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  • If <img alt="\\\mathrm{f(x) = x^2 + 2(a-1)x + (a+5)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5C%5C%5Cmathrm%7Bf%28x%29%20%3D%20x%5E2%20&plus;%202%28a-1%29x%20&

If \\\mathrm{f(x) = x^2 + 2(a-1)x + (a+5)} , then the values of ‘a’ for which f(x) = 0  have two real and equal roots is 

Option: 1

a \epsilon [1, 4]


Option: 2

a = -4, 1


Option: 3

a = -1, 4


Option: 4

4, -1


Option: 5

4, -1


Option: 6

4, -1


Answers (1)

best_answer

As we have learnt in

 


 

If  D = 0 and a, b,c is from real, then we have equal and real roots so since both root will coincide hence the equation will touch the x- axis at one point only, and it will open upward or downward depending upon the value of ‘a’. Graph is shown below. 

 

 

                         

 

-

 

 

f(x) will  have  real and equal roots only when D = 0

Let’s find the discriminant of the function, so

D  = 4(a-1)2 - 4(a+5)

After further solving this equation, we get

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

So a = 4, -1

 

correct option is (d)

Posted by

Riya

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