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Find all the values of a for which 0 and 5 lie between the roots of the equation (a-6)x^2-4ax+10=0 

Option: 1

a\in(-\infty, 0)\cup (4,\infty)


Option: 2

a\in(-\infty, 0)


Option: 3

a\in(6, \infty)


Option: 4

None of these


Answers (1)

best_answer

Required conditions: \\\mathrm{ af(k_1) < 0 \; and \; af(k_2) < 0 \;}  

 

Now,

k1 = 0 and k2 = 5

(a-6)x^2-4ax+10=0

i) a.f( k_1)<0

\\(a-6)[(a-6 )0^2-4a(0)+10]<0\\ (a-6)<0\\ a<6\\ a\in(-\infty,6)

 

ii) a( k_2)<0

\\(a-6)[(a-6 )5^2-4a(5)+10]<0\\ (a-6)[25(a-6 )-20a+10]<0\\ (a-6)[5a-140]<0\\(a-6)(a-28)<0 \\a\in(6,28)

 

Now from above, there is no value of a in the intersection.

Posted by

seema garhwal

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