f5數學MC求答案&求解

kx² + 2x + 1 > 0
k[x² + (2/k)x] + 1> 0
k[x² + (2/k)x + (1/k)²] ‒k(1/k)² + 1> 0
k[x + (1/k)]² + [1 ‒ (1/k)] > 0

For all real values of x and k, [x + (1/k)]² ≥0
Hence, k > 0 and [1 ‒ (1/k)] > 0
k > 0 and (k ‒ 1)/k > 0
k > 0 and k(k ‒ 1)/k² > 0
k > 0 and k(k ‒ 1) > 0
k > 0 and (k < 0 or k > 1)

Hence, range of k: k > 1...... The answer is D.

2015-06-28 13:30:50 補充：
Alternative answer :

kx² + 2x + 1 > 0
(k ‒ 1)x² + x² + 2x + 1 > 0
(k ‒ 1)x² + (x + 1)² > 0

For all real values of x and k, x² and (x + 1)² ≥ 0
Hence, k ‒ 1 > 0
k > 1 ...... The answer is D.

2015-06-29 00:20:08 補充：
One more alternative solution :

For a quadratic equation ax² + bx + c = 0 :
When a > 0 and Δ < 0, the graph y = ax² + bx + c lies above the x-axis and does not meet the x-axis. In other words, ax² + bx + c > 0 for all real values of x.

2015-06-29 00:20:21 補充：
When kx² + 2x + 1 is positive for all real values of x :
k > 0 and Δ < 0
k > 0 and 2² ‒ 4k < 0
k > 0 and ‒4k < ‒4
k > 0 and k > 1

Range of x : k > 1

土扁 大哥，對於一般的中學題，或許可以再加上一個常用方法以供同學參考，就是

　"(coeff. of x²) > 0 AND Δ < 0"

可以請同學想想圖像得知。