1.
FIND ALL VAlUES OF m for which mx2 + (m-3)x + 1 <0 for atleast one positive real x
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3 Answers
11X < [ 3-m + √(m2+9-6m -4m)]/2m
X <
[ 3-m + √(m2+9-6m -4m)]/2m>0
3-m+√m2+9-10m >0
m2+9-10m>m2+9-6m
m = 0
11VIRAG U SAID
X < [ 3-m + √(m2+9-6m -4m)]/2m
X <
[ 3-m + √(m2+9-6m -4m)]/2m>0
3-m+√m2+9-10m >0
m2+9-10m>m2+9-6m
m = 0
BUT THERE IS A CONCEPTUAL MISTAKE IN UR ANSWER AND HENCE UR ANSWER IS WRONG
CONCEPT :U CAN NEVER SQUARE AN EXPRESSION WHILE TAKING INEQUALITY
THIS IS NOT AN EASY QUES U HAVE TO THINK GRAPHICALLY ALSO
MAKE CASES OF m>0 AND m<0 AND SOLVE TO GET THE ANSWER
62two cases
m<0
then it is true by defalult for some x due to shape of the parabola
m>0
then D>0.. that is all...
(m-3)2>4m
m2-10m+9>0
(m-1)(m-9)>0
m>9 or m<1
thus from above we get
(-∞,0) U (0,1) Ï
(9,∞)
