If atleast one of the root of the equation x2−(a−3)x+a=0 is greater than 2, then a lies in the interval
A
[7,9]
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B
[7,∞)
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C
[9,∞)
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D
[7,9)
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Solution
The correct option is C[9,∞) Let f(x)=x2−(a−3)x+a
Case 1: Both roots are greater than 2,
Required conditions are (i)D≥0 ⇒D=(a−3)2−4a≥0 ⇒a2−10a+9≥0⇒(a−1)(a−9)≥0⇒a∈(−∞,1]∪[9,∞)⋯(1)
f(2)>0⇒4−(a−3)2+a>0⇒a<10⋯(2) Now, x coordinate of the vertex will be greater than 2. ∴a−32>2⇒a>7⋯(3) From equation (1),(2) and (3), ⇒a∈[9,10)⋯(4)
Case 2: One root is greater than 2 and other root is less than 2 ∵2 lies in between the roots ⇒f(2)<0 ⇒4−(a−3)2+a<0 ⇒a>10⋯(5)
Case 3: One root is greater than 2 and other root (lowest root) =2 ∵ Lowest root is 2. Required conditions are (i)f(2)=0 and (ii)−b2a>0 f(2)=0⇒a=10...(6) −b2a>0⇒a>7...(7) From (6) and (7) a=10...(8)