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7 years ago

85 Points
```										Dear  Nikhil ,
f'(x)=3x2+6(a−7)x+3(a2−9)
For f (x) to have a maximum at some point, f'(x)=0 and f"(x)<0 for that point
Now, f'(x)= 0
or 3x2+6(a−7)x+3(a2−9)=0
or, x2+2(a−7)x+(a2−9)=0
x=−(a−7)± sqrt.(58−14a)
For f'(x) to have real roots,
sqrt.(58−14a) > 0     a< 29/7       ...............1
Now we determine which of the roots of f '(x) in (i) will give a local maximum and which will give a local minimum.
f"(x)=6x+6(a−7) =6(x+a−7)
At  x= −a−7 + sqrt.(58−14a)      or , f"x=6 sqrt.(58−14a)> 0
At  x=−a−7 − sqrt.(58−14a)        or,  f"x=6 sqrt.(58−14a)<0
Therefore, x = −a−7 − sqrt.(58−14a) is a point of local maximum. Since we want this to be positive.
−(a−7)− sqrt.(58−14a)>0
or, 7−a>sqrt.(58−14a)
Upon squaring, we get
a2−9>0
⇒a<−3 or a>3  ................2
From 1 and 2..

All the best.
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Anil Pannikar
IIT Bombay
```
7 years ago
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