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Grade:12

1 Answers

Anil Pannikar AskiitiansExpert-IITB
85 Points
10 years ago

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..

Answer is (b)

 

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Askiitians Expert

Anil Pannikar

IIT Bombay

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