Grade:12

## 1 Answers

Anil Pannikar AskiitiansExpert-IITB
85 Points
13 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)

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Anil Pannikar

IIT Bombay

## ASK QUESTION

Get your questions answered by the expert for free