QUAD DOUBT 1

1.
FIND ALL VAlUES OF m for which mx2 + (m-3)x + 1 <0 for atleast one positive real x

3 Answers

11
virang1 Jhaveri ·

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

11
Mani Pal Singh ·

VIRAG 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

62
Lokesh Verma ·

two 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,∞)

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