Given roots of x2−mx+4=0 are distinct and lies in (1,5), Following conditions must be true
(i) D>0⇒m2−16>0
⇒(m−4)(m+4)>0
⇒m∈(−∞,−4)∪(4,∞)
(ii) f(1)>0⇒m<5
(iii) f(5)>0⇒m<529
(iv) 1<−2ab<5⇒2<m<10
Taking intersection of all conditions, we get m∈(4,5)
If both the roots of the quadratic equation x2−mx+4=0 are real and distinct and they lie in the interval (1,5), then m lies in the interval:
Note: In the actual JEE paper interval was [1,5]
Held on 9 Jan 2019 · Verified 6 Jul 2026.
(−5,−4)
(3,4)
(5,6)
(4,5)
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