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Question

If both the roots of the quadratic equation x26nx+(9n22n+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: x26nx+(9n22n+2)=0 Roots are greater than 3
To find: Range of n
Step-1: Draw the graph for the given expression y=x26nx+(9n22n+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).D0
ii).f(k)>0
iii).b2a>k

Now, solving for all the conditions,
i). D0
D=b24ac0(6n)24.1.(9n22n+2)036n236n2+8n808n808n8n1n[1,)(A)

ii).f(k)>0, where k is 3
f(3)>0
326n×3+(9n22n+2)>0
918n+9n22n+2>0
9n220n+11>0
(9n11)(n1)>0
n<1 or n>119
n(,1) or n(119,)
n(,1)(119,)(B)

iii.)b2a>k
6n2.1>3
3n>3n>1
n(1,)(C)
Range of n, is ABC
[1,)(,1)(119,)(1,)
ABC=(119,)
n(119,)

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