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Question

# If both the roots of the quadratic equation x2âˆ’6nx+(9n2âˆ’2n+2)=0 are greater than 3, then the range of n is:

A
(,1)
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B
(,119)
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C
[1,)
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D
(119,)
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Solution

## The correct option is D (119,∞)Given: x2−6nx+(9n2−2n+2)=0 Roots are greater than 3 To find: Range of n Step-1: Draw the graph for the given expression y=x2−6nx+(9n2−2n+2) Step-2: Write the applicable conditions. Step-3: Solve all the conditions. Step-4: Take intersection and combine the results for n. Now, the applicable conditions that needs to be satisfied are: i).D≥0 ii).f(k)>0 iii).−b2a>k Now, solving for all the conditions, i). D≥0 D=b2−4ac≥0⇒(−6n)2−4.1.(9n2−2n+2)≥0⇒36n2−36n2+8n−8≥0⇒8n−8≥0⇒8n≥8⇒n≥1⇒n∈[1,∞)⋯(A) ii).f(k)>0, where k is 3 ⇒f(3)>0 ⇒32−6n×3+(9n2−2n+2)>0 ⇒9−18n+9n2−2n+2>0 ⇒9n2−20n+11>0 ⇒(9n−11)(n−1)>0 ⇒n<1 or n>119 ⇒n∈(−∞,1) or n∈(119,∞) ⇒n∈(−∞,1)∪(119,∞)⋯(B) iii.)−b2a>k ⇒−6n2.1>3 ⇒3n>3⇒n>1 ⇒n∈(1,∞)⋯(C) ∴ Range of n, is A∩B∩C ⇒[1,∞)∩(−∞,1)∪(119,∞)∩(1,∞) ⇒A∩B∩C=(119,∞) ⇒n∈(119,∞)

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