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Question

# Find the values of m such that both the roots of the quadratic equation x2−(m−3)x+m=0 (m∈R) lies in the interval (1,2).

A
ϕ
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B
None of the above.
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C
(5,7)
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D
(9,)
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Solution

## The correct option is A ϕLet, f(x)=x2−(m−3)x+m When both the roots of f(x) lies in the interval (1,2) Condition : (i) D≥0 (ii) f(1)>0 & f(2)>0 (iii) 1<−b2a<2 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) f(1)>0 ⇒12−(m−3).1+m>0 ⇒1−m+3+m>0 ⇒4>0⇒m∈R & f(2)>0 ⇒22−(m−3).2+m>0 ⇒4−2m+6+m>0 ⇒m<10⇒m∈(−∞,10) ∴f(1)>0,f(2)>0 m∈(−∞,10) (iii) 1<−b2a<2 ⇒1<m−32<2 ⇒2<m−3<4 ⇒5<m<7 m∈(5,7) Now, taking intersection of all the three above conditions, we get There is no common solution, ∴m∈ϕ

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