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Manipal MET2020MathematicsQuadratic Equation

The least value of a , for which the function 4 x + 1 1- x =a has atleast one solution in the interval (0, 2 ) , is

Options

  1. A9
  2. B4
  3. C5
  4. D1

Correct answer

A. 9

Step-by-step solution

Given, 4 x + 1 1- x =a 4(1- x)+ x=a x(1- x) 4-4 x+ x=a x-a ^2 x a ^2 x-(3+a) x+4=0 ....(i) It is a quadratic equation in x , so D 0 array cr & (3+a)^2-4 4 a 0 & 9+a^2+6 a-16 a 0 & a^2-10 a+9 0 & (a-1)(a-9) 0 array a 9 or a -9 Now, at a = 9, Eq. (i) becomes 9 ^2 x-12 x+4=0 (3 x-2)^2=0 x= 2 3 0 x (0, 2 ) Hence, least value of a is 9 .

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