)}}{2 (k^{2} - 1)}$ $2 . \frac{s i n (x)}{1 + c o s (x)} = \frac{1 - c o s (x)}{s i n (x)} \Leftrightarrow$ ${s i n}^{2} (x) = 1 - {c o s}^{2} (x)$ ${s i n}^{2} (x) + {c o s}^{2} (x) = 1$ $3 . Q (x) = x + 11$ $P (x) = Q (x) R (x) + u$ $P (- 11) = Q (- 11) R (- 11) + u$ $Q (- 11) = - 11 + 11 = 0$ $P (- 11) = u$ $(- 2 x^{3} - 2 x^{2} - x + 2409) = (x + 11) (- 2 x^{2} + 20 x - 221) + 4840$ $P (- 11) = 4840$
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