Question
If x is real and k=x2−x+1x2+x+1, then
Answer: Option A
:
A
Fromk=x2−x+1x2+x+1
We have x2(k−1)+x(k+1)+k−1=0
As given, x is real ⇒ (k+1)2−4(k−1)2≥0
⇒ 3k2−10k+3≤0
Which is possible only when the value of k lies between the roots of the equation3k2−10k+3=0
That is, when 13≤k≤3 {Since roots are 13 and 3}
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:
A
Fromk=x2−x+1x2+x+1
We have x2(k−1)+x(k+1)+k−1=0
As given, x is real ⇒ (k+1)2−4(k−1)2≥0
⇒ 3k2−10k+3≤0
Which is possible only when the value of k lies between the roots of the equation3k2−10k+3=0
That is, when 13≤k≤3 {Since roots are 13 and 3}
Was this answer helpful ?
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