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  • If x2 + 2(a-1)x + (a+5) > 0  for all x in real numbers  then a lies betweenOption: 1 <img alt="(-1,4)" src="https://learn.car

If x2 + 2(a-1)x + (a+5) > 0  for all x in real numbers  then a lies between

Option: 1

(-1,4)


Option: 2

a\epsilon R


Option: 3

(-\infty,1)\cup (4,\infty )


Option: 4

(-\infty,-1)\cup (4,\infty )


Answers (1)

best_answer

 

 

Sign of Quadratic Expression -

Let y = ax2 + bx +c = 0  be the quadratic equation, such that a is non-zero and a,b,c are form real numbers, then the condition for y to be +ve, -ve and zero are as following:

 

3. If  D > 0, then the quadratic equation y will have two real solution ? and ?, so if a > 0 then y = 0 on ? and ?, and between the solution (? < x < ?),, y will be -ve and left (for x < ? ) and right (x > ?) part of the solution will give +ve value of y

If a < 0, exactly the opposite will happen,  y = 0 on ? and ?, and between the solution (? < x < ?), y will be +ve and left (for x < ? ) and right (x > ?) part of the solution will give -ve value of y.

               

-

For equation x2 + 2(a-1)x + (a+5) > 0 to satisfy, equation should not have any real roots for all x in real, for that to happen D < 0 and 1 > 0 so a > 0 condition is already satisfied. So 

D < 0 ⇒ b2 - 4ac < 0  

Here a=1, b = 2(a-1) and c = a + 5 putting all these value in discriminant inequality, we have

\\\mathrm{4 (a-1)^2 - 4\cdot 1 \cdot (a+5) < 0}

After solving this inequality we get

(a-4) (a+1) < 0

So -1 < a < 4 is the answer

 

correct option is (a)

Posted by

Gaurav

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