InterviewSolution
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InterviewSolution
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Let `f(x) =4x^2-4ax+a^2-2a+2` be a quadratic polynomial in x,a be any real number.If x-coordinate ofd vertex of parabola y =f(x) is less thna 0 and f(x) has minimum value 3 for `x in [0,2]` then value of a isA. `a le 0 or a ge 4`B. `0 le a le 4`C. `a ge 0`D. none of these |
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Answer» Correct Answer - 1 `f(x) = 4x^(2) - 4ax + a^(2)-2a + 2` represents a parabola whose vertex is at `((a)/(2), 2 - 2a)` Case-I : `0 lt(a)/(2) lt 2` In this case , `f(x)` attains the global minimum value at `x = (a)/(2)` . Thus , `f((a)/(2)) = 3` `rArr 3 = -2a + 2` `rArr a = - (1)/(2)` (Rejected) Case II : `(a)/(2)le2` In this cases, f(x) attains the global minimun value at x = 0 . Thus, `f(0) = 3` `rArr 3 = a^(2) - 2a + 2` `rArr a = 5 pm sqrt(10)` So, `a = 5 + sqrt(10)` Case III: `(a)/(2) le 0` In case,f(x) attains the global minimum value at x = 0. `rArr 3=a^(2) - 2a + 2 ` `rArr a = 1 p, sqrt(2)`. So, `a= 1 - sqrt(2)`. Hence, premissible values of a are `1-sqrt(2)` and `5+ sqrt(10)`. `f(x) = 4x^(2)-4ax + a^(2) -2a+ 2` is monotonic in `[0,2]` Hence, point of minima of funciton should not lie in `[0,2]`. Now, `f(x) = 0` `rArr 8x - 4a =0` If `(a)/(2)in [0,2]`, then `a in [0,4]`. For f(x) to be montomic in `[0,2], a notin [0,4]`. So , `a le 0 or a ge 4`. |
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