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Question

# Find the values of m such that both roots of the quadratic equation x2−(m−3)x+m=0 (m∈R) are greater than 2

A
(7,9]
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B
(10,)
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C
[9,10)
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D
[1,9]
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Solution

## The correct option is C [9,10)Let, f(x)=x2−(m−3)x+m When both roots of f(x)=0 are greater than 2. Condition : (i) D≥0 (ii) −b2a>2 (iii) f(2)>0 Now, on solving it, (i) D≥0 ⇒(m−3)2−4m≥0 ⇒m2−6m+9−4m≥0 ⇒m2−10m+9>≥ ⇒(m−1)(m−9)≥0 m∈(−∞,1]∪[9,∞) (ii) −b2a>2 ⇒m−32>2 ⇒m−3>4⇒m>7 m∈(7,∞) (iii) f(2)>0 ⇒(2)2−(m−3)(2)+m>0 ⇒4−2m+6+m>0 ⇒−m+10>0⇒m<10 ∴m∈(−∞,10) Now, taking intersection of all the above three conditions, we get m∈[9,10)

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